text:
\section{Introduction}

Fractional and anomalous diffusions have a long history. The terms fractional and anomalous have been considered with different meaning and in different contexts. By fractional diffusion we mean a diffusion in a medium with fractional dimension (fractals, for instance) whereas, by anomalous diffusions, according to the most significant literature, we refer  to a motion whose mean squared displacement is proportional to a power of time. The anomalous dynamics is considered in many fields of research and many practical applications, for example in finance, physics, ecology, biology, hydrology:  the literature is huge, we mention here only few works, for example, \cite{Clark, FGIS, GGPS, GoldCox}.  Our aim is to pay exclusive attention to the probabilistic models for anomalous dynamics.

The well-known theory  of time-changed processes considers Markov and non-Markov random times. As usual we refer to subordination if the random time is Markovian. Subordinated processes are associated to subordinated semigroups (in the sense of Bochner for instance) and the generators can be represented by considering the Phillips formula (not necessarily for the generator of a Markov process). Processes obtained through non-Markovian time changes can be considered in order to solve fractional Cauchy problems. In this case, in fact, time-changed processes turn out to be non-Markovian and  the fractional operators in time are convolution operators with kernels associated with inverses of subordinators. A mathematical approach has been introduced by \cite{Baz} and further investigations have been considered by many researchers  (see, for example, \cite{Koc,Kolok,Meer, ORs}). Given a Markov process $X$ with generator $(A,D(A))$ and the inverse $L$ to a stable subordinator $H$ of order $\alpha \in (0,1)$, then $u(t,x) = \mathbf{E}_x[f(X_{L_t})]$ solves
\begin{align*}
\partial^\alpha_t u= A u, \quad f \in D(A)
\end{align*}
where $\partial^\alpha_t$ is the Caputo fractional derivative. Since the mean squared displacement of the process $X_L$ is non linear in time (proportional to a power of time, $t^\alpha$, with $\alpha<1$) we say that $X_L$ exhibits a subdiffusive behaviour. In the literature, such a process has been also associated to a delayed process in the sense that the trajectories of $L$ may have plateaus. Since $L$ is the new clock for $X$, then the process $X_L$ is usually termed \lq\lq delayed\rq\rq .

On the other hand, if $H$ is an $\alpha$-stable subordinator, due to the fact that the trajectories of $H$ exhibit jumps, the subordinated process $X_H$ may have jumps. This should suggest that the process $X_H$ runs faster than $X$. The generator of $X_H$ is the fractional power of $-A$ given by $-(-A)^\alpha$. Here we assume that $-A$ is a non-negative definite operator. We point out that non-local operators in space may introduce non-local boundary conditions and non-local operators in time may introduce fractional initial conditions.

Recently, new fractional operators in time have been introduced in \cite{chen, toaldo}. The probabilistic representation of the solutions to the associated fractional Cauchy problems is  still obtained through time changes. This class of new fractional equations brings our attention to a new characterization of the corresponding motions. Beside the anomalous behaviour of the time-changed processes, interesting aspects are given by the comparison between the lifetimes of the base processes and those of the time-changed processes. Surprisingly, our analysis reveals that inverse processes are not necessarily related to delayed processes.

The aim of the present paper is therefore to investigate the underlying dynamics for the time changes. We introduce a precise definition of delayed and rushed processes and provide some examples which are, in some cases, counterintuitive.

\section{Time changes and operators}
\label{secTF}

In this section, we  introduce the operators and the equations governing $X_L$ and $X_H$ for general time changes $H$ and $L$ characterized by the symbol $\Phi$.

Let $E$ be a locally compact, separable metric space   and $E_\partial= E \cup \{\partial\}$ be the one-point compactification of $E$. Denote by $\mathcal{B}(E)$ the $\sigma$-field of the Borel sets in $E$ ($\mathcal{B}_\partial$ is the $\sigma$-field in $E_\partial$).  Let $X=\{(X_t)_{t\geq 0}, (\mathbf{P}_x)_{x \in E}\}$ with infinitesimal generator $(A, D(A))$ be the continuous and symmetric Markov process on $(E, \mathcal{B}(E))$ with transition function $p(t, x, B)$ on $[0, \infty) \times E \times \mathcal{B}(E)$. The point $\partial$ is the cemetery point for $X$ and a function $f$ on $E$ can be extended to $E_\partial$ by setting $f(\partial)=0$. The associated semigroup is uniquely defined by
\begin{align*}
P_t f(x):= \int_E p(t,x,dy) f(y) = \mathbf{E}_x[f(X_t)], \quad f \in C_\infty(E)
\end{align*}
where $\mathbf{E}_x$ denote the mean value with respect to $\mathbf{P}_x$ with $X_0=x \in E$ and $C_\infty$ is the set of continuous function $C(E)$ on $E$ such that $f(x)\to 0$ as $x\to \partial$. Let $\mathcal{E}(u,v)= (\sqrt{-A}u, \sqrt{-A}v)$ with domain $D(\mathcal{E})= D(\sqrt{-A})$ be the Dirichlet form associated with (the non-positive definite, self-adjoint operator) $A$. Then $X$ is equivalent to an $m$-symmetric Hunt process whose Dirichlet form $(\mathcal{E}, D(\mathcal{E}))$ on $L^2(E,m)$ is local (see  the books \cite{Chen-book, Fuk-book}). Without restrictions we assume that the form is regular. Then, $X$ is a continuous strong Markov process (we say Feller diffusion).\\

We now introduce the Bernstein function
\begin{equation}
\Phi(\lambda) = \int_0^\infty \left( 1 - e^{ - \lambda z} \right) \Pi(dz), \quad \lambda \geq 0
\label{LevKinFormula}
\end{equation}
where $\Pi$ on $(0, \infty)$ with $\int_0^\infty (1 \wedge z) \Pi(dz) < \infty$ is the associated L\'evy measure.  We also recall that
\begin{align}
\label{tailSymb}
\frac{\Phi(\lambda)}{\lambda} = \int_0^\infty e^{-\lambda z} \overline{\Pi}(z)dz, \qquad \overline{\Pi}(z) = \Pi((z, \infty))
\end{align}
and $\overline{\Pi}$ is the so called \emph{tail of the L\'evy measure}. For details, see the book  \cite{BerBook}. The symbol $\Phi$ can be associated with the Laplace exponent of a subordinator $H$, that is
$$\mathbf{E}_0[\exp( - \lambda H_t)] = \exp(- t \Phi(\lambda)).$$
We introduce the inverse process
$$L_t = \inf \{s \geq 0\,:\, H_s >t \}$$
and define the time-changed process
$$X^{L}_t :=X_{L_t}, \quad t \geq 0$$
as the composition $X \circ L$ and the time-changed process
$$X^H_t := X_{H_t}, \quad t>0$$
as the composition $X \circ H$. We do not consider step-processes with $\Pi((0, \infty)) < \infty$ and therefore we focus only on strictly increasing subordinators with infinite measures. Thus, the inverse process $L$ turns out to be a continuous process with non-decreasing paths. By definition, we also can write
\begin{align}
\label{relationHL}
\mathbf{P}_0(H_t < s) = \mathbf{P}_0(L_s>t), \quad s,t>0.
\end{align}

Let $M>0$ and $w\geq 0$. Let $\mathcal{M}_w$ be the set of (piecewise) continuous function on $[0, \infty)$ of exponential order $w$ such that $|u(t)| \leq M e^{wt}$. Denote by $\widetilde{u}$ the Laplace transform of $u$. Then, we define the operator $\mathfrak{D}^\Phi_t : \mathcal{M}_w \mapsto \mathcal{M}_w$ such that
\begin{align*}
\int_0^\infty e^{-\lambda t} \mathfrak{D}^\Phi_t u(t)\, dt = \Phi(\lambda) \widetilde{u}(\lambda) - \frac{\Phi(\lambda)}{\lambda} u(0), \quad \lambda > w
\end{align*}
where $\Phi$ is given in \eqref{LevKinFormula}. Since $u$ is exponentially bounded, the integral $\widetilde{u}$ is absolutely convergent for $\lambda>w$.  By Lerch's theorem the inverse Laplace transforms $u$ and $\mathfrak{D}^\Phi_tu$ are uniquely defined. We note that
\begin{align}
\label{PhiConv}
\Phi(\lambda) \widetilde{u}(\lambda) - \frac{\Phi(\lambda)}{\lambda} u(0) = & \left( \lambda \widetilde{u}(\lambda) - u(0) \right) \frac{\Phi(\lambda)}{\lambda}
\end{align}
and thus, $\mathfrak{D}^\Phi_t$ can be written as a convolution involving the ordinary derivative and the inverse transform of \eqref{tailSymb} iff $u \in \mathcal{M}_w \cap C([0, \infty), \mathbb{R}_+)$ and $u^\prime \in \mathcal{M}_w$. That is,
\begin{align*}
\mathfrak{D}^\Phi_t u(t) = \int_0^t u^\prime(s) \overline{\Pi}(t-s)ds.
\end{align*}
By Young's inequality we also observe that
\begin{align}
\label{YoungSymb}
\int_0^\infty |\mathfrak{D}^\Phi_t u |^p dt \leq \left( \int_0^\infty |u^\prime |^p dt \right) \left( \lim_{\lambda \downarrow 0} \frac{\Phi(\lambda)}{\lambda} \right)^p, \qquad p \in [1, \infty)
\end{align}
where $\lim_{\lambda \downarrow 0} \Phi(\lambda) /\lambda$ is finite only in some cases (\cite{CapDovVIA}) .
We notice that when $\Phi(\lambda)=\lambda$ (that is we deal with the ordinary derivative) we have that $H_t = t$ and $L_t=t$ a.s. and in \eqref{YoungSymb} the equality holds. For explicit representation of the operator $\mathfrak{D}^\Phi_t$ see also the recent works \cite{chen, toaldo}.

\begin{remark}
\label{remark:telegr}
We notice that for $\Phi(\lambda)=\lambda^\beta$, the symbol of a stable subordinator, the operator $\mathfrak{D}^\Phi_t$ becomes the Caputo fractional derivative
\begin{align*}
\mathfrak{D}^\Phi_t u(t) = \frac{1}{\Gamma(1-\beta)} \int_0^t \frac{u^\prime(s)}{(t-s)^\beta}ds
\end{align*}
with $u^\prime(s)=du/ds$.
A further example is given by the symbol $\Phi(\lambda)=\lambda^{2\beta} + \lambda^\beta$ for $\beta \in (0, 1/2)$, that is, $\mathfrak{D}^\Phi_t$ becomes the telegraph fractional operator (\cite{DOTtelegraph}).

The case $\beta=1$ is an interesting case of exponential velocity correlation. If $\mathcal{T}_t$ is the telegraph process, then $
\mathcal{T}_t = \int_0^t V_s\, ds \quad \textrm{with} \quad V_s = V_0  (-1)^{N_s}$ where $N_s$, $s\geq 0$ is an homogeneous Poisson process of rate $\lambda$ independent from $V_0$ with $\mathbf{P}(V_0=+c)=\mathbf{P}(V_0=-c)$, $c>0$. The process on the real line is performed by a random motion with finite speed $c$. The second moment is proportional to $t\,c^2/\lambda$ for large $t$ (\cite{orsT}). For the first passage time $\tau_a$ through the level $a$ we have that $\mathbf{E}_x[\tau_a] = \infty$ (see \cite{Foong, orsT}). This well accords with the fact that $\mathcal{T}_t$ is a process with finite velocity, the process spends an infinite (mean) amount of time in a set.
\end{remark}

We now introduce the Phillips representation (\cite{Phillips})
\begin{align*}
-\Phi(-A) f(x) = \int_0^\infty \big( P_z f(x) -  f(x) \big) \Pi(dz)
\end{align*}
where $\Phi$ has been given in \eqref{LevKinFormula}. Let $\Psi$ be the Fourier multiplier of $A$. Then the Fourier symbol of the semigroup $P_t = e^{tA}$ is written as $\widehat{P_t} = e^{-t \Psi}$. For a well-defined function $f$, from \eqref{LevKinFormula} we have that
\begin{align*}
\int_{\mathbb{R}} e^{i\xi x} \left( -\Phi(-A) f(x) \right) dx = \left( \int_0^\infty \left( e^{-z \Psi(\xi)} -1 \right) \Pi(dz) \right) \widehat{f}(\xi)= -\Phi \circ \Psi(\xi)\, \widehat{f}(\xi).
\end{align*}
We also recall that composition of Bernstein functions is a Bernstein function.

\begin{remark}
We note that if $H$ is the stable subordinator with symbol  $\Phi(\xi)=\xi^\alpha$ and  $X$ is the Brownian motion with $\Psi(\xi)= \xi^2$,  the process $X^H$ is the symmetric stable process with symbol $|\xi |^{2\alpha}$ driven by the fractional Laplacian
\begin{align}
\label{fracLaplacianS}
-\Phi(- A) = -(-\Delta)^{\alpha}, \quad \alpha\in (0,1).
\end{align}
\end{remark}

\section{Time-changed process $X^L_t$}
\label{secSF}

In this section we consider the time fractional equation
\begin{align}
\label{time-frac-problem}
\mathfrak{D}^\Phi_t u = A u, \qquad u_0=f \in D(A).
\end{align}

The probabilistic representation of the solution to \eqref{time-frac-problem} is written in terms of the time-changed process $X^{L}$, that is
\begin{align}
\label{sol-time-frac-problem}
u(t,x) = \mathbf{E}_x[f(X^{L}_t)]= \mathbf{E}_x[f({^*X}^{L}_t), t < \zeta^{L}]
\end{align}
where $\zeta^{L}$ is the lifetime of $X^{L}$, the part process of ${^*X}^{L}$ on $E$. The fact that $L$ is continuous implies that
\begin{align*}
\mathbf{E}_x[f({^*X}^{L}_t), t < \zeta^{L}] = \mathbf{E}_x[f({^*X}_{L_t}), \, L_t < \zeta]
\end{align*}
(see for example \cite[Corollary 3.2]{Meer} or \cite[Section 2]{Svond}). Moreover, we have that
\begin{align}
\label{semigPhi-L-continuity}
\mathbf{E}_x[f({^*X}^{L}_t), t < \zeta^{L}] =  \int_0^\infty P_s f(x) \mathbf{P}_0(L_t \in ds) :=P^{L}_t f(x)
\end{align}
where $P_tf(x) = \mathbf{E}_x [f(X_t)] = \mathbf{E}_x [f({^*X}_t), t < \zeta]$ and $\zeta$ is the lifetime of $X$, the part process of ${^*X}$ on $E$. Then, for the  time-changed processes we have that
\begin{align}
\label{semigPhi-1}
P^{L}_t \mathbf{1}_E (x) = \mathbf{P}_x(t < \zeta^{L}) = \mathbf{P}_x(L_t < \zeta) = \int_0^\infty \mathbf{P}_x(s < \zeta) \mathbf{P}_0(L_t \in ds).
\end{align}
We notice that $P_t^{L}$ is not a semigroup (indeed the random time is not Markovian). We also introduce the following $\lambda$-potentials ($\lambda>0$)
\begin{align*}
R_\lambda f(x) = \mathbf{E}_x \left[ \int_0^\infty e^{-\lambda t} f(X_t)dt \right], \qquad R^{L}_\lambda f(x) = \mathbf{E}_x \left[ \int_0^\infty e^{-\lambda t} f(X^{L}_t)dt \right].
\end{align*}

The following result concerns the relation between the lifetime of the time-changed process  $X^L$ with the lifetime of the random time $H$.
\begin{tm}
\label{meanLzeta}
Let $\Phi$ be the Bernstein function in \eqref{LevKinFormula}. Let $H$ be the subordinator with symbol $\Phi$. We have
\begin{align}
\mathbf{E}_x[\zeta^L] = \mathbf{E}_x[H_\zeta], \quad x \in E.
\end{align}
\end{tm}
\begin{proof}
From \eqref{semigPhi-1}
\begin{align}
\label{NoSemig}
\mathbf{E}_x[\zeta^L]
= & \lim_{\lambda \to 0} R^L_\lambda \mathbf{1}_E(x) = \int_0^\infty  P^L_t \mathbf{1}_E(x)\, dt  = \int_0^\infty \mathbf{E}_x[L_t < \zeta] \, dt.
\end{align}
Since
\begin{align*}
 \mathbf{E}_x[L_t < \zeta]
 = & \int_0^\infty \mathbf{P}_x(\zeta >s) \frac{d}{d s} \mathbf{P}_0(L_t < s) ds \\
 = & \mathbf{P}_x(\zeta>s) \mathbf{P}_0(L_t < s) \bigg|_{s=0}^{s=\infty} + \int_0^\infty \mathbf{P}_x(\zeta \in ds) \mathbf{P}_0(L_t < s) \\
 = & \int_0^\infty \mathbf{P}_x(\zeta \in ds) \mathbf{P}_0(L_t < s),
\end{align*}
from the relation \eqref{relationHL},
 \begin{align*}
 \mathbf{E}_x[L_t < \zeta] =  \int_0^\infty \mathbf{P}_x(\zeta \in ds) \mathbf{P}_0(H_s >t)= \mathbf{E}_x[ t < H_\zeta].
 \end{align*}
Then,
\begin{align*}
\int_0^\infty  \mathbf{E}_x[L_t < \zeta] \, dt = \int_0^\infty \mathbf{P}_x(H_\zeta >t)dt = \mathbf{E}_{x}[H_\zeta].
\end{align*}
\end{proof}

Now we recall the following result given in \cite{CapDovVIA} (see also \cite{chen}) that relates the lifetime of the time changed process $X^L$ with the lifetime of the base process $X$.
\begin{tm}
\label{B}
Let $\Phi$ be the Bernstein function in \eqref{LevKinFormula}. Let $L$ be the inverse of $H$ with symbol $\Phi$. We have
  \begin{equation}
\mathbf{E}_x[\zeta^{L}]=
\lim_{\lambda \downarrow 0} \frac{\Phi(\lambda)}{\lambda} \mathbf{E}_x[\zeta].\end{equation}
\end{tm}
\begin{proof}
From \eqref{semigPhi-1} and \eqref{relationHL}, for the $\lambda$-potential above we obtain
\begin{align}
R^{L}_\lambda \mathbf{1}_{E}(x) = & \int_0^\infty \mathbf{P}_x(s < \zeta) \int_0^\infty e^{-\lambda t} \frac{d}{ds} \mathbf{P}_0(H_s >t) dt\, ds \notag \\
= & \frac{1}{\lambda} \int_0^\infty \mathbf{P}_x(s < \zeta) \frac{d}{ds}\left(1 - \mathbf{E}_0[e^{-\lambda H_s}] \right)ds
= \frac{1}{\lambda} \mathbf{E}_x[1 - e^{-\zeta \Phi(\lambda)}].\label{potential-equality}
\end{align}
By considering \eqref{potential-equality} we have that, as $\lambda \to 0$,
\begin{align}
R^{L}_\lambda \mathbf{1}_{E}(x) = \frac{1}{\lambda} \mathbf{E}_x[1 - e^{-\zeta \Phi(\lambda)}] \to \left( \lim_{\lambda \downarrow 0} \frac{\Phi(\lambda)}{\lambda} \right) \mathbf{E}_x[\zeta] = \mathbf{E}_x[\zeta^{L}]
\end{align}
which is the mean lifetime of the time-changed process $X^{\Phi}$.
\end{proof}

\section{Time-changed process $X^H_t$}
\label{secTSF}

For the process $X$ with generator $(A, D(A))$ introduced before and the independent subordinator $H$ with symbol \eqref{LevKinFormula}, the process $X^H_t=X_{H_t}$, $t\geq 0$ (the composition $X \circ H$) can be considered in order to solve the equation
\begin{align}
\label{space-frac-problem}
\frac{\partial u}{\partial t} = - \Phi( - A) u, \quad u_0=f \in D(\Phi( - A)) \subset D(A).
\end{align}
More precisely, the probabilistic representation of the solutions to \eqref{space-frac-problem} is given by
\begin{align*}
P^{H}_t f(x):=\mathbf{E}_x [f(X^{H}_t)]= \mathbf{E}_x [f(X_{H_t})]
\end{align*}
and (see \cite[Section 2]{Svond} or \cite[Section 3]{SVrel} for a detailed discussion)
\begin{align*}
\mathbf{E}_x [f(X^{H}_t)] = \mathbf{E}_x [f({^*X}^{H}_t), t < \zeta^{H}]
\end{align*}
where $\zeta^{H}$ is the lifetime of $X^{H}$, that is the part process of ${^*X}^{H}$ on $E$. We recall that the multiplicative functional $M_t = \mathbf{1}_{(t<\zeta^H)}$ characterizes uniquely the semigroup $P^H_t$ (see \cite[pag. 100, Proposition 1.9]{BluGet68}). We also introduce the following $\lambda$-potential ($\lambda>0$)
\begin{align*}
R^{H}_\lambda f(x) = \mathbf{E}_x \left[ \int_0^\infty e^{-\lambda t} f(X^{H}_t)dt \right].
\end{align*}
We notice that if $H$ is a stable subordinator with symbol $\Phi(\lambda)=\lambda^\alpha$ and $X$ is a Brownian motion, then $-\Phi(-A)$ is the fractional Laplacian (see, for example, \cite{DovJMA, MK} for details and connections between fractional Laplacians, subordinators and random walks continuous in time and space).

The next theorem relates the lifetime of the time-changed process $X^H$ with the lifetime of the random time $L$.

\begin{tm}
\label{meanHzeta}
Let $\Phi$ be the Bernstein function \eqref{LevKinFormula}. Let $L$ be the inverse to a subordinator $H$.  We have that
\begin{align}
\mathbf{E}_x[\zeta^H] = \mathbf{E}_x[L_\zeta], \quad x \in E.
\end{align}
\end{tm}
\begin{proof}
We observe that, as $\lambda\to 0$,
\begin{align*}
R^H_\lambda \mathbf{1}_E(x) \to& \int_0^\infty \int_0^\infty \mathbf{P}_x(X_s \in E) \mathbf{P}_0(H_t \in ds)\, dt\\
= & \int_0^\infty \int_0^\infty \mathbf{P}_x(\zeta >s) \mathbf{P}_0(H_t \in ds)\, dt
=  \int_0^\infty \mathbf{P}_x(\zeta >H_t)\, dt
\end{align*}
where
\begin{align*}
\mathbf{P}_x(\zeta >H_t)
= & \int_0^\infty \mathbf{P}_x(\zeta >s) \frac{d}{d s} \mathbf{P}_0(H_t < s) ds\\
= & \int_0^\infty \mathbf{P}_x(\zeta \in ds) \mathbf{P}_0(L_s >t)
=  \mathbf{P}_x(L_\zeta >t)
\end{align*}
and therefore
\begin{align*}
R^H_\lambda \mathbf{1}_E(x) \to &  \int_0^\infty \mathbf{P}_x(L_\zeta >t)dt = \mathbf{E}_x[L_\zeta].
\end{align*}
Since
\begin{align*}
R^H_\lambda \mathbf{1}_E(x) \to \mathbf{E}_x[\zeta^H], \quad \textrm{as} \quad \lambda\to 0
\end{align*}
we obtain the result.
\end{proof}

Let us focus on the case $\Phi(\lambda)=\lambda^\alpha$. We first recall that $L$, the inverse to a subordinator $H$, can be characterized as follows
\begin{align}
\label{Lap-mittag-L-F}
\mathbf{E}_0 [\exp (-\lambda L_t)] = E_\alpha(-\lambda t^\alpha), \quad \lambda > 0
\end{align}
where, for $\alpha, \gamma>0$,
\begin{align*}
E_{\alpha, \gamma}( -z) = \sum_{k \geq 0} \frac{(-z)^k}{\Gamma(\alpha k + \gamma)}, \quad z\geq 0
\end{align*}
is the generalized Mittag-Leffler function. For $\gamma=1$,  $E_\alpha = E_{\alpha, 1}$ which is the so called Mittag-Leffler function.

The next theorem gives a precise characterization of the lifetime of $X^H$ in terms of the symbol $\Phi$.

\begin{tm}
\label{A}
Let $\Phi(\lambda) = \lambda^\alpha$. Let $H$ be the subordinator with symbol $\Phi$. Then
\begin{align*}
\mathbf{E}_x[\zeta^{H}] = & \frac{1}{\Gamma(\alpha +1)} \mathbf{E}_x \left[ \Phi(\zeta) \right] =  \lim_{\lambda \downarrow 0} \frac{\mathbf{E}_x \left[ \Phi(\zeta) \right]}{\mathfrak{D}^\Phi_\lambda \Phi(\lambda)}
\end{align*}
and
\begin{align*}
\lim_{s\to \infty} \Phi(s) \mathbf{P}_x(\zeta >s) =0.
\end{align*}
Moreover, if
$$\mathbf{P}_x(\zeta > s) \leq  e^{-\omega s}, \quad \textrm{for some } \; \omega>0$$
then
\begin{align*}
\mathbf{E}_x[\zeta^{H}] \leq \frac{1}{\gamma^\alpha} \mathbf{E}_x[ e^{\gamma \zeta}] , \quad \gamma < \omega
\end{align*}
\end{tm}
\begin{proof}
We have that
\begin{align}
P^{H}_t \mathbf{1}_{E}(x) = \int_0^\infty \mathbf{P}_x(s < \zeta) \mathbf{P}_0(H_t \in ds)
\label{semigH-1}
\end{align}
is a semigroup (the random time is Markovian) with the associated  resolvent
\begin{align}
R^{H}_\lambda \mathbf{1}_E(x) = & \mathbf{E}_x \left[\int_0^\infty e^{-\lambda t} \mathbf{1}_E(X^H_t) dt \right], \quad \lambda>0.
\end{align}
Since $\mathbf{P}_0(H_t < x) = \int_0^x h(t, s)ds$ where
\begin{align*}
\int_0^\infty e^{-\lambda t} h(t,s) dt = & \int_0^\infty e^{-\lambda t} \frac{d}{d s} \mathbf{P}_0(H_t < s)dt
=  \frac{d}{d s} \int_0^\infty e^{-\lambda t} \mathbf{P}_0(L_s > t)dt \\
= & \frac{d}{d s} \left( \frac{1}{\lambda} - \frac{1}{\lambda} \int_0^\infty e^{-\lambda t} \mathbf{P}_0(L_s \in dt) \right)
=  \frac{d}{d s} \left( \frac{1}{\lambda} - \frac{1}{\lambda} E_\alpha(- \lambda s^\alpha) \right)
\end{align*}
and
\begin{align}
\frac{d}{d s} E_\alpha(-\lambda s^\alpha) = - \lambda s^{\alpha -1} E_{\alpha, \alpha}(- \lambda s^\alpha),
\label{derivML}
\end{align}
we obtain
\begin{align}
R^{H}_\lambda \mathbf{1}_E(x) = & \mathbf{E}_x \left[\int_0^\infty e^{-\lambda t} \mathbf{1}_E(X_{H_t}) dt \right]
= \int_0^\infty e^{-\lambda t} \int_0^\infty \mathbf{P}_x(X_{s} \in E) \mathbf{P}_0 (H_t \in ds)\, dt \notag \\
= &  \int_0^\infty \mathbf{P}_x(s < \zeta) E_{\alpha,\alpha}(- \lambda s^\alpha) \, s^{\alpha-1} ds \label{potentialML}\\
= & \frac{1}{\lambda} \mathbf{E}_x [1 - E_\alpha(-\lambda (\zeta)^\alpha)]. \label{potentialML-2}
\end{align}
Thus, from \eqref{potentialML-2} we have that, as $\lambda\to 0$,
\begin{align}
\label{limLAMBDA1}
R^{H}_\lambda \mathbf{1}_E(x) \to \mathbf{E}_x [\zeta^{H}]
\end{align}
where
\begin{align*}
\mathbf{E}_x [\zeta^{H}] = \frac{1}{\Gamma(\alpha +1)} \mathbf{E}_x\left[ (\zeta)^\alpha \right].
\end{align*}
Moreover, due to the fact that $\Phi(0)=0$, the Riemann-Liouville and the Caputo derivatives of $\Phi$ coincides. In particular,
$$\mathfrak{D}^\Phi_\lambda \Phi(\lambda) = \frac{\Gamma(1+ \alpha)}{\Gamma(1+\alpha - \alpha)} \lambda^{\alpha - \alpha}={\Gamma(\alpha +1)} .$$
Passing to the limit in \eqref{potentialML} we have that
\begin{align*}
\lim_{\lambda \to 0} R^{H}_\lambda \mathbf{1}_E(x) = & \frac{1}{\Gamma(\alpha +1)} \left( s^\alpha \mathbf{P}_x(\zeta >s) \bigg|_{s=0}^{s=\infty} + \mathbf{E}_x[(\zeta)^\alpha] \right)
\end{align*}
and, from \eqref{limLAMBDA1}, it holds that
\begin{align*}
\lim_{s\to \infty} s^{\alpha} \mathbf{P}_x(\zeta >s) =0
\end{align*}
and this concludes the first part of the proof.\\

Since we have that
\begin{align}
R^{H}_\lambda \mathbf{1}_E(x) = & \int_0^\infty \frac{s^\alpha}{s} E_{\alpha, \alpha} (-\lambda s^\alpha) \mathbf{P}_x(\zeta >s) ds \label{firstRpsi}\\
\leq & \int_0^\infty \frac{s^\alpha}{s} E_{\alpha, \alpha} ( - \lambda s^\alpha) \frac{\mathbf{E}_x[g(\zeta)]}{g(s)} ds, \qquad \forall\, \lambda \geq 0 \notag
\end{align}
for $g$ increasing and positive, for $g(s)=e^{\gamma s}$, under the assumption that
\begin{align*}
\mathbf{P}_x(\zeta > s) \leq e^{-\omega s}, \quad \textrm{for some } \; \omega>0,
\end{align*}
we obtain that, as $\lambda \to 0$,
\begin{align*}
R^{H}_\lambda \mathbf{1}_E(x) \leq \frac{1}{\gamma^\alpha + \lambda} \mathbf{E}_x[ e^{\gamma \zeta}] \to \frac{1}{\gamma^\alpha} \mathbf{E}_x[ e^{\gamma \zeta}] , \quad \gamma < \omega
\end{align*}
where we used (\cite{HMS})
\begin{align*}
\int_0^\infty e^{-\gamma s} s^{\alpha-1} E_{\alpha, \alpha}(-\lambda s^\alpha) ds = \frac{1}{\gamma^\alpha + \lambda}.
\end{align*}
\end{proof}

Let us consider now the Dirichlet boundary condition on $\partial D$ for which the Brownian motion $X$ on $D$ has lifetime $\zeta=\tau_D$, the first exit time from $D$. A killed subordinate Brownian motion on the bounded domain $D\subset \mathbb{R}^d$ is associated with the Dirichlet form on $L^2(D, dx)$ given by
\begin{align*}
\int_D \int_D \big( u(x) - u(y) \big)^2 J(x,y)dx\, dy + \int_D |u(x)|^2\, \kappa^*(x)\, dx, \quad u \in H^\alpha_0(D)
\end{align*}
with the jumping measure $J(x, y)=C(d,-2\alpha)/2 |x-y|^{-(d+2\alpha)}$ and the killing measure
\begin{align*}
\kappa^*(x) = C(d,-2\alpha) \int_{D^c} \frac{dy}{|x-y|^{d+2\alpha}}
\end{align*}
($C(d,-2\alpha)$ is a constant depending on $d$ and $\alpha \in (0,1)$). In \cite{Svond} the authors provide an interesting discussion comparing the killed subordinate Brownian motion (with killing measure $\kappa^*$) with the subordinate killed Brownian motion (in our notation $X^H$, $H$ is a stable subordinator of order $\alpha \in (0,1)$). In particular, they show that such processes have comparable killing measures. Denote by $\kappa(\cdot)$ the killing measure of $X^H$ where $H$ has symbol $\Phi(\lambda)=\lambda^\alpha$ and $X$ is a killed Brownian motion on $D \subset \mathbb{R}^d$. Than, we have that (\cite[Lemma 3.1]{Svond})
\begin{align*}
\kappa(x) = \frac{1}{\Gamma(-\alpha + 1)} \mathbf{E}_x[(\zeta)^{-\alpha}], \quad x \in \mathbb{R}^d.
\end{align*}
We notice that for the subordinate killed Brownian motion $X^H$ on $D \subset \mathbb{R}^d$ we have that  (by applying Theorem \ref{A})
\begin{align*}
\mathbf{E}_x[\zeta^H] = \frac{1}{\Gamma(\alpha + 1)} \mathbf{E}_x[(\zeta)^\alpha], \quad x \in \mathbb{R}^d
\end{align*}

We think that the previous theorem can be generalized  for an arbitrary $\Phi$. In the following proposition we generalize the Lemma 3.1 in \cite{Svond}.
\begin{prop}
Let $X$ be the $d-$dimensional killed Brownian motion on $D$. Let $H$ be the subordinator with symbol $\Phi$ defined in \eqref{LevKinFormula}. Let $\Phi$ be regularly varying at $+\infty$. If
\begin{align}
\label{condZERO}
\lim_{s\to 0} \Phi(1/s) \mathbf{P}_x(\zeta \leq s ) = 0
\end{align}
then, the killing measure of $X^H$ is written as
\begin{align}
\label{killingMeas}
\kappa(x) = \mathbf{E}_x[\overline{\Pi}(\zeta)], \quad x \in \mathbb{R}^d
\end{align}
where $\overline{\Pi}$ has been defined in \eqref{tailSymb}.
\end{prop}
\begin{proof}
We follows the proof of Lemma 3.1 given in \cite{Svond} starting for the fact that
\begin{align*}
d \overline{\Pi}(s) = d\Pi((s,\infty)) = - \Pi(s) ds
\end{align*}
and
\begin{align*}
\mathbf{E}_x[\overline{\Pi}(\zeta)] = \int_0^\infty \overline{\Pi}(s) \,d\mathbf{P}_x(\zeta \leq s) = - \int_0^\infty \mathbf{P}_x(\zeta \leq s)\, d\overline{\Pi}(s)
\end{align*}
where, in the integration by parts, we used \cite[Proposition 1.5]{BerBook} and \eqref{condZERO}.
\end{proof}

We now provide a further result obtained in a fruitful discussion with Prof. Zoran Vondra\v{c}ek about the previous proposition. He noticed that for every subordinate semigroup it holds that
\begin{align*}
\kappa(x) = \int_0^\infty (1-P_t \mathbf{1}(x)) \Pi(dt).
\end{align*}
Thus, by exploiting the fact that $1-P_t\mathbf{1}(x) = \mathbf{P}_x(\zeta  \leq t)$,
\begin{align*}
\kappa(x) = \int_0^\infty \mathbf{P}_x(\zeta \leq t) \, \Pi(dt) = \int_0^\infty \mathbf{E}_x[1_{(\zeta > t)}] \Pi(dt)
\end{align*}
and, by Fubini's theorem,
\begin{align*}
\kappa(x) = \mathbf{E}_x \left[ \int_0^\infty 1_{(\zeta > t)} \Pi(dt)\right] = \mathbf{E}_x [\Pi((\zeta, \infty))] = \mathbf{E}_x[\overline{\Pi}(\zeta)],
\end{align*}
that is, we obtain \eqref{killingMeas} without any conditions about $\Phi$. Then, we state the following result.
\begin{prop}
Let $X$ be the $d-$dimensional killed Brownian motion on $D$. Let $H$ be the subordinator with symbol $\Phi$ defined in \eqref{LevKinFormula}. Let $\Phi$ be regularly varying at $+\infty$. Then
\begin{align*}
\lim_{s\to 0} \Phi(1/s) \mathbf{P}_x(\zeta \leq s ) = 0.
\end{align*}
\end{prop}
\begin{proof}
From the previous proof we immediately see that \eqref{killingMeas} implies \eqref{condZERO}. The observation by Prof. Vondra\v{c}ek says that
\begin{align*}
\lim_{s\to 0} \overline{\Pi}(s) \mathbf{P}_x(\zeta \leq s ) = 0
\end{align*}
for any L\'{e}vy measure $\Pi$. Thus, from \cite[Proposition 1.5]{BerBook} we obtain a characterization in terms of the symbol $\Phi$.
\end{proof}

\begin{remark}
We observe that the results obtained in Section \ref{secTF}, Section \ref{secSF} and Section \ref{secTSF} still hold for the wide class of continuous Markov processes.
\end{remark}

\section{Delayed and Rushed  anomalous diffusions}

Normal and anomalous diffusions differ for the rate at which velocity correlation decreases to zero. Normal diffusion occurs if the velocity correlation decreases rapidly whereas anomalous diffusion is concerned with processes moving coherently for long times with no (or not frequent) changes of direction. This can be associated with the tail behaviour of the autocorrelation function: if the correlation function decays exponentially, then there is normal diffusion, whereas if the correlation function decays algebraically, then there is the possibility of anomalous diffusion. Thus, anomalous diffusion should be related with non-Markovian dynamics. However, the definition commonly used is based on the second moment (or mean square displacement) proportional to a power of time, say $t^\gamma$, for which we say that the process exhibits a subdiffusive behaviour if $\gamma<1$ (long tailed distribution in time) or a superdiffusive behaviour if $\gamma>1$ (long tailed distribution in space).
Some other characterizations have been also considered, for example ultra-slow and strong anomalous diffusions. The ultra-slow diffusive behaviour is described by a mean square displacement proportional to $(\log t )^\gamma$ (Sinai-like diffusion). Strong anomalous diffusion has been investigated in the interesting paper \cite{CastVulp}.

Throughout
we dealt with anomalous diffusion just considering the anomalous behaviour given by the second moment (or the mean square displacement). \\

 Anomalous diffusions have been also considered in terms of the second moment proportional to $t^{2/d_w}$ where $d_w$ is the random walk dimension. This scaling behaviour occurs in a variety of circumstances and have been observed with exponents $2/d_w$ smaller and larger than $1$, that is the ordinary Brownian motion on $\mathbb{R}^d$. For $d_w>2$ we refer to the process as subdiffusive (charge carrier transport in amorphous semiconductors, the motion of a bead in a polymer network, diffusion on fractals, see for example \cite{Mosco}). For $d_w < 2$ we refer to the process as superdiffusive (diffusion of two particles in turbulent flows, diffusion of tracer particles in vortex arrays in a rotating flow, layered velocity fields). \\

We would like to recall that the non-linear diffusion equation
$$\partial_t u = \Delta u^m = \textsl{div}(m \,u^{m-1} \nabla u)$$
gives rise to a diffusion called normal (if $m=1$), slow (if $m>1$), fast (if $m<1$), ultra-fast (if $m<0$) depending on the diffusivity $D(u) = m \, u^{m-1}$ (see, for example, \cite{Vasq}). In the present paper, we consider slow/fast diffusion meaning subdiffusive/superdiffusive behaviour in the anomalous motion given by the mean square displacement.\\

In this scenario, the main aspect to be analysed seems to be how the anomalous behaviour modifies first passage times. More precisely, how does the time change modify first passage times?

We start with the following definition given in \cite[formulas (1.1) and (1.2)]{BCM} for the Brownian motion.
\begin{defin}
\label{def1}
Let $E \subset \mathbb{R}^d$ be an open connected set with finite volume. Let $B \subset E$ be a closed ball with non-zero radius. Let $X$ be a reflected Brownian motion on $\overline{E}$ and denote by $T_B= \inf\{t\geq 0\,:\, X_t \in B\}$ the first hitting time of $B$ by $X$. We say that $E$ is a \emph{trap domain} for $X$ if
\begin{align*}
\sup_{x \in E} \mathbf{E}_x[T_B] = \infty.
\end{align*}
Otherwise, we say that $E$ is a \emph{non-trap domain} for $X$.
\end{defin}
In Definition \ref{def1}, the random time $T_B$ plays the role of lifetime for the Brownian motion on $\bar{E} \setminus \bar{B}$ reflected on $\partial E \setminus \partial B$ and killed on $\partial B$. On the other hand, $\mathbf{E}_x[T_B]< \infty$ for any $x \in E$ (\cite[Lemma 3.2]{BCM}). This aspect suggests to pay particular attention on the boundary. A process  may have an infinite lifetime depending on the regularity of the boundary $\partial E$ where it can be trapped.\\

Further on we denote by $\zeta$ (possibly with some superscript) the lifetime of a process $X$, that is for the process $X_t$ in $E$ with $X_0=x \in E$ (denote by $E^c$ the complement set of $E$),
\begin{align*}
\zeta := \inf\{t >0\,:\, X_t \in E^c\}.
\end{align*}
Let $T$ be a random time and denote by $X^T := X \circ T$ the process $X$ time-changed by $T$. It is well-known that $X^T$ is Markovian only for a Markovian time change $T$, otherwise from the Markov process $X$ we obtain a non-Markov process $X^T$.

We introduce the following characterization of $X$ introduced in Section \ref{secTF} and the time-changed process $X^T$ in terms of the corresponding lifetimes. Denote by $\zeta^T$ the lifetime of $X^T$.
\begin{defin}
\label{defDelRus}
Let $E \subset \mathbb{R}^d$.
\begin{itemize}
\item[-] We say that $X$ is delayed by $T$ if $\mathbf{E}_x[\zeta^T] > \mathbf{E}_x[\zeta]$, $\forall \, x \in E$.
\item[-] We say that $X$ is rushed by $T$ if $\mathbf{E}_x[\zeta^T] < \mathbf{E}_x[\zeta]$, $\forall \, x \in E$.
\end{itemize}
Otherwise, we say that $X$ runs with its velocity.
\end{defin}

\begin{remark}
Let $X$ be a Brownian motion. If $X$ is killed on $\partial E$ we notice that $\mathbf{E}_x[\zeta] < \infty$. We underline the fact that if $X$ is reflected on $\partial E \setminus \partial B$ and killed on $\partial B$ with $B \subset E$, we have that $\mathbf{E}_x[\zeta] < \infty$ only if $E$ is non-trap for $X$.

Apart from smooth domains, examples of non-trap domains are given by snowflakes or more in general scale irregular fractals as in figures \ref{fig1} and \ref{fig2} (see \cite{CAP} for details). Figures \ref{fig1} and \ref{fig2} are realizations of the random domain obtained by choosing randomly the contraction factor step by step in the construction of the pre-fractal.

\begin{figure}
\includegraphics[scale=.8]{KochPic-outside.pdf}
\caption{Koch curves outside the square. We have Neumann condition
on $\partial E \setminus \partial B$ and Dirichlet condition on $\partial B$ where $B$ is the Ball inside $E$. The domains are non-trap for the Brownian motion.
}
\label{fig1}
\end{figure}

\begin{figure}
\includegraphics[scale=.5]{KochPic-con.pdf}
\caption{Koch curves inside the square. We have Dirichlet condition on the boundary $\partial E$. The domains are non-trap for the Brownian motion.
}
\label{fig2}
\end{figure}

\end{remark}

We immediately see that, if $X^T$ is a delayed process,
\begin{align*}
\mathbf{P}_x(\zeta >t) < \mathbf{P}_x(\zeta^T > t) \leq \frac{1}{t} \mathbf{E}_x[\zeta^T],
\end{align*}
whereas, if $X_T$ is a rushed process,
\begin{align*}
\mathbf{P}_x(\zeta^T>t) < \mathbf{P}_x(\zeta > t) \leq \frac{1}{t} \mathbf{E}_x[\zeta].
\end{align*}

\begin{remark}
We observe that, from Theorem \ref{B},
\begin{align*}
\mathbf{E}_x[(\zeta^L - \zeta)] = \mathbf{E}_x[(H_\zeta - \zeta)] = \int_0^\infty (\mathbf{E}_0[H_s] - s) \mathbf{P}_x(\zeta \in ds) = (\Phi^\prime (0) - 1) \mathbf{E}_x[\zeta]
\end{align*}
(which is negative/positive depending on $\Phi^\prime$) whereas, from Theorem \ref{meanHzeta},
\begin{align*}
\mathbf{E}_x[(\zeta^H - \zeta)] = & \mathbf{E}_x[(L_\zeta - \zeta)] = \int_0^\infty (\mathbf{E}_0[L_s] - s) \mathbf{P}_x(\zeta \in ds).
\end{align*}
There are explicit representations of $\mathbf{E}_0[L_s]$ only in some case.
\end{remark}

Some interesting consideration have been given in \cite{GDLS} where analytical and numerical results on exit times for stable processes have been investigated. The authors observed that the mean exit time depends on the domain size and the index $\alpha$. Some comparison has been given in terms of $\alpha_1, \alpha_2 \in [0,1]$. They argued that the mean exit time can help us to establish the dominant component between the jump frequency or the the jump size.

\section{Examples and Applications}

In this section we
consider some special examples for the time change $T$ which are of interest in our analysis. We write $f\sim g$ if $f(z)/g(z) \to 1$ as $z \to \infty$. We denote by $C$ any positive constant appearing in the following examples.

\begin{example}
Let us consider $\Phi(\lambda)=\lambda^\alpha$ with $\alpha \in (0,1)$.
\begin{itemize}
\item[i)] By using the Laplace transform \eqref{Lap-mittag-L-F} of the density of $L$, we have that
\begin{align*}
\sum_{k \geq 0} \frac{(-\lambda)^k}{k!} \mathbf{E}_0[(L_t)^k] = \mathbf{E}_0[e^{-\lambda t}]= E_\alpha (-\lambda t^\alpha) = \sum_{k\geq 0} \frac{(- \lambda t^\alpha)^k}{\Gamma(\alpha k + 1)}
\end{align*}
from which
\begin{align*}
\mathbf{E}_0[L_t] = \frac{ t^\alpha}{\Gamma(\alpha + 1)}
\end{align*}
where we denoted by $\Gamma(z) = \int_0^\infty e^{-s} s^{z-1} ds$ the gamma function. Thus, for the standard Brownian motion $X$ on $\mathbb{R}^d$ we have that $X^L$ is a diffusion with
\begin{align*}
\mathbf{E}_x[(X\circ L_t)^2] = \mathbf{E}_0[L_t] = \frac{ t^\alpha}{\Gamma(\alpha + 1)}.
\end{align*}
\item[ii)] Let us consider the killed Brownian motion $X$ in $E \subset \mathbb{R}^d$ with the first exit time
\begin{align*}
\tau_E = \inf \{ t> 0\,:\, X_t \not \in E\}
\end{align*}
such that $\mathbf{E}_x[\tau_E] < \infty$. From Theorem \ref{B}
\begin{align*}
\infty = \mathbf{E}_x[\zeta^L] > \mathbf{E}_x[\zeta] = \mathbf{E}_x[\tau_E]
\end{align*}
and we say that $X$ in $E$ is delayed by $L$. The asymptotic behaviour of the delayed Brownian motion have been investigated also in the interesting paper \cite{Mschilling}.
\end{itemize}

The time-changed process $X_{L_t}$, $t\geq 0$ is usually termed delayed process in the sense that the new clock $L_t$ is a process whose trajectories have plateaus. Actually, $X$ is delayed by $L$ in the sense of Definition \ref{defDelRus}.
\end{example}

\begin{example}
\label{commentsZV}
Now we focus on the stable subordinator $H$ with symbol $\Phi(\lambda)=\lambda^\alpha$, $\alpha \in (0,1)$. Let $\mathcal{B}_r= \{x \in \mathbb{R}^d\,:\, |x| \leq r\}$ be the ball of radius $r>0$. Let $X$ be a Brownian motion on $\mathbb{R}^d$, $d\geq 1$.
\begin{itemize}
\item[i)] First we recall that (see \cite[formula (4.7)]{dov2010} for example)
\begin{align}
\mathbf{E}_0 [(H_t)^\gamma] = \frac{\Gamma(1- \gamma/\alpha)}{\Gamma(1-\gamma)} t^{\gamma/\alpha}, \quad -\infty < \gamma < \alpha
\label{momH}
\end{align}
Then,
\begin{align*}
\mathbf{E}_x[(X \circ H_t)^2] = \mathbf{E}_0[H_t]
\end{align*}
which is infinite. We should say that $X^H$ behaves like a superdiffusion.
\item[ii)] Let us consider symmetric stable process of order $2\alpha$, then the mean exit time from the ball $\mathcal{B}_r$ is given by (see \cite[formula (24)]{BGR})
\begin{align}
\mathbf{E}_x[\tau_{\mathcal{B}_r}] = \frac{1}{2^{2\alpha}} \frac{1}{\Gamma(\alpha + 1)} \frac{\Gamma(d/2)}{\Gamma(\alpha + d/2)} \left( r^2 - |x|^2 \right)^\alpha.
\label{hittingSS}
\end{align}
Let us consider a Brownian motion $X$ in $\mathbb{R}^d$ with $X_0=x \in \mathcal{B}_r$ killed on $\partial \mathcal{B}_r$. Then $X^H$ is identical in law to a symmetry stable process of order $2\alpha$. That is, $X^H$ is a subordinated killed Brownian motion (see the interesting paper \cite{KSvond}). Let $\zeta^H$ be the lifetime of $X^H$. We have that $\tau_{\mathcal{B}_r} = \zeta^H$ almost surely.
Let $\zeta$ be the lifetime of the base process $X$.

For $X_0=0$, that is, $X$ is a Brownian motion on $\mathcal{B}_r$ started at zero, formula \eqref{hittingSS} says that
\begin{align}
\label{rmkZV}
 \mathbf{E}_0[\zeta^H] < \mathbf{E}_0[\zeta]
\end{align}
only if
$$r > 2 \left( \frac{1}{\Gamma(\alpha + 1)} \frac{\Gamma(1+d/2)}{\Gamma(\alpha + d/2)} \right)^{1/(2-2\alpha)} = \rho(\alpha, d)= \rho > 0. $$
In particular, if $d=1$, then formula \eqref{rmkZV} holds for $r  > e^{\frac{3}{2}- \gamma} \approx 2.52$ $\forall \, \alpha$ where $\gamma \approx 0.577$ is the Euler-Mascheroni constant. If $d=2$, then formula \eqref{rmkZV}  holds for $r> 2 e^{1-\gamma} \approx 3.05$, $\forall\, \alpha$.

Let us  consider, for $r>\rho$, the annulus $\Sigma=\mathcal{B}_r \setminus \mathcal{B}_{\rho^*}$ and the set $\mathcal{B}_r = \mathcal{B}_{\rho^*} \cup \Sigma$ where $\rho^*=\sqrt{r^2 - \rho^2}$. For the Brownian motion $X$ started at $x \in \mathcal{B}_r$, we obtain
\begin{align*}
\mathbf{E}_x[\zeta^H] < \mathbf{E}_x[\zeta] \quad \forall\, x \in \mathcal{B}_{\rho^*}
\end{align*}
and
\begin{align*}
\mathbf{E}_x[\zeta^H] > \mathbf{E}_x[\zeta] \quad \forall\, x\in \mathcal{B}_r \setminus \mathcal{B}_{\rho^*}
\end{align*}
that is, $X$ on $\mathcal{B}_{\rho^*}$ is rushed by $H$ and $X$ on $\Sigma$ is delayed by $H$. Notice that, as $r\to \infty$, $X$ on $\mathbb{R}^d$ is rushed by $H$.\\

Let us consider now the ball $\mathcal{B}_r$ with $r<\rho$. For the Brownian motion $X$ started at $x \in \mathcal{B}_r$ we have that
\begin{align*}
\mathbf{E}_x[\zeta^H] > \mathbf{E}_x[\zeta] \quad \forall\, x \in \mathcal{B}_r
\end{align*}
and $X$ on $\mathcal{B}_r$ is delayed by $H$.

\begin{figure}
\includegraphics[scale=.6]{exitT.pdf}
\caption{Mean exit times \eqref{hittingSS} from the set (ball) $\mathcal{B}_r \subset \mathbb{R}$ with $r=1$ (first picture) and $r=3$ (second picture). The dashed line refers to $\alpha=0.6$ whereas the solid line refers to $\alpha=1$ (the Brownian motion case). We have that $\rho=2.101$. In the second picture we have that $\rho^*=  2.1356$ determines the set (annulus) $\Sigma= \mathcal{B}_3 \setminus \mathcal{B}_{\rho^*}$ in which $\mathbf{E}_x[\zeta^H] > \mathbf{E}_x[\zeta]$, $\forall\, x \in \Sigma$.}
\end{figure}
\end{itemize}
\end{example}

\begin{example}
Let us consider $\Phi(\lambda)=\lambda^\alpha$ and the exit time from $(-\infty, a)$ for a Brownian motion $X$ on the real line.
\begin{itemize}
\item[i)] From \eqref{momH}, the mean square displacement of $X \circ H$ is infinite. Furthermore, we immediately see that, for $\gamma \in (0, \alpha)$, the process $X \circ (H)^\gamma$ is a subdiffusion.
\item[ii)] It is well known that (see \cite[page 30]{BatWay})
\begin{align}
\mathbf{P}_{x}(\tau_{(-\infty, a)} \in dt) = \frac{|a-x|}{\sqrt{4\pi t^3}} e^{- \frac{(a-x)^2}{4t}} dt, \quad x<a
\label{densityHittingBM}
\end{align}
is the density of the exit time from $(-\infty, a)$ of a Brownian motion with generator $\Delta$. Formula \eqref{densityHittingBM} gives the density of a $1/2$-stable subordinator  for which
\begin{align*}
\mathbf{E}_x[(\tau_{(-\infty, a)})^\alpha] = \frac{\Gamma(1/2 - \alpha)}{2^\alpha\, \Gamma(1/2)} |a -x|^{2\alpha }, \quad \alpha < 1/2
\end{align*}
and is infinite for $\alpha>1/2$ (as stated also in \eqref{momH}). Thus, for $\alpha < 1/2$
\begin{align*}
\mathbf{E}_x[(\tau_{(-\infty, a)})^\alpha] < \mathbf{E}_x[\tau_{(-\infty, a)}] = \infty
\end{align*}
This means that for the killed Brownian motion $X_t$ on $E=(-\infty, a)$, from Theorem \ref{meanHzeta}, we obtain for $\alpha<1/2$
$$\mathbf{E}_x[\zeta^H] = \frac{1}{\Gamma(\alpha + 1)} \mathbf{E}_x[(\zeta)^\alpha] < \mathbf{E}_x[\zeta]= \mathbf{E}_x[\tau_E]$$
and we say that $X$ is rushed by $H$ (only if $\alpha < 1/2$).
\end{itemize}
\end{example}

\begin{example}
(Sierpinski gasket)
Consider $\Phi(\lambda)=\lambda^\alpha$. Let $X$ the Brownian motion on a Sierpinski gasket with walk dimension $d_w= \log(5)/\log(2) \simeq 2.32$.
\begin{itemize}
\item[i)] With formula \eqref{momH} at hand, from  \cite[Corollary 1.6]{BarPerk}, we know that there exists $C$ such that, $\forall\, t>0$,
\begin{align*}
\mathbf{E}_x[(X \circ H_t)^2] = C\, \mathbf{E}_0[(H_t)^{2/d_w}] = \left\lbrace
\begin{array}{ll}
\displaystyle C\frac{\Gamma(1-2/(\alpha\, d_w))}{\Gamma(1-2/d_w)} t^{2/(\alpha\, d_w)}, & 2/d_w <\alpha\\
\displaystyle \infty, & 2/d_w \geq \alpha
\end{array}
\right. .
\end{align*}
From this we argue that, if $H$ is of order $\alpha > 2/d_w$, the Brownian motion on the gasket time-changed by $H$, $X^H_t$,  maintains the subdiffusive behaviour. On the other hand, the behaviour for $\alpha < 2/d_w$ may be related, roughly, to a superdiffusive behaviour.\\

We notice that the previous argument applies for $d_w>2$, that is for Brownian motions on various fractal domains.
\item[ii)] Focus on the random time given by the inverse $L$ of the stable process $H$. We have that
\begin{align*}
\mathbf{E}_x[(X \circ L_t)^2] = C\, \mathbf{E}_0[(L_t)^{2/d_w}] \sim t^{2\alpha / d_w}.
\end{align*}
Thus, $\forall\, \alpha$, $X^L_t$ exhibits a subdiffusive behaviour.
\end{itemize}
\end{example}

\begin{example}
\label{exampleGammaInverso}
Let us focus on the symbol $\Phi(\lambda)= a \ln (1+ \lambda/b)$, $a>0$, $b>0$, that is $L$ is an inverse to a gamma subordinator $H$.

\begin{itemize}
\item[i)] It is well-known that, for an inverse process $L$, there exist two positive constants $c_1, c_2$ such that (\cite{BerBook})
\begin{align}
c_1 / \Phi(1/t) <  \mathbf{E}_0[L_t] < c_2 / \Phi(1/t).
\end{align}
Than, for the symbol of a gamma subordinator it holds that
\begin{align*}
\Phi(1/t) = \frac{1}{t} \ln \left( 1+ \frac{1}{b t} \right)^{a t} \sim  \frac{a}{b} \frac{1}{t} \quad \textrm{as } \, t \to \infty.
\end{align*}
Thus, the inverse $L$ has the property
\begin{align*}
\mathbf{E}_0[L_t]  \sim C \frac{b}{a} t \quad \textrm{as } \, t \to \infty
\end{align*}
with $a,b>0$. We have a process with natural diffusivity, that is
\begin{align*}
\mathbf{E}_x[(X\circ L_t)^2] \sim C\,  t, \quad as\ t \to \infty.
\end{align*}
\item[ii)] By considering the time-changed process $X \circ L_t$ we have that (see Theorem \ref{B})
\begin{align*}
\mathbf{E}_x[\zeta^L] = \frac{a}{b} \mathbf{E}_x[\zeta]
\end{align*}
that is, if $a<b$ then $X$ is rushed by $L$ whereas, if $a>b$ then $X$ is delayed by $L$.
\end{itemize}

\end{example}

 \begin{example}
 \label{exampleGammaH}
Let us consider now the gamma subordinator $H$ introduced in the previous example.
\begin{itemize}
\item[i)] From
\begin{align}
\mathbf{E}_0[H_t] = \frac{d}{d\lambda} \left(1 - e^{- t \Phi(\lambda)} \right) \bigg|_{\lambda=0} = t \Phi^\prime(0)
\label{phiH}
\end{align}
we obtain
\begin{align*}
\mathbf{E}_0[H_t] = \frac{a}{b} t
\end{align*}
with $a,b>0$. We still have a process with natural diffusivity, that is
\begin{align*}
\mathbf{E}_x[(X\circ H_t)^2] \sim C\,  t, \quad as\ t \to \infty.
\end{align*}
\item[ii)] For $X \circ H$ , from Theorem \ref{meanHzeta} we have that
\begin{align}
\mathbf{E}_x[\zeta^H] = \mathbf{E}_x[L_\zeta].
\label{lifeH}
\end{align}
We show that
\begin{align}
\label{tmp-a-coeff}
\mathbf{E}_x[L_\zeta] > \mathbf{E}_x[\zeta]
\end{align}
and the base process $X$ is delayed by the random time $H$ only if $a<b$. For the sake of simplicity we set $b=1$. By using the explicit representation of $\mathbf{E}_0[L_t]$  given in \cite[formula  (21)]{kumarPS} we write
\begin{align*}
\mathbf{E}_x[L_\zeta] = \frac{1}{a}\mathbf{E}_x[\zeta] + \frac{1}{2a} - \frac{1}{a} \mathbf{E}_x\left[ e^{-\zeta} \int_0^\infty \frac{e^{-y\zeta}}{(1+y)\, [(\ln y)^2 + \pi^2]} dy \right].
\end{align*}
By considering the integral representation (\cite[pag 2130]{BergPed}, \cite[pag. 989]{QiZhang})
\begin{align}
\label{formulaLN}
\frac{1}{\ln(1+z)} = \frac{1}{z} + \int_1^\infty \frac{1}{(z+y)[(\ln(y-1))^2 + \pi^2]} dy, \quad z \in \mathbb{C} \setminus (-\infty , 0]
\end{align}
and the fact that $\mathbf{E}_x[e^{-(y+1)\zeta}] < 1$, $\forall\, y$ we get that
\begin{align*}
\mathbf{E}_x\left[ e^{-\zeta} \int_0^\infty \frac{e^{-y\zeta}}{(1+y)\, [(\ln y)^2 + \pi^2]} dy \right] < \int_0^\infty \frac{1}{(1+y)\, [(\ln y)^2 + \pi^2]} dy.
\end{align*}
From \eqref{formulaLN}, we have
\begin{align*}
\int_0^\infty \frac{1}{(1+y)\, [(\ln y)^2 + \pi^2]} dy = \lim_{z\to 0} \left( \frac{1}{\ln (1+z)} - \frac{1}{z} \right) = \frac{1}{2}
\end{align*}
and we obtain that
\begin{align*}
\mathbf{E}_x[L_\zeta] > \frac{1}{a} \mathbf{E}_x[\zeta].
\end{align*}
The inequality \eqref{tmp-a-coeff} follows only if $a<1$. Then, we say that $X$ is delayed by $H$.
\end{itemize}
\end{example}

\begin{example}
\label{ExTSS}
A further important example is given by $\Phi(\lambda) = (\lambda + \eta)^\alpha - \eta^\alpha$ for the relativistic stable subordinator $H$. The associated time operator $\mathfrak{D}^\Phi_t$ is usually called tempered fractional derivative (\cite{Beghin}). Tempered stable distribution is particularly attractive in modelling transition from the initial subdiffusive character of motion to the standard diffusion for long times (for the applications of tempered stable distributions see, for example, \cite{jeon,Stani}). Let us consider the inverse $L$ to the subordinator $H$ and the time-changed process $X \circ L$ where $X$ is the Brownian motion.
\begin{itemize}
\item[i)] From Proposition 3.1 in \cite{KumVell}, for large $t$,
\begin{align*}
\mathbf{E}_x[(X \circ L_t)^2] = \mathbf{E}_0[L_t] \sim \frac{1}{\alpha \eta^{\alpha - 1}} t
\end{align*}
and we have a natural diffusion.
\item[ii)] From Theorem \ref{B} we get that
\begin{align*}
\mathbf{E}_x[\zeta^L] =\alpha \eta^{\alpha -1} \mathbf{E}_x[\zeta]
\end{align*}
that is, if $\alpha \eta^{\alpha-1} < 1$ then the process $X$ is rushed by $L$,  whereas if $\alpha \eta^{\alpha-1} > 1$ then the process $X$ is delayed by $L$.
\end{itemize}
\end{example}

\begin{example}
We now consider a further instructive example which does not completely fit the previous scenarios. More precisely, we deal with the one-dimensional fractional Brownian motion $B^\mathcal{H}$ whose density solves
\begin{align*}
\partial_t u = \mathcal{H} t^{2\mathcal{H}-1} \partial_{xx} u, \quad t >0,\, x \in \mathbb{R}, \;\; \mathcal{H} \in (0,1)
\end{align*}
where $\mathcal{H}$ is the Hurst exponent. Although this process is not in the class introduced in Section 2 we consider such a process because of the second moment $\mathbf{E}_x[(B^\mathcal{H}_t)^2] = t^{2\mathcal{H}}$. We provide the following characterizations as in the previous analysis.
\begin{itemize}
\item[i)] We have that
\begin{align*}
\mathbf{E}_x[(B^\mathcal{H} \circ H_t)^2] = \mathbf{E}_0[(H_t)^{2\mathcal{H}}], \quad \mathbf{E}_x[(B^\mathcal{H} \circ L_t)^2] = \mathbf{E}_0[(L_t)^{2\mathcal{H}}].
\end{align*}
Let $L$ be an inverse to $H$ with symbol $\Phi(\lambda) = a\ln (1+ \lambda/b)$. Then,
\begin{align*}
\mathbf{P}_0(H_t \in dx)= \frac{b^{at} x^{at -1}}{\Gamma(at)} e^{-bx}dx
\end{align*}
and, for large $t$,
\begin{align*}
\mathbf{E}_0[(H_t)^\gamma] = \frac{1}{b^\gamma} \frac{\Gamma(at + \gamma)}{\Gamma(at)} \sim \left(\frac{a}{b}t \right)^\gamma.
\end{align*}
Using the relation $\mathbf{P}_0(L_t<x) = \mathbf{P}_0(H_x >t)$ we obtain density and moments for $L_t$. By following  \cite{kumarPS} we are able to obtain
\begin{align*}
\int_0^\infty e^{-\lambda t} \mathbf{E}_0[(L_t)^\gamma] dt =  \frac{\Gamma(1 + \gamma)}{\lambda \left( a \ln (1+ \lambda/b)\right)^\gamma} \sim \frac{b^\gamma}{a^\gamma} \frac{\Gamma(1+\gamma)}{\lambda^{\gamma +1}} \quad \textrm{as}\; \lambda \to 0
\end{align*}
from which we get the $\gamma$-moment of the inverse gamma subordinator
\begin{align*}
\mathbf{E}_0[(L_t)^\gamma] \sim \left( \frac{b}{a}t \right)^\gamma.
\end{align*}
Therefore, we have that, for large $t$,
\begin{align*}
\mathbf{E}_x[(B^\mathcal{H} \circ L_t)^2] \sim \left( \frac{b}{a}t \right)^{2\mathcal{H}}, \quad \mathbf{E}_x[(B^\mathcal{H} \circ H_t)^2] \sim \left( \frac{a}{b}t\right)^{2\mathcal{H}}
\end{align*}
and the subordination leads to anomalous diffusion as well as the time change by inverse subordinator.
\item[ii)] By generalizing the examples \ref{exampleGammaInverso} and \ref{exampleGammaH} we obtain
\begin{align}
\label{nnnggg}
\mathbf{E}_x[\zeta^L] = \frac{a}{b} \mathbf{E}_x[\zeta]
\end{align}
where $\zeta$ is the lifetime of $B^\mathcal{H}$ and $\zeta^L$ is the lifetime of $B^\mathcal{H} \circ L$. Notice that formula \eqref{NoSemig} still holds if $P_t$ is not a semigroup (that is the case here for $B^\mathcal{H}$). Thus, Theorem \ref{meanLzeta} leads to \eqref{nnnggg}.
\end{itemize}

In particular, for the (time-changed) process $B^\mathcal{H} \circ L$ we can write the following table
\begin{center}
\begin{tabular}{c|c|c|c}
 & sub & super & natural\\
 \hline
rushed & $a<b$, $\mathcal{H} < 1/2$ & $a<b$, $\mathcal{H} > 1/2$ & $\mathcal{H}=1/2$\\
\hline
delayed & $a>b$, $\mathcal{H} < 1/2$ & $a>b$, $\mathcal{H} > 1/2$ &$\mathcal{H}=1/2$
\end{tabular} .
\end{center}
The time change for the fractional Brownian motion has been recently considered in \cite{KGWP}. The authors considered the tempered stable subordinator introduced in Example \ref{ExTSS}.
\end{example}

\textbf{Acknowledgement.} The authors would like to thank Prof. Zoran Vondra\v{c}ek for precious discussions and valuable comments.

\textbf{Grant.} The authors are members of GNAMPA (INdAM) and are partially supported by Grants Ateneo \lq\lq Sapienza" 2017.
\bigskip

redpajama_set_name: RedPajamaArXiv


text:
Rad Listed – Three month edition!
February 14, 2016 Rad Comments 0 Comment
Wow! Has it been 3 months since my last post!? Too busy playing games, not enough time writing about them. Fortunately, I have a lot of amazing games to talk about, and a ton of great recommendations to make. There's also a lot of exciting games coming out in the next few months to look forward to as well!
Before I get started, I wanted to mention that the Chats with Rad Teamspeak server is still up and running and we are pretty active on there. Feel free to join us anytime if you're looking for a group to play PC games with. We're mostly playing Heroes of the Storm every night, along with a smattering of games on Friday and Saturday nights–currently, Rocket League, Dying light, or GTA: Online. You can find all the information about the server in the above 'Teamspeak Voice Server' link.
Prepare yourself: this is going to be a big ol' Rad Listed!
With so many great games coming out in the coming months, I have been trying to spare my wallet. X-Com 2 just came out this last Friday, and even though I tried my best to resist its call, I eventually caved and bought the game the following Sunday. The game is great, and I'm having a lot of fun with it.
The edition of the custom character pool is brilliant and makes it easy to start a new game and still have your custom-made characters available. I like to make all of my custom characters the names of my friends and family– it adds a whole new depth to the game (It also gives me an excuse to yell at them on occasion for missing crucial shots, and angrily texting them to get their act together).
I'm only a few missions in, but this new iteration seems to be much harder and more unforgiving than the last X-Com. However, this is a difficulty I relish in, and look forward to when I learn how to overcome it. Stay tuned for a potential full write-up on X-Com 2 in the future.
If you ever played Dead Island, think of that game with improved combat mechanics and parkour. Got it? Now you have Dying Light, a melee-heavy zombie survival parkour game (I can't make this stuff up!). I, along with my usual crew, have recently been playing this game cooperatively and having an absolute ball with the combination of first-person platforming and melee-based combat. The fun melee and rag-doll physics occasionally creates hilarious situations that have us rolling on the floor laughing.
I can't speak to how good the story-line is, as we just skip all dialogue to get back to the action, but the gameplay is top-notch! As always, fun is multiplied with friends, but I'm sure Dying Light would be a fun single-player experience, if not for different reasons.
We like to yell 'parkour!' occasionally while jumping and climbing over buildings.
Yup! Still playing Heroes of the Storm, on an almost-daily basis with a group of friends. While 'salty' is a flavor I choose often, I still have a blast playing. The store-wide 50% off sale over the holidays was a blow to my wallet, but I managed to nab myself quite a few heroes. I have unlocked essentially all of the heroes I'm interested in playing, along with some neat skins. While I lament my abandonment of DotA sometimes, I'm quite content with HotS' shorter game times and less-complicated gameplay, allowing your team to focus more on the intricacies of team fights.
I had been eagerly awaiting Fallout 4's release for quite a while, and I was not disappointed! I thoroughly enjoyed the new mechanics they introduced, most especially the weapon modification options. While the base-building mechanics weren't terribly engaging, I did have fun participating in it a little bit– I just wish that, mechanically, it had more impact on the game world. I look forward to the additional content that Bethesda will be inevitably releasing down the road.
Fallout 4's release may or may not explain my sudden absence. =P
I may have mentioned Divinity: Original Sin in the past. It's a single player and/or cooperative top-down, turn-based RPG, not dissimilar to the gameplay of Baldur's Gate. A friend and I have been playing it cooperatively and having quite a bit of fun in the process. The way that elemental effects can impact the battlefield make magic-user classes very enjoyable. Gas clouds explode when fire meets them, wet enemies have a higher chance of being frozen by ice-based attacks, and oil can be lit on fire to slowly burn your enemies. The developer recently released a free 'Enhanced Edition' update for all Original Sin owners, which has added a plethora of new activities and features!
Shadowrun: Hong Kong
The Shadowrun games are narrative, turn-based, tactical RPGs, similar to the X-Com combat. There is a lot of emphasis on player dialogue, choice, story-telling and character skill progression.
I absolutely love the Shadowrun games that Hare Brained Schemes has revitalized in the last few years! I have played through all of the Shadowrun games they have made so far: Shadowrun Returns, Shadowrun: Dragonfall, and now Shadowrun: Hong Kong. I even joined in on the kickstarter on Hong Kong, which i did not regret in the slightest. They even recently added a free epilogue campaign to all current owners… I know what I'll be doing in the near future!
What I like most about the Shadowrun games is the world they've created. It's exceedingly interesting, as the games take place in a fantasy and tech-heavy dystopian future, where dragons rule the leading world corporations, pulling the strings from the shadows.
It took way too dang long, but Batman: Arkham Knight finally became relatively playable near the end of 2015–pretty sad considering it had been out since June! Regardless of my misgivings for the awful PC release of Arkham Knight, I had a ton of fun with the game itself!
I found the Batmobile combat pretty enjoyable, dodging tank shots and trading fire, with explosions everywhere! As usual, the Batman melee combat is solid, and by the end of the game you find yourself chaining together a ridiculous amount of blows without even getting touched, like the true Batman. In the end though, it feels like more of the same, so if you think you've had enough of the Batman combat, puzzle solving, and Riddler trophies, this may be something you'll want to pass on.
I've had my eye on this physics-based multiplayer brawler for awhile. I finally managed to get four willing participants just the other day, and man, did we have a lot of fun! We were laughing uncontrollably the entire duration. The gelatinous movements of your character and the silly rag doll physics, combined with the ridiculous levels really comes together into a very entertaining experience. My favorite level was one with two trucks driving alongside each other, where you must jump across to the other to engage your opponents whilst dodging passing road signs.
I found myself playing an old classic on my Nexys 7 during the Christmas Holiday while I was away from my computer. Knights of the Old Republic is still a fantastically fun RPG, and has actually aged pretty well. While controlling it via the touch screen is a battle sometimes, after a bit of practice, I became pretty proficient with it. Even when I try to build my character as a ranged damage dealer, I always end up switching to melee. There's something about bashing someone in the face with a Vibro-Blade that is just…so…satisfying.
I finally caved. While I will always primarily be a PC gamer in every way, shape, and form, I determined that the PS4 had enough console exclusives I was interested in to justify the purchase. I will still prioritize PC for all multi-platform releases, but the PS4 has some heavy hitters coming to the console in the future. I'm most excited about the PS4 exclusive; Horizon Zero Dawn.
While Destiny is not a PS4 exclusive, it is a console exclusive. Since Bungie has given no indication in bringing Destiny over to PC, I decided to get the game on my new shiny PS4. I never played Destiny until well after The Taken King's release, so I missed out on the rocky transition of the original game. However, what I'm playing now is a great first person shooter MMO-ish RPG. I say MMO-ish because it has almost all of the tropes of an MMO, but can't seem to decide on what it really is. That's okay though– the game is quite enjoyable the way it is, despite a feeling of segmentation due to a large number of load screens when transitioning between areas.
One of the primary reasons I wanted a PS4 was so I could get a chance at Bloodborne. While it was not my only reason for purchasing a PS4, it was a big factor in the equation, as it was another game that was seemingly never going to find its way to PC. As a huge Dark Souls fan, I already knew I'd love Bloodborne. I didn't expect to be as floored by it as I have been, though. I love the quicker combat that rewards the player with quick dodging and poking attacks. Just don't over commit to a combo, like I do!
Expect another post next week, where I plan on talking about all the exciting upcoming games in the next couple months!
Batman Arkham Knight, Bloodborne, Destiny, Divinity: Original Sin, Dying Light, Fallout 4, Gangbeasts, Heroes of the Storm, PS4, Shadowrun: Hong Kong, Star Wars: Knights of the Old Republic, Video Games, XCOM 2
Woot Post – Primal Post Apocolyptic Souls in the Necropolis
Metal Gear Solid: The Phantom Pain

redpajama_set_name: RedPajamaCommonCrawl


A Mendelian disease is one where a single genetic mutation decides whether you get the disease or not. For instance, PRNP mutations are wholly responsible for fatal familial insomnia, and CFTR mutations are wholly responsible for cystic fibrosis.
By contrast, complex traits and complex diseases are determined by a whole bunch of different genetic factors. Height, for instance: if your parents are tall, you're likely to be tall too, but it's not as simple as 'you got the tall gene.' Though we don't know the full genetic architecture of any complex diseases, it is believed that hundreds of genomic loci contribute to many complex diseases as well – such as schizophrenia (I couldn't find any good recent reviews; one relevant paper is International Schizophrenia Consortium 2009).
The concept of 'heritability' is pretty simple for Mendelian diseases themselves because they follow Mendelian inheritance. If the disease is dominant, then an affected person's kids are 50/50 to have the disease too, and if the disease is recessive then each child of two heterozygous carriers has a 25% chance of having the disease. For complex traits and complex diseases, a great deal of effort goes into trying to figure out how 'heritable' they are – how much of the phenotypic variance in the population can be explained by genetics, as opposed to environmental factors or stochastic variation (see heritability post).
However, even Mendelian diseases can have genetic modifiers, and those modifiers can be heritable. Cutting 2011 reviews what we know about genetic modifiers in cystic fibrosis. Even though everyone with both CFTR alleles gets cystic fibrosis, some people get very sick very young and others stay healthy much longer. Because cystic fibrosis is a loss-of-function disease, it can be caused by a wide variety of loss-of-function mutations in CFTR: Cutting says over 1800 mutations have been identified so far. Not surprisingly the biggest genetic modifier of many cystic fibrosis phenotypes (especially in the pancreas and intestines) – is simply which CFTR mutation(s) you have. Some are complete loss-of-function while others are able to preserve a little bit of functionality in some tissues.
However it has long been suspected that there are other genetic modifiers in cystic fibrosis as well. Cutting reviews a number of studies that have tried to estimate the heritability of some of the disease phenotypes (quantitative measurements of lung function, body mass index, etc.). The methods used are largely the same as the methods for estimating heritability in complex traits – MZ vs DZ twin studies, sibling correlation, etc. I didn't get any sense of a consensus from Cutting's article, though: the heritability estimates for cystic fibrosis phenotypes vary dramatically, all the way from 0.0 (the first MZ/DZ twin study found MZ twins no more similar than DZ twins for individual phenotypes, though there was heritability of one composite measure [Mekus 2000]) to 1.0 [Vanscoy 2007]. The only methodology new to me here was a study of phenotypic concordance between twins before and after they moved out of the home [Collaco 2010], a methodology purported to separate out the effects of shared vs. individual environment.
Cystic fibrosis is a bit of a special case: it is one of the most common Mendelian diseases. For more rare diseases you aren't going to find enough twins to do twin studies. So I also spent some time looking at how heritability has been estimated in very rare diseases.
But apparently there was a global correlation with r = .25 – which, as I interpret it, would mean an upper limit on heritability of 2r = 50%. This correlation seemed to be driven mostly by the P102L cases (black dots in above), which had a significant r = .47, though the other mutations also trended towards correlation.
Because Webb's study used parent-offspring pairs, it's biased towards underestimating heritability if anything. Webb does not go into great detail about how the clinical data were processed and how/whether not-yet-affected carrier offspring were considered in the analysis, so it is hard to know how big an issue this might be.
Moving on, for prion disease in particular another source of evidence towards the existence of genetic modifiers is that scrapie incubation times vary about twofold between different inbred mouse lines [reviewed in Lloyd 2011]. This is no guarantee that there are similar modifiers in humans, but it's fairly suggestive. Lloyd reviews several studies which have used mice for QTL mapping – a concept reviewed by Peters 2007. In brief, two inbred mouse lines are crossed to get highly heterozygous F1s, then the F1s are crossed again to get mixed F2s in which you can look for which genomic locations correlate best with the phenotype. For prion diseases these studies have revealed some candidate loci, but no conclusive modifier genes yet.
There is also some interesting literature on heritability of age of onset in spinocerebellar ataxia 2 (SCA2), which is caused by polyQ repeat expansion in the ATXN2 gene. Just like in Huntington's Disease, the Q repeat length explains much of the variance in age of onset (more repeats is worse). So just like in HD, people who study SCA2 consider a residual age of onset – the difference between actual onset and predicted onset based on polyQ length. There are a number of other SCAs, each caused by polyQ expansion in a different gene, and Pulst 2005 (ft) hypothesized that some of the residual age of onset in SCA2 might be explained by polyQ length in the other SCA-causing genes. Pulst genotyped polyQ length in those genes in ~400 unrelated people and sure enough, it turned out that polyQ length polymorphisms in CACNA1A, the gene that causes SCA6, explained about 6% of the residual (not total) variance in age of onset in SCA2. None of the SCA2 patients had a CACNA1A allele long enough to cause SCA6 – but the early onset people had slightly longer alleles than late onset.
That's quite an interesting finding and the results were significant even after Bonferroni correction. However, the heritability analysis in that paper was somewhat less awesome. The heritability calculations used a separate set of patients from the above analyses, and simply doubled the sibling correlation to obtain a figure of 55%. Pulst correctly describes this as an upper bound and not an estimate of heritability (no control to rule out shared environment contribution). But still, there is no control here for the ascertainment bias that Squitieri 2000 discusses – age onset studies are simply more likely to include concordant sib pairs than discordant sib pairs. If Pulst did anything to control for this problem, it's not discussed in the paper – all it says about the sib pair data is that "For the analysis of residual variance and heritability, information from 148 SCA2 patients from 57 sibships was used." This ascertainment bias will tend to inflate Pulst's figure, so I would take 55% as a rather loose upper bound.
In myotonic dystrophy, the CTG repeat length explains most of the age onset variance. The repeat length also explains most of the level of somatic instability, but the residual instability explains part of the age of onset [Morales 2012]. Morales also suggests that the level of somatic instability itself is itself ~40% heritable, but here the analysis is not transparent. Morales feeds the data into QTDT, a software tool introduced by Abecasis 2000. Abecasis' formula for partitioning variance into genetic, shared environment and individual environment components depends upon "the proportion of alleles shared identical by descent (IBD) between siblings j and k in family i" – to my reading, this is sibling IBD regression, similar to that used by Visscher 2006 in calculating height heritability. Visscher actually had genome-wide SNP markers to determine sibling IBD (which ranges from ~40% to ~60% – not all sibling pairs are equally alike!) Yet people have for years been using QTDT to do heritability estimates (both Morales above and U.S.–Venezuela Collaborative Research Project 2004) without any empirical sibling IBD estimates. Perhaps this involves an assumption that all sib pairs share 50% IBD, but if so, neither Morales nor the USVCRP paper mention this assumption, and it is not clear to me from Abecasis' paper how the shared environment and genetic components can possibly be partitioned if IBD doesn't vary at all. I also checked the QTDT online tour and, indeed, the variance partitioning relies on IBD from sibling pedigree files.
Maybe there is something I'm missing here (if so, please leave a comment to let me know) but, even after reading Abecasis 2000 and looking at the QTDT online tour, I am still at a loss for understanding how authors are using QTDT to create heritability estimates without any genome-wide SNP genotyping to determine IBD.
Heritability estimation is difficult to begin with [see Visscher 2008 for a review]. Doubly so in diseases that are too rare for twin studies to be possible, and doubly so again for age of onset phenotypes, where family studies suffer from often unexamined ascertainment bias. It is clear that some Mendelian diseases do have heritable modifiers of age of onset. Pulst 2005's discovery of a modifier for SCA2 is convincing and (to my knowledge) has never been contested in the literature; and Webb 2009's finding of heritability in prion diseases, because it uses parent-offspring regression, is more likely to underestimate than overestimate heritability. But point estimates of heritability in these diseases should probably be taken with a large dose of salt – even more than for heritability studies generally.


Adelaide's newest endurance challenge ride - Hot on the tyre tracks of the Tour Down Under!
Adelaide is hosting the newest on road endurance cycle challenge to kick off the Millennium, hot on the tyre tracks of the Tour Down Under. This brilliant ride will take you from the city of Adelaide through to the tranquil towns and spectacular scenery of the Adelaide Hills. Watch out for the whales as you ride along the southern coastline to complete the challenge with a champion's finish at the resort town of Victor Harbor.
This event also provides for the family with a fun carnival at Victor Harbor to welcome in the finishing cyclists.
The entry fee includes some great incentive prizes for teams and individuals as well as plenty of refreshment and fuel stops, detailed signage, mechanical and medical support as well as a well earned massage at the finish.
The distance is 105 km. Get ready for the challenge.


text:
OneBadger CRM Initiative
Constituent Relationship Management for UW–Madison
CRM Adoption
Program Spotlight
Spotlight Program
Strategic Partnerships "Org" In Salesforce Paints Clearer Picture Of External Relations
Hiring our students. Seeking professional development and training for employees. Sponsoring a research partnership. Establishing a scholarship fund.
There are many reasons why a business might want to develop a relationship with a world-class institution like UW–Madison. And it's the mission of the Office of Business Engagement to connect companies of all sizes with the resources they need to build successful relationships across—and with—the university.
But with so many industry contacts engaging with faculty, staff and researchers across campus at any given time, keeping track of all that activity is a big challenge.
That's where a constituent relationship management platform, or CRM, comes into play. In implementing and using the Salesforce CRM, part of the campus-wide OneBadger Initiative, the Office of Business Engagement (OBE) and its campus collaborators sought to more efficiently and effectively manage the strategic partnerships that OBE cultivates and nurtures.
"By connecting UW external relations professionals and providing a platform for collaboration and information-sharing, the result is less overlap and more development of strategic plans that span schools, colleges, institutes and other UW units—all of whom are looking to engage more with external partners," explains OBE Managing Director Amy Achter.
Bridging silos, across UW and partner companies
From the outside looking in, a large, complex institution like UW–Madison can appear to be siloed and disparate across a vast campus. That's where the capabilities of Salesforce can help form bridges between university entities. At the same time, large corporations can operate in a siloed manner, too.
Take, for example, a company that's already in talks with the Wisconsin School of Business about recruiting graduating students for its workforce. While that relationship progresses, another company leader initiates talks with the College of Engineering about a technology collaboration to develop a new product. When those new opportunities come up, OBE is often engaged to help take those external partners "up and out," exposing the companies to potentially beneficial relationships they may want to form across the university, OBE leaders say.
"It became clear to us that there's an opportunity to begin to pull all of this information into one place, so there was visibility across our schools, colleges and units and the opportunity to collaborate and think about our external partners more holistically," Achter explains.
Additionally, having all the information in one place provides UW users with context about a company's prior campus engagements, including creating connections with campus contacts in other units who may be able to help with introductions or background about an existing relationship.
From fishing expeditions to dashboards and metrics
Prior to housing all that information in Salesforce, OBE staff would have to go on a fishing expedition to find information in multiple systems and applications, or spend a lot of time on the phone or sending emails to uncover all the activity across campus tied to a specific company.
"There was a lot of digging," Achter said. "The only thing I knew for sure was something was probably missing somewhere."
Perhaps the most well-known example of a multi-faceted, comprehensive partnership with industry is the university's relationship with American Family Insurance. OBE manages this wide-ranging relationship with AmFam, which spans across multiple schools and colleges, data sciences, marketing, athletics, community relations and talent acquisition.
By using Salesforce to track all aspects of the relationship—with AmFam or other partners—campus external relations staff are better positioned to track the outcomes of the engagements, from revenue generation to the hiring of students.
"In Salesforce, we can watch the engagements, see how we're doing and how they're working," Achter explains. "There's a lot of ability to look at where we are and what's in our pipeline."
A virtual endeavor
The Office of Business Engagement's Salesforce implementation began in March, led by Business Engagement Director Megan O'Rourke. Working with consultants from Huron, O'Rourke and a core team of campus stakeholders configured and developed UW–Madison's first external-facing instance of Salesforce, which launched in June.
It was a complex task made even trickier by the COVID-19 pandemic: Two days after the project began, the university went virtual.
But even amid the complexities of working remotely, O'Rourke says several common themes immediately emerged in the discussions—involving OBE staff along with representatives from the SuccessWorks career center in the College of Letters & Science, the School of Human Ecology career center, the Wisconsin School of Business and the College of Engineering.
"When we bring companies to campus, these are the folks we're calling first to meet with these companies," O'Rourke explains. "They're so often the ones who can offer the solution that company is looking for."
In a half-day retreat, Division of Information Technology Senior Business Analyst Tamra Dagnon took the core team through an exercise to develop a variety of "user stories," to help articulate how different Salesforce configuration options might provide value to the end users of the CRM.
"It was so valuable, having her expertise to tease out what we needed and what was important to us," O'Rourke recalls.
The core group also made several key decisions about which data sources to pull into Salesforce, to help tell stories about recruitment, research and philanthropic activity.
"It was a fun challenge," O'Rourke says, adding that the flexibility within Salesforce to be "endlessly configurable" made that particular CRM a valuable solution to meet the needs identified by OBE and its campus partners.
"We could create a system for us," O'Rourke emphasized. "With other systems, it can be a lot more 'out of the box.' Whereas in our work with Salesforce, we could truly configure all of the nuts and bolts in setting it up."
Growing the ranks
After getting past the initial implementation hurdle, O'Rourke says one of the big post-launch goals is to grow the number of users across campus of the Strategic Partnerships Salesforce "org," or instance. Other Salesforce "orgs" that have gone live across campus as part of the OneBadger CRM Initiative include a Clinical Placement org in the School of Education and a shared Division of Continuing Studies/Office of Admissions and Recruitment org that includes UW–Madison Online and Graduate School Recruitment.
Initial users of the Strategic Partnerships org included about 35 campus entities representing 14 schools, colleges, units and research institutes in functional areas including career services, employer relations, community and state relations, data science and analytics, professional development and entrepreneurship. A second wave of users was onboarded in August, bringing the total to about 56.
When considering even more potential user growth for the Strategic Partnerships org, O'Rourke cites the many research centers across campus doing fee-for-service work with industry, or seeking out research collaborations with companies.
"We're excited to really grow the number of people that want to use the system, input their data and share that information across campus," O'Rourke says, noting the level of technological sophistication afforded by Salesforce.
Adds O'Rourke: "It'll be nice to get rid of a lot of the spreadsheets people are using."
Posted in Spotlight Program
Previous post: Meet Megan O'Rourke, Lead Salesforce Admin For OBE
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4-Step Guide to Supporting Students During COVID-19
Salesforce Powers Multiple Recruiting Efforts Across UW–Madison
OneBadger CRM: 2020 Year in Review
Meet Megan O'Rourke, Lead Salesforce Admin For OBE
Feedback, questions or accessibility issues: webchanges@doit.wisc.edu.


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General Hospital Episodes That Feel Like Deja Vu
Kathy Hutchins/Shutterstock
By Hope Campbell/Jan. 3, 2022 8:07 pm EST
When a show is on the air for more than half a century, 5 days a week, 52 weeks a year, it's bound to repeat stories — or at least types of stories — and even have episodes we feel like we have seen before. "General Hospital" premiered on April 1, 1963 and throughout it's nearly 60 years on the air, as of this writing, it has only had one makeshift rerun season and that didn't come until 2020 when Hollywood productions were forced to shut down at the start of the coronavirus pandemic. Rather than not have any soap opera programming during the day, ABC gave fans theme weeks that included encore episodes of some of "GH's" best hours (via Soap Central).
While those theme weeks certainly gave fans a feeling of deja vu, many still get that sense watching first-run episodes that seem eerily similar to plot points and story vehicles we have seen before. In 2021, we even saw characters doing and saying the same thing over and over again for days on end, making fans wonder if the soap is back in its only rerun season again.
Mob shootouts are the norm on General Hospital
ABC/YouTube
Since 1993, one of "General Hospital's" most popular characters has been Sonny Corinthos, played by Maurice Benard, who eventually became the mob boss of the fictional New York town of Port Charles. Sonny is always defending his territory from one interloping mob family or another, making sure drugs and guns stay off the streets (Fans have yet to figure out what Sonny really traffics through the town, although he claims it's just Corinthos Coffee).
Despite Sonny not wanting Port Charles to be the wild west with guns in the hands of all citizens, he has plenty of them, as do his hitmen and mafia enemies. So shootouts on Pier 54 happen every so often. New fan-favorite Brando Corbin was introduced in the 2020 shootout that introduced the character to the show (via Soaps In Depth), but it still seemed like any other "GH" gunfire sequence through the years, including the one that put in bullets in Kiki Jerome (Hayley Erin) and Nathan West (Ryan Paevey) in 2016 (via Michael Fairman TV).
Dr. Finn to the rescue when General Hospital characters are poisoned
It was 2019 when put-upon heroine Sasha Gilmore (Sofia Mattsson) was poisoned by Cassandra Pierce (Jessica Tuck) — and this poison introduced a version of the bird flu to Sasha's system, according to Soaps.com. As Sasha lay near death, Dr. Hamilton Finn (Michael Easton) had to race against time to save the character, which he eventually did.
Too bad Dr. Finn couldn't save Sasha's newborn baby, who died just in time for Christmas in 2021, enraging fans, according to Soaps.com. However, he did manage to save his brother, Harrison Chase (Josh Swickard), in 2021, when Chase was poisoned by the cartoon villain, Peter August (Wes Ramsey). Once again, Dr. Finn had to race against the clock to concoct any antidote for Chase, which led him to figuring out that Chase really was his brother after all and not his son, as a false DNA test had led him to believe (via Soaps In Depth). The poison left Chase paralyzed and married Willow Tait (Katelyn MacMullen), a woman who didn't love him but married him because she thought he was on his deathbed. And deathbed weddings? As a soap trope, every single one makes fans feel an enormous sense of deja vu.
All of Port Charles mourns Jason Morgan — again
It was only 2012 when mob enforcer Jason Morgan (Steve Burton) was shot on Pier 54, kicked into the river, and presumed dead. Carly Corinthos (Laura Wright) and Sam McCall (Kelly Monaco) walked around in tears for months and Carly almost married a man she thought was Jason's brother, Franco Baldwin (Roger Howarth), who eventually turned out to be his cousin. Jason was deemed a hero by all who knew him, despite the fact that he was a hitman for nearly 20 years at the time. At least he turned out to be alive in 2017.
Flash forward a decade and Jason is presumed dead once again after a tunnel collapse on Cassadine Island (via Soaps In Depth). Sam isn't as broken up about it this time, but Carly is once again apoplectic and turning to Jason's actual brother, Drew Cain (Cameron Mathison), for comfort. And, just like before, Jason the perpetual hitman is being hailed as a saint by the entire town. The more things change in Port Charles, the more they stay the same.


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MICHELLE CONNOR'S A GONNER?
Coronation Street's Michelle Connor to become Pat Phelan's next victim in gruesome wedding day scenes
The shocking scene will be so gory that ITV bosses have decided to air the episode after 9pm so they do not break any broadcasting rules
Updated: 8 Jun 2018, 11:36
DURING his reign of terror in Weatherfield, Pat Phelan has killed off four characters but none of them have been household names.
But his next victim could not be any more well-known — one of Coronation Street's biggest characters Michelle Connor.
Michelle Connor, played by Kym Marsh, will become Pat Phelan's fifth victim after getting caught-up in violent scenes at her weddingMichelle, played by Kym Marsh, will be shot on her wedding day in scenes so graphic they are to be aired after the 9pm watershed.
She will be caught in the cross-fire as dodgy builder Phelan returns — having been presumed dead after being pushed off a cliff by wife Eileen.
Evil Phelan (Connor McIntyre), will crash Michelle and Robert's big day at the Bistro and wreak havoc.
My Cobbles source said: "The scenes were shot last week and are more like a horror film than a soap.
Corrie villain Pat Phelan was thought to be dead after wife Eileen pushed him off a cliffCredit: ITV
Michlle Connor and Robert are competely unaware Pat Phelan plans to crash their big day
"Phelan will come back to Weatherfield and crash Michelle and Robert's wedding day. In all the carnage, she will end up being shot with Robert trying desperately to save her life before she bleeds out.
"Inevitably, the other guests will be involved in the madness too in the aftermath of him shooting her.
"The scenes are so gruesome ITV bosses have decided to play them after 9pm so they don't fall foul of any rules."
And as I'm relaunching Britain's best TV column today, I thought I'd bring you a second Corrie exclusive.
Michelle Connor's wedding day scene will be gruesome that ITV bosses have decided to air it after the watershedCredit: Rex Features
I am told soap bosses plan to bring back Nick Tilsley. The character, introduced in 1981, will return this summer with Ben Price reprising the role.
It's the first time Nick will have been seen on screen since he said goodbye to mum Gail Platt last June and drove off into the sunset. In a heart-to-heart just a few days after his near-death beach quicksand experience, he told his mum he needed "a reboot".
My Corrie source added: "Exactly how Nick will be brought back into the fray is still to be decided.
"But the Platts, and in particular Nick, are a massive part of the Corrie family and always at the heart of some of the show's biggest storylines.
Corrie bosses have decided to bring back Nick Tilsley, played by Ben Price, who left last JuneCredit: Handout
"Producers are currently coming up with a bombshell way for him to come back to Weatherfield."
Given Nick has been married three times — twice to the same woman — I'd put money on it involving a new love interest.
Coronation Street fans obsessed Pat Phelan has returned to kill Eileen Grimshaw
Not lou-sing sleep on Vic
Louis Walsh has forced Simon Cowell into having him back on The X Factor next year – and now he's trying to decide who gets to sit next to him.
The outspoken Irishman wants Mel B or Rita Ora to replace outgoing Nicole Scherzinger. But one person he doesn't want is Victoria Beckham.
Louis Walsh has convinced Simon Cowell to keep him on The X Factor judging panel after Nicole Scherzinger was given the axeCredit: Rex Features
Talking on Irish TV's The Ray D'Arcy Show, he said: "I don't know who will fill Nicole's shoes. Mel B maybe or Rita Ora? I like Beyonce's sister Solange.
"It could be anybody from Little Mix. It could be anybody from the Spice Girls – except Posh. She does not talk. I want somebody who is fun."
And to be fair to Louis, he is that if nothing else.
WHAT? Fergie Vs Wenger: The Feud, Channel 5, 10pm.
WHY? Arsene Wenger might have taken Arsenal backwards in recent years, but the Premier League will be two greats shorter without him and his rival Sir Alex Ferguson at Manchester United. Here's a chance to relive their famous battles in this docu.
Sunny outlook
A NEW version of Love Island that gives older folk the chance to find love has made its first signing – legendary ex-weather presenter Sian Lloyd.
ITV series My Shirley Valentine Summer – produced by the makers of hit reality show Love Island – will see Sian and seven other women aged 40-plus jet off to a Greek island in a bid to land a new man.
Sian Lloyd will be jetting off to Greece to find love on ITV series My Shirley Valentine SummerCredit: CPG Ltd
A source said: "Sian has been pretty unlucky in love over the past few years. It's been three years since her divorce and she believes it's time to settle down again.
"She thinks the show will be the ideal way to move on with the support of like-minded women."
Sian was "traumatised" after racing entrepreneur Jonathan Ashman left her in 2015 following eight years of marriage.
She was also devastated when her relationship with MP fiancé Lembit Opik broke down in 2006.
Sian Lloyd previously revealed that 'inappropriately young' people were asking her out on dates following her divorceCredit: PA:Press Association
But now Sian has her eye on younger men, as she recently revealed: "Inappropriately young people ask me out. It's great and it's flattering."
She would have been better suited for the standard version of Love Island.
Gary Lineker has spent nearly two decades on the box. And now the crisp-loving pundit's eldest son could soon be following him on to the TV.
I can reveal that George is in talks to appear in E4's Made In Chelsea after becoming close pals with cast member and I'm A Celebrity winner Georgia Toffolo.
Gary Lineker's son George is in talks to join the cast of reality show Made In ChelseaCredit: Getty - Contributor
A show insider said: "George is never far from Chelsea – and is constantly out partying with the show's cast members.
"He's really close with Alex Mytton, and him and Sam Prince are always joined at the hip.
"He's also got a real penchant for the reality TV ladies, as he's been getting pretty cosy with Toff recently, and once dated Big Brother winner Sophie Reade.
"Just like his dad, George is very straight-talking too and doesn't mind getting involved in a debate so producers thought he would be a great signing in that aspect. They think it would be an explosive series if he were to get on board."
George Lineker was approached by E4's Made In Chelsea after getting close to star Georgia ToffoloCredit: The Mega Agency
George has been linked to a host of Made In Chelsea beauties in the past, including Jess Woodley and Jamie Laing's ex Tara Keeney. He's also regularly seen with Kate Moss's half-sister Lottie, who lives with cast member Frankie Gaff.
I'm sure they're all hanging out with him for his sparkling personality and not the fact his dad does Match Of The Day.
Mad about Christina
IT'S fair to say Christina Hendricks came out of her shell at the Tribeca Film Festival debut of her new movie Egg.
The voluptuous former Mad Men star wore this eye-popping top at the SVA Theatre screening in New York on Saturday.
Christina Hendricks showcased her famous curves at the Tribeca Film FestivalCredit: Getty Images - Getty
DEADLY END
Corrie's Katie McGlynn to exit soap after Sinead Tinker loses cancer battle
'I CRIED & CRIED'
Kate Winslet's 'dying mum urged her to go on Who Do You Think You Are?'
BRAVE VIN BACK
Vinnie Jones returns to work on X Factor days after death of wife Tanya
Christina, who plays housewife Karen in the comedy directed by Marianna Palka, joined co-star Alysia Reiner, who plays adventurous Tina, on the red carpet.
The festival says Egg tackles the "personal, social, professional, and ideological implications of bringing a new life into the world."
Sounds, er, cracking.
DAVID TENNANT has a 2ft tall sparkly pink penis in his house. He said it was a "trophy" wife GEORGIA MOFFETT nabbed from a TV production she was in. I've always thought he was a bit of a c**k, so it'll be in good company.
Kym Marsh


Acid reflux is a condition characterized by the uncontrollable reverse flow of gastric or intestinal fluids into the tube connecting the throat and the stomach (esophagus). This may be due to a brief relaxation of the muscular opening at the base of the esophagus (referred to as the sphincter), as well as chronic vomiting. Gastroesophageal reflux is fairly common in dogs, and may occur at any age, although younger dogs are at greater risk.
As a result, the gastric stomach acid and other components of the gastrointestinal juices cause damage to the protective mucus lining the esophagus. This can result in inflammation of the esophagus (esophagitis).
What are the symptoms and types of acid reflux?
Gastroesophageal reflux can cause esophagitis with varying amounts of damage. Mild esophagitis is limited to a mild inflammation of the esophageal lining, while more severe ulcerative esophagitis causes damage to the deeper layers of the esophagus.
The dog's behavioral history can reveal symptoms such as spitting up (regurgitation) of food, evidence of pain (whining or howling, for example) while swallowing, lack of appetite, and weight loss. A physical exam will often not reveal any concrete findings. Severe esophagitis may include symptoms of fever and extreme salivation.
Most treatment is done at home, by withholding food for one to two days, and thereafter following a dietary regimen of low-fat, low-protein meals given in small, frequent feedings. Dietary fat and protein should be limited, as fat decreases the strength of the muscle between the stomach and esophagus, while protein stimulates the secretion of gastric acid.
(Famotidine/Pepcid is available in 10 mg, 20 mg, and 40 mg tablets as well as in the form of oral powder. The recommended dose of Famotidine to be administered to pets is 0.25 to 0.5 mg per pound (0.5 to 1.0 mg/kg) every 12 to 24 hours.If you miss a dose while giving your pet Famotidine, administer is as soon as your remember during the same day. However, if you do not remember until the next day, skip the missed dose and only give your pet the next regularly scheduled dose. Do not give your pet a double dose of Famotidine.
Overdose of Famotidine is rare, but if you suspect your pet may have overdosed, seek emergency veterinary medical treatment. Symptoms of Famotidine overdose may include vomiting, pale gums, rapid heart rate, restlessness, or collapse.
Medications are an additional option. Drugs known as gastrointestinal pro-kinetic agents improve the movement of stomach contents through the intestines and also strengthens the gastroesophageal sphincter. Regardless of any medications, a change in diet is advisable.
Pepcid Dosage for Dogs: For both dogs and cats, the dosage is one 10-milligram tablet for a 20-pound dog every 12-to-24 hours, says Dr. Klein. It is best to give this medication one hour before meals. Check with a veterinarian to verify the dosage is accurate for your pet. Also, if purchasing Pepcid, make sure to buy Pepcid Original Strength (10 milligram tablets). Pepcid Complete contains additional active ingredients, and Pepcid Maximum Strength contains more medication per tablet.


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A Look into The Winnipeg Brutalist Manifesto
In celebration of Gimli Film Festival's brand new Manitoba Movie Night, we're taking a look at Winnipeg filmmaker's Ryan McKenna's The Winnipeg Brutalist Manifesto. McKenna co-wrote this love/hate memo with his fellow Winnipeg-Montreal transplant and editor Matthew Rankin. The manifesto is a great way to set the tone for McKenna's seminal film The First Winter which plays during GFF's Manitoba Movie Night's double feature Wednesday, May 20.
McKenna shot The First Winter during a particularly rough winter in 2012 fuelled by the desire to make a film about Winnipeg. While talking about his film with Malcom Fraser for Cult MTL McKenna said " You trash your city, but you love it, and you get defensive if anyone else trashes it. And part of what's interesting about it are its flaws, so it's important to bring those things up."
Manitoba Movie Night will be available online for free every month! This series of classic Manitoba-made movies will screen on GFF's website and GFF's Facebook page on the 20th of each month, in celebration of GFF's 20th anniversary, and Manitoba's 150th anniversary in 2020.
May 20 | Double Feature: The Last Winter &
The First Winter
7 PM CDT" – The Last Winter
The Last Winter by Aaron Kim Johnston
9 PM CDT" – The First Winter.
The First Winter by Ryan McKenna
THE WINNIPEG BRUTALIST MANIFESTO
1. Winnipeg Brutalism is a new cinema of Winnipeg. Stark and austere, it is like a Québec cinema, but with jokes.
2. Winnipeg Brutalism is a cinema of winter and of darkness. Exterior shots are to be filmed only between the Winter Solstice and the Vernal Equinox, and only at night.
3. Winnipeg Brutalism is very strict. Each Brutalist film must contain at least one (unfaked) blizzard.
4. Winnipeg Brutalism is an urban nightmare. The Brutalist film allows no beauty upon the Winnipeg landscape. The streets are empty, the buildings are abandoned, human drama is devoid of any warmth or compassion.
5. Winnipeg Brutalism is an outsider cinema. It must be pulled out of your own blood, uncontaminated by the neutering dogma of any Canadian "Institutes" or "Film Centres" or any other forced sterilization campaigns.
6. Winnipeg Brutalism becomes more brutal still if the director is kept in a perpetual state of discouragement. Living in squalour, avoiding sunlight, and eating only processed foods will only heighten the success of a Brutalist film production.
7. Winnipeg Brutalism is stark and it is real. The viewer must be Brutally exposed to the elements – there can be no reassuring lamb's wool of artifice, formalism or phantasmagoria.