text:
\section{Introduction}

Sequences of independent random variables have no memory at all, Markovian processes remember their past
through their present value only.
In the case of processes with longer memory the entire past may influence the current evolution of the given
stochastic system, e.g. in the case of fractional Brownian motion and related processes.

Econometric time series exhibit various degrees of influence of the past on the present,
depending on the sampling frequency.
High-frequency
volatility has long-range dependence while asset prices may or may not have this property, \cite{baillie}.
The principal motivating question of our research is the following: how does the memory of an asset's price
influence the satisfaction attainable from investing into this asset ?

The present paper concentrates on a Markovian setting. It
precisely characterizes the dependence of performance on memory in a concrete model class where
the price follows a Gaussian autoregressive processes. In the case of investors with exponential utility we
find the optimal trading strategy for each finite time horizon and analyse what happens when the horizon tends
to infinity. We determine the exact dependence of the asymptotic performance on the autoregression parameter and
hence make the first step towards general results linking investment performance to the memory length
of the underlying
security price.

The present paper continues previous investigations of \cite{dokuchaev,fs,martin}, where asymptotic
arbitrage in the utility sense was considered, i.e. the speed of the expected utility growth
when the time horizon tends to infinity. The first two references concentrated on
continuous-time models, \cite{martin} treated a model where borrowing and short-selling were
forbidden and utility functions were defined on the positive axis only.

The possibly negative prices of the model we consider may be acceptable in certain contexts (e.g. futures trading).
Its parameters may also be tuned such that negative prices practically never occur.
We nonetheless stress that our purpose is to exhibit a theoretical model whose qualitative conclusions are
hoped to extend
to a broader class of processes in the future so we are not bothered by the eventual negativity of prices.

We stress that it occurs very rarely that optimal strategies can be determined in closed form
for discrete-time investment problems. As far as we know our paper is the first to have found
the explicit solution for the case of autoregressive Gaussian processes.

In the present section we explain our model and the optimisation problem in consideration. In Section 2 we
present our
results,  Section 3 contains the proofs.

We are working with a financial market in which two assets are traded: a riskless asset with price constant $1$,
and a single risky asset whose price $X_t$ is an $\mathbb{R}$-valued stochastic process governed by
the equation
\begin{equation}\label{80}
X_{t+1}=\alpha X_t +\sigma \varepsilon_{t+1},\ t\geq 0,
\end{equation}
where $\alpha\in\mathbb{R}$, $\sigma>0$ are parameters and $\varepsilon_t$ are i.i.d.
standard Gaussian, independent of $X_0$. Introducing $\beta:=\alpha-1$, we may rewrite
\eqref{80} as
\begin{equation}
X_{t+1}-X_t=\beta X_t +\sigma\varepsilon_{t+1},\ t\geq 0.
\end{equation}
The information flow is given by
\begin{equation}
\mathcal{F}_t :=\sigma(X_s, 0\leq s \leq t).
\end{equation}
We interpret $\alpha$ (or, equivalently, $\beta$) as a ``memory parameter'' indicating how
previous values of the process $X$influence its present value. Eventually, our purpose is to
find the dependence of maximal achievable utility on this parameter.

A trading strategy is described by the number of units in the risky asset at $t$, denoted
by $\phi_t$ for $t\geq 1$. Trading strategies are assumed $(\mathcal{F}_t)_{t\geq 0}$-predictable
$\mathbb{R}$-valued
processes (i.e. $\phi_t$ is $\mathcal{F}_{t-1}$-measurable for all $t$), in particular, short-selling is
allowed. The totality of trading strategies is denoted by $\Phi$.

The wealth process corresponding to a given trading strategy $(\phi_t)_{t\geq 1}$ is
\begin{equation}\label{logwealth}
L_t^{\phi}:=L_{t-1}^{\phi}+\phi_t(X_t-X_{t-1}), \qquad t\geq 1,
\end{equation}
where $L_0^{\phi}:=L_0$ is the initial capital of the investor.
In other words, the terminal wealth of the investor is given by
\begin{equation}\label{logwealth2}
L_T^{\phi}=L_0+\sum_{j=1}^T\phi_j(X_j-X_{j-1}),
\end{equation}
where $T\geq 1$ is a time horizon.

We focus on a finite horizon utility maximization problem and look for the optimal strategy $({\phi}_t^*)_{1\leq t\leq T}$
which satisfies
\begin{equation}\label{90}
\sup_{\phi\in\Phi}\mathbb{E}U\left(L_T^{\phi}\right)=\mathbb{E}U\left(L_T^{{\phi}^*}\right),
\end{equation}
where $U:\mathbb{R}\rightarrow\mathbb{R}$ is the utility function $U(x)=-e^{-x}$. Note that
the expectations exist but may be $-\infty$.
We are going to give an explicit solution for this problem.

In order to make a comparison, we also consider an investor who is not using the accumulated
past information,
i.e. we define $\Phi_0$ as the set of trading strategies for which $\phi_t$ is $\mathcal{F}_0$-measurable for
all $1\leq t\leq T$. We wish to find ${\eta}^*\in\Phi_0$ such that
$$
\sup_{\phi\in\Phi_0}\mathbb{E}U\left(L_T^{\phi}\right)=\mathbb{E}U\left(L_T^{{\eta}^*}\right).
$$
We may and will suppose $L_0=0$ in the sequel.

\section{Results}

\begin{theorem}\label{maintheorem}
\begin{enumerate}
	\item  The optimal strategies for time horizon $T\in\mathbb{N}$ are
				$({\phi}^*_t)_{1\leq t\leq T}:=(\hat{\phi}_t^T(X_{t-1}))_{1\leq t\leq T}$ when past information
				is used and
				$({\eta}^*_t)_{1\leq t\leq T}:=(\hat{\phi}_t^T(X_{0}))_{1\leq t\leq T}$ when past
				information is ignored
				where
				\begin{equation}\label{strategies}
					\hat{\phi}_t^T(z)=\frac{\beta z}{\sigma^2}\theta_t^T \qquad \textrm{for all }1\leq t\leq T\mbox{ and }z\in\mathbb{R},
				\end{equation}
				and  $\theta_t^T=1-(T-t)\beta$.
	\item Using these strategies, the maximal conditional expected utilities are
				\begin{equation}\label{condexp}
				\mathbb{E}\left[U\left(L_T^{{\phi}^*}\right)|X_0=z\right]=-\frac{1}{\sqrt{\gamma_\beta(T)}}e^{-\frac{\beta^2z^2}{2\sigma^2}T}
						\textrm{, and}
				\end{equation}
				and
				\begin{equation}\label{condexpnomem}
					\mathbb{E}\left[U\left(L_T^{{\eta}^*}\right)|X_0=z\right]=-e^{-\frac{\beta^2z^2}{2\sigma^2}T},
				\end{equation}
				respectively, where $\gamma_\beta$ is given b
				\begin{equation}\label{gamma}
					\gamma_\beta(T)=\left\{\begin{array}{cl}
					\frac{\beta^{2T}\Gamma(\frac{1}{\beta^2}+T)}{\Gamma(\frac{1}{\beta^2})} & \textrm{if }\beta\neq 0, \\
					 1 & \textrm{if }\beta=0, \end{array}\right. \\
				\end{equation}
				and $\Gamma$ is the well-known gamma function, $\Gamma(x):=\int_0^{+\infty}t^{x-1}e^{-t}\mathrm{d}t$.
	\item In case of a stable autoregressive process (when $|\alpha|<1$, or, equivalently, $\beta\in (-2,0)$)
		assuming that $\mathrm{var}(X_t)=1$, and $X_0$ is $N(0,1)$, the maximal expected utility is
				\begin{equation}\label{exp}
					\mathbb{E}\left[U\left(L_T^{{\phi}^*}\right)\right]=-
					\sqrt{\frac{\beta+2}{\left(2-(T-1)\beta\right)\gamma_{\beta}(T)}}\textrm{, and}
				\end{equation}
				\begin{equation}\label{expnomem}
					\mathbb{E}\left[U\left(L_T^{{\eta}^*}\right)\right]=-\sqrt{\frac{\beta+2}{\left(2-(T-1)\beta\right)}}.
				\end{equation}
	\item The asymptotic behaviour of $\gamma_\beta$ is
				$\lim_{T\rightarrow+\infty}\frac{\gamma_\beta(T)}{h_\beta(T)}=1$
				where
				\begin{equation}\label{asymp}
					h_\beta(T)=\left\{\begin{array}{cl}
					\Gamma\left(\frac{1}{\beta^2}\right)\left(\beta^2\right)^{1-\frac{1}{\beta^2}}\sqrt{2\pi\left(T-1+\frac{1}{\beta^2}\right)}
					\left(\frac{1+(T-1)\beta^2}{e}\right)^{T-1+\frac{1}{\beta^2}} & \textrm{if }
					\beta\neq 0, \\ 1 & \textrm{if }\beta=0. \end{array}\right. \\
				\end{equation}
	\end{enumerate}
\end{theorem}

Our conclusion is that using accumulated past information leads to a qualitatively better strategy
(i.e. the expected utility using past information is the expected utility without past information
times a factor tending to $0$ much faster than that).
In order to make a meaningful comparison, however, we should normalize our processes.
Hence in 3. above we assume stability and choose $X_0$ in such a way that
$X_t$, $t\in\mathbb{N}$ becomes stationary with $\mathbb{E}X_t^2=1$.
We obtain the maximal expected utilities
both with and without past information.
They show that an increase in $\beta$ leads to an increase in \eqref{exp} and \eqref{expnomem},
so more information  from the
past (represented by
lager $|\beta|$) leads to better performance. This platitude is, however, supported by precise formulae in the present case.
Note also that $\gamma_{\beta}$ is continuous at $\beta=0$ as well.

\section{Computations and proofs}

The following lemma will help to construct the optimal strategy $\phi^*$ inductively.

\begin{lemma}\label{argumentofphistatement} Let $\tilde{\phi}_t$, $t=1,\ldots,T-1$ be a trading
strategy up to time $T-1$ given by $\tilde{\phi}_t=f_t(X_0,\ldots, X_{t-1})$ with Borel functions $f_t$. If $\tilde{\phi}$ is optimal up to $T-1$, i.e.
for all strategies $\phi_t$, $t=1,\ldots,T-1$ one has
\begin{equation}\label{porous}
\mathbb{E}\left[U(L^{\phi}_{T-1})\vert\mathcal{F}_{0}\right]\leq \mathbb{E}\left[U(L^{\tilde{\phi}}_{T-1})\vert\mathcal{F}_{0}\right],
\end{equation}
then the strategy defined by $\bar{\phi}_t:=f_{t-1}(X_1,\ldots,X_{t-1})$, $t=2,\ldots,T$ is optimal between $1$ and $T$, i.e. for all
strategies $\psi_t$, $t=2,\ldots,T$ one has
\begin{equation}\label{pp}
\mathbb{E}\left[U\left(\sum_{t=2}^T {\psi}_t (X_t-X_{t-1})\right)\vert\mathcal{F}_{1}\right]\leq \mathbb{E}\left[U\left(\sum_{t=2}^T \bar{\phi}_t (X_t-X_{t-1})\right)\vert\mathcal{F}_{1}\right].
\end{equation}
\end{lemma}
\begin{proof}
Let us denote by $\ell(dy_T,\ldots,dy_2|y_1,y_0)$ the conditional law of $(X_T,\ldots,X_2)$ w.r.t.
to $X_1=y_1$, $X_0=y_0$. We fix a regular version (see III. 70--73 in \cite{dm}).
As $X$ is a homogeneous Markov chain, $\ell$ does not depend on $y_0$ (hence we will write, with a slight abuse of notation, $\ell(dy_T,\ldots,dy_2|y_1)$)
and $\ell(dy_{T-1},\ldots,dy_1|y_0)$ is also the density function of  $(X_{T-1},\ldots,X_1)$ conditional to $X_0=y_0$

Let $\psi_t=g_t(X_0,\ldots,X_{t-1})$, $t=2,\ldots,T$ with some Borel functions $g_t$. Define for each $z\in\mathbb{R}$
the strategy $\phi_t:=\phi_t(z)=g_{t+1}(z,X_0,\ldots,X_{t-1})$, $t=1,\ldots,T-1$. By \eqref{porous},
\begin{eqnarray}\label{ppp}
\int_{\mathbb{R}^{T-1}}U\left(\sum_{t=1}^{T-1} g_{t+1}(z,y_0,\ldots,y_{t-1})(y_t-y_{t-1})\right)
\ell(dy_{T-1},\ldots,dy_1|y_0)
\leq\\
\nonumber
\int_{\mathbb{R}^{T-1}}U\left(\sum_{t=1}^{T-1} f_t(y_0,\ldots,y_{t-1})(y_t-y_{t-1})\right) \ell(dy_{T-1},\ldots,dy_1|y_0),
\end{eqnarray}
for a.e. $y_0$ (with respect to the law of $X_0$). Note that the right-hand side can be rewritten as
\begin{eqnarray}\label{pppp}
\int_{\mathbb{R}^{T-1}}U\left(\sum_{t=2}^{T} f_{t-1}(y_1,\ldots,y_{t-1})(y_t-y_{t-1})\right) \ell(dy_{T},\ldots,dy_2|y_1)
=\\ \nonumber
\mathbb{E}\left[U\left(\sum_{t=2}^T \bar{\phi}_t (X_t-X_{t-1})\right)|X_1=y_1,X_0=y_0\right],
\end{eqnarray}
for a.e. $(y_1,y_0)$ w.r.t. the law of $(X_0,X_1)$, note the Markov property again.
Similarly, the left-hand side of \eqref{ppp} equals
\begin{eqnarray}\label{ppppp}
\int_{\mathbb{R}^{T-1}}U\left(\sum_{t=2}^{T-2} g_{t}(z,y_1,\ldots,y_{t-1})(y_t-y_{t-1})\right) \ell(dy_{T},\ldots,dy_2|y_1) =\\ \nonumber \mathbb{E}\left[U\left(\sum_{t=2}^T {\psi}_t (X_t-X_{t-1})\right)|X_1=y_1,X_0=z\right].
\end{eqnarray}
Now plugging in $z:=y_0$ we obtain \eqref{pp} from \eqref{ppp}, \eqref{pppp} and \eqref{ppppp}.
\end{proof}

After these preparations we are able to give an explicit solution for the optimal strategies of the wealth process in case of the price is an autoregressive process.

In this Section we prove \autoref{maintheorem}, first we focus on the case where the investor
uses past information. We consider the case $T=1$, so the wealth process according to
\eqref{logwealth} takes the form
\begin{equation}
L_1^{\phi}=\phi_1(X_1-X_0)=\phi_1(\beta X_0+\sigma\varepsilon_1).
\end{equation}

We have
\begin{equation}
	\mathbb{E}\left[e^{-\phi_1(\beta X_0+\sigma \varepsilon _1)} | X_0=z\right]
	= e^{-\phi_1\beta z} \mathbb{E}\left[e^{-\phi_1 \sigma \varepsilon_1}\right]=
	e^{-\phi_1\beta z}e^{\frac{\phi_1^2\sigma^2}{2}},
\end{equation}
hence we get
\[
\mathrm{arg}\min _{\phi_1}e^{-\phi_1\beta z+\frac{(\phi_1\sigma)^2}{2}}
 = \mathrm{arg}\min_{\phi_1}e^{\frac{1}{2}\left(\phi_1\sigma-\frac{\beta z}{\sigma}\right)^2-\frac{\beta^2 z^2}{2\sigma^2}}
 =\frac{\beta z}{\sigma^2}=\frac{\beta z}{\sigma^2}\theta_1^1=\hat{\phi}_1^1(z), \]
 because $\theta_1^1=1$. So we proved the first part of \autoref{maintheorem} for $T=1$. Now let's assume that
 \eqref{strategies} is true for $T-1$, i.e.
 \begin{equation}\label{opp}
\phi^*_t:=\hat{\phi}_t^{T-1}(X_{t-1})\mbox{ with }\hat{\phi}_t^{T-1}(z)=\frac{\beta z}{\sigma^2}\theta_t^{T-1} \quad 1\leq t\leq T-1
 \end{equation}
satisfies \eqref{90} for all $\phi\in\Phi$.
 We will prove that \eqref{strategies} also holds for $T$. By Lemma \ref{argumentofphistatement},
 for all $\psi\in\Phi$,
 \begin{eqnarray}\nonumber
\mathbb{E}[e^{-L^{\psi}_T}|\mathcal{F}_0] = \mathbb{E}[e^{-\psi_1(X_1-X_0)}
\mathbb{E}[e^{-\sum_{j=2}^T \psi_j(X_j-X_{j-1})}|\mathcal{F}_1]\vert\mathcal{F}_0] &\geq &\\ \nonumber
\mathbb{E}[e^{-\psi_1(X_1-X_0)}
\mathbb{E}[e^{-\sum_{j=2}^T \hat{\phi}_{j-1}^{T-1}(X_{j-1})(X_j-X_{j-1})}|\mathcal{F}_1]\vert\mathcal{F}_0]&=&\\
\mathbb{E}[e^{-\psi_1(X_1-X_0)}
\mathbb{E}[e^{-\sum_{j=2}^T \hat{\phi}_{j}^{T}(X_{j-1})(X_j-X_{j-1})}|\mathcal{F}_1]\vert\mathcal{F}_0],\label{parr} & &
 \end{eqnarray}

since $\phi_{j-1}^{T-1}=\phi_j^T$. Now define the trading strategy
$\omega=(\phi, \hat{\phi}_2^T(X_1), \dots , \hat{\phi_T^T}(X_{T-1}))$ and the function
$Q_T\,:\,\mathbb{R}^{T+2} \rightarrow\mathbb{R}$ such that
\begin{equation}
Q_T(\phi,X_0,\mathbf{\varepsilon}):=L_T^{\omega},\quad\textrm{where }\mathbf{\varepsilon}=(\varepsilon_1, \dots \varepsilon_T)^{\mathbf{T}}.
\end{equation}
Hence, according to \eqref{parr}, it remains to find $\phi$ which minimizes
\begin{equation}\label{phi1def}
\mathbb{E}\left[e^{-Q_T(\phi,X_0,\mathbf{\varepsilon})}|X_0=z\right].
\end{equation}
If we prove that $\phi=\hat{\phi}_1^{T}(z)$ does the job then we will be able to conclude that the optimal
strategy for time horizon $T$ is indeed as given in \eqref{opp} for $T-1$.

To compute the minimiser $\phi$ we will write $Q_T(\phi, X_0, \varepsilon)$
in a sum of a quadratic, a linear and a constant function of $\mathbf{\varepsilon}$.

\begin{eqnarray*}
 	Q_T(\phi,X_0,\mathbf{\varepsilon})&=&\phi(X_1-X_0)+\sum_{j=2}^T\hat{\phi}_j^T(X_{j-1})(X_j-X_{j-1}) \\
 	   &=&\phi(X_1-X_0)+\sum_{j=2}^T\frac{\beta\theta_j^T}{\sigma^2}\underbrace{X_{j-1}(X_j-X_{j-1})}_{A_j}
\end{eqnarray*}

For $A_j$ we have
\begin{eqnarray*}
 	A_j&=&\left(\alpha^{j-1}X_0+\sigma\sum_{i=1}^{j-1}\alpha^{j-i-1}\varepsilon_i\right)\left(\alpha^{j}X_0+
 		\sigma\sum_{i=1}^{j}\alpha^{j-i}\varepsilon_i - \alpha^{j-1}X_0-\sigma\sum_{i=1}^{j-1}\alpha^{j-i-1}\varepsilon_i\right) \\
 	&=&\left(\alpha^{j-1}X_0+\sigma\sum_{i=1}^{j-1}\alpha^{j-i-1}\varepsilon_i\right)\left(\alpha^{j-1}\beta X_0+\sigma\varepsilon_j +
 		\sigma\sum_{i=1}^{j-1}\alpha^{j-i-1}\beta\varepsilon_i\right) \\
 	&=&\alpha^{2j-2}\beta X_0^2+\sigma\beta X_0\sum_{i=1}^{j-1}\alpha^{2j-i-2}\varepsilon_i+\sigma X_0\alpha^{j-1}\varepsilon_j+
 		\sigma^2\sum_{i=1}^{j-1}\alpha^{j-i-1}\varepsilon_i\varepsilon_j+\sigma\beta X_0\sum_{i=1}^{j-1}\alpha^{2j-i-2}\varepsilon_i \\
 	&\,& +\sigma^2\beta\left(\sum_{i=1}^{j-1}\alpha^{j-i-1}\varepsilon_i\right)^2 \\
 	&=&\underbrace{\alpha^{2j-2}\beta X_0^2}_{B_1(j)}+ \underbrace{2\sigma\beta X_0\sum_{i=1}^{j-1}\alpha^{2j-i-2}\varepsilon_i}_{B_2(j)}
 		+\underbrace{\sigma X_0\alpha^{j-1}\varepsilon_j}_{B_3(j)}
 		+ \underbrace{\sigma^2\sum_{i=1}^{j-1}\alpha^{j-i-1}\varepsilon_i\varepsilon_j}_{B_4(j)}
 		+\underbrace{\sigma^2\beta\sum_{i=1}^{j-1}\sum_{k=1}^{j-1}\varepsilon_i\varepsilon_k\alpha^{2j-i-k-2}}_{B_5(j)}. \\
\end{eqnarray*}

Let's substitute this into  $Q_T(\phi,X_0,\mathbf{\varepsilon})$.
\begin{eqnarray*}
 	Q_T(\phi,X_0,\mathbf{\varepsilon})&=&\phi(X_1-X_0)+\sum_{j=2}^T\frac{\beta\theta_j^T}{\sigma^2}(B_1(j)+B_2(j)+B_3(j)+B_4(j)+B_5(j)) \\
 		 &=&\phi(\beta X_0+\sigma\varepsilon_1)+\underbrace{\sum_{j=2}^T\frac{\beta\theta_j^T}{\sigma^2}B_1(j)}_{C_1}
 		 	+ \underbrace{\sum_{j=2}^T\frac{\beta\theta_j^T}{\sigma^2}B_2(j)}_{C_2}
 		 	+\underbrace{\sum_{j=2}^T\frac{\beta\theta_j^T}{\sigma^2}B_3(j)}_{C_3}+ \\
 		 &+&\underbrace{\sum_{j=2}^T\frac{\beta\theta_j^T}{\sigma^2}B_4(j)}_{C_4}
 		 	+\underbrace{\sum_{j=2}^T\frac{\beta\theta_j^T}{\sigma^2}B_5(j)}_{C_5}
\end{eqnarray*}

We compute each $C_n$ separately.
\begin{eqnarray*}
 	C_1&=&\frac{\beta^2 X_0^2}{\sigma^2}\sum_{j=2}^T\theta_j^T\alpha^{2j-2}. \\
 	C_2&=&\sum_{j=2}^T\frac{\beta\theta_j^T}{\sigma^2}2\sigma\beta X_0\sum_{i=1}^{j-1}\alpha^{2j-i-2}\varepsilon_i
 	   =\sum_{j=1}^{T-1}\sum_{i=1}^j\frac{2\beta^2X_0}{\sigma}\theta_{j+1}^T\alpha^{2j-i}\varepsilon_i \\
 	   &=&\sum_{i=1}^{T-1}\sum_{j=i}^{T-1}\frac{2\beta^2X_0}{\sigma}\theta_{j+1}^T\alpha^{2j-i}\varepsilon_i
 	   =\sum_{i=1}^{T-1}\varepsilon_i\left(\sum_{j=i}^{T-1}\frac{2\beta^2X_0}{\sigma}\theta_{j+1}^T\alpha^{2j-i}\right).\\
 	C_3&=&\sum_{j=2}^T\frac{\beta\theta_j^T}{\sigma^2}\sigma X_0\alpha^{j-1}\varepsilon_j
 		 =\sum_{j=2}^T\varepsilon_j\left(\frac{\beta X_0}{\sigma}\theta_{j}^T\alpha^{j-1}\right).\\
 	C_4&=&\sum_{j=2}^T\frac{\beta\theta_j^T}{\sigma^2}\sigma^2\sum_{i=1}^{j-1}\alpha^{j-i-1}\varepsilon_i\varepsilon_j
 		 =\sum_{j=2}^T\sum_{i=1}^{j-1}\varepsilon_i\varepsilon_j\left(\beta\theta_j^T\alpha^{j-i-1}\right).\\	C_5&=&\sum_{j=2}^T\frac{\beta\theta_j^T}{\sigma^2}\sigma^2\beta\sum_{i=1}^{j-1}\sum_{k=1}^{j-1}\varepsilon_i\varepsilon_k\alpha^{2j-i-k-2}
 		  = \beta^2\sum_{j=1}^{T-1}\sum_{i=1}^j\sum_{k=1}^j\theta_{j+1}^T\varepsilon_i\varepsilon_k\alpha^{2j-i-k} \\
 	&=&\beta^2\sum_{i=1}^{T-1}\sum_{j=1}^{T-1}\sum_{k=1}^j\theta_{j+1}^T\varepsilon_i\varepsilon_k\alpha^{2j-i-k}
 		= \beta^2\sum_{i=1}^{T-1}\left(\sum_{k=1}^{i}\sum_{j=i}^{T-1}\theta_{j+1}^T\varepsilon_i\varepsilon_k\alpha^{2j-i-k}
 		+ \sum_{k=i+1}^{T-1}\sum_{j=k}^{T-1}\theta_{j+1}^T\varepsilon_i\varepsilon_k\alpha^{2j-i-k}\right) \\
 	&=&\beta^2\sum_{i=1}^{T-1}\left(\sum_{k=1}^{i}\varepsilon_i\varepsilon_k\sum_{j=i}^{T-1}\theta_{j+1}^T\alpha^{2j-i-k}
 		+ \sum_{k=i+1}^{T-1}\varepsilon_i\varepsilon_k\sum_{j=k}^{T-1}\theta_{j+1}^T\alpha^{2j-i-k}\right).
\end{eqnarray*}

According to these, we can write $Q_T(\phi,X_0,\mathbf{\varepsilon})$ as
\begin{equation}\label{quadraticform}
 		Q_T(\phi,X_0,\mathbf{\varepsilon})=
 		\mathbf{\varepsilon^T}\left(\mathbf{A}_T-\frac{1}{2}\mathbf{I}\right)\mathbf{\varepsilon}+\mathbf{b^T}(\phi,X_0)\mathbf{\varepsilon}+c(\phi,X_0),
\end{equation}

 	where $\mathbf{A}_T=[a_{ik}]\in\mathbb{R}^{T\times T}$ is a symmetric matrix with
 	\begin{eqnarray*} a_{ii}&=&\frac{1}{2}+\beta^2\sum_{j=i}^{T-1}\theta_{j+1}^T\alpha^{2j-2i}, \qquad 1\leq i\leq T-1, \\
 	 a_{TT}&=&\frac{1}{2}, \\
 	 a_{ik}&=&\frac{\beta\theta_i^T\alpha^{i-k-1}}{2}+\beta^2\sum_{j=i}^{T-1}\theta_{j+1}^T\alpha^{2j-i-k}, \qquad 1 \leq k<i\leq T-1,\\
 	 a_{Tk}&=&\frac{\beta\alpha^{T-k-1}}{2} \qquad 1\leq k\leq T-1;
 	\end{eqnarray*}

 	 $\mathbf{b}\,:\,\mathbb{R}^2\rightarrow\mathbb{R}^T$ is a vector with
 	\begin{eqnarray*} b_1(\phi, X_0)&=&\phi\sigma+\frac{2\beta^2X_0}{\sigma}\sum_{j=1}^{T-1}\theta_{j+1}^T\alpha^{2j-1}, \\
 		 b_i(\phi, X_0)&=&\frac{\beta X_0}{\sigma}\theta_i^T\alpha^{i-1}+
 		 \frac{2\beta^2X_0}{\sigma}\sum_{j=i}^{T-1}\theta_{j+1}^T\alpha^{2j-i}, \qquad 2\leq i\leq T-1, \\
 		 b_T&=&\frac{\beta X_0}{\sigma}\alpha^{T-1};
 		 \end{eqnarray*}

 	and $c\,:\,\mathbb{R}^2\rightarrow\mathbb{R}$ where
 	\[c(\phi,X_0)=\phi\beta X_0+\frac{\beta^2 X_0^2}{\sigma^2}\sum_{j=2}^T\theta_j^T\alpha^{2j-2}.\]

 We need to compute the conditional expected utility given by
 \begin{equation}\label{integralformofexp}
 \mathbb{E}\left[e^{-Q_T(\phi,X_0,\mathbf{\varepsilon})}|X_0=z\right]= \frac{1}{\left(\sqrt{2\pi}\right)^T}\int_{\mathbb{R}^T}e^{-\mathbf{x^TA}_T\mathbf{x}-\mathbf{b^T}(\phi,z)\mathbf{x}-c(\phi,z)} \mathrm{d}\mathbf{x}
 \end{equation}
 In order to evaluate this integral we need some preparation.

 We know that
 \begin{equation}\label{integral1}
 	\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-ax^2-bx}\mathrm{d}x=\frac{1}{\sqrt{2a}}e^{\frac{b^2}{4a}},
 \end{equation}
for all  $b\in \mathbb{R}$ and $a>0$.

\begin {lemma}\label{integral2lemma}
Let $\mathbf{D}$ be a positive
definite diagonal matrix with diagonal entries $d_1, d_2, \dots, d_n$, and let $\mathbf{b}\in\mathbb{R}^n$. Then
\begin{equation}\label{integral2}
	\frac{1}{\left(\sqrt{2\pi}\right)^n}\int_{\mathbb{R}^n}e^{-\mathbf{x^TDx}-\mathbf{b^Tx}}\mathrm{d}\mathbf{x}
	=\frac{1}{\sqrt{2^n\det \mathbf{D}}}e^{\frac{\mathbf{b^TD^{-1}b}}{4}}.
\end{equation}
\end{lemma}

\begin{proof} Using \eqref{integral1}
\begin{eqnarray*}
\frac{1}{\left(\sqrt{2\pi}\right)^n}\int_{\mathbb{R}^n}e^{-\mathbf{x^TDx}-\mathbf{b^Tx}}\mathrm{d}\mathbf{x}&=&
\frac{1}{\left(\sqrt{2\pi}\right)^n}\int_{\mathbb{R}^n}e^{-\sum_{i=1}^nd_ix_i^2-\sum_{i=1}^nb_ix_i}\mathrm{d}\mathbf{x}\\
&=&\prod_{i=1}^n\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-d_ix_i^2-b_ix_i}\mathrm{d}x_i \\
&=& \prod_{i=1}^n\frac{1}{\sqrt{2d_i}}e^{\frac{b_i^2}{4d_i}}
 =\frac{1}{\sqrt{2^n\det \mathbf{D}}}e^{\frac{\mathbf{b^TD^{-1}b}}{4}}.
\end{eqnarray*}
\end{proof}

\begin{lemma}\label{integral3lemma}
Let $\mathbf{A}\in\mathbb{R}^{n\times n}$ be a symmetric, positive definite matrix , and $\mathbf{b}\in\mathbb{R}^n$. Then
\begin{equation}\label{integral3}
\frac{1}{\left(\sqrt{2\pi}\right)^n}\int_{\mathbb{R}^n}e^{-\mathbf{x^TAx}-\mathbf{b^Tx}}\mathrm{d}\mathbf{x} = \frac{1}{\sqrt{2^n\det \mathbf{A}}}e^{\frac{\mathbf{b^TA^{-1}b}}{4}}.
\end{equation}
\end{lemma}

\begin{proof} Since $\mathbf{A}$ is symmetric, there is an $\mathbf{S}$ orthonormal, and a $\mathbf{D}$ diagonal matrix for which $\mathbf{SDS^{-1}}=\mathbf{SDS^{T}}=\mathbf{A}$ and $|\det \mathbf{S}|=1$.
Using Lemma \ref{integral2lemma} and setting $\mathbf{y}:=\mathbf{{S^Tx}}$
\begin{eqnarray*}
\frac{1}{\left(\sqrt{2\pi}\right)^n}\int_{\mathbb{R}^n}e^{-\mathbf{x^TAx}-\mathbf{b^Tx}}\mathrm{d}\mathbf{x}
	&=& \frac{1}{\left(\sqrt{2\pi}\right)^n}\int_{\mathbb{R}^n}e^{-\mathbf{x^TSDS^Tx}-\mathbf{b^Tx}}\mathrm{d}\mathbf{x}
	=\frac{1}{\left(\sqrt{2\pi}\right)^n}\int_{\mathbb{R}^n}e^{-\mathbf{(S^Tx)^TDS^Tx}-\mathbf{b^Tx}}\mathrm{d}\mathbf{x} \\
	&=&\frac{1}{\left(\sqrt{2\pi}\right)^n}\int_{\mathbb{R}^n}e^{-\mathbf{y^TDy}-\mathbf{b^TSy}}|\det \mathbf{S}|\mathrm{d}\mathbf{y}
	=\frac{1}{\left(\sqrt{2\pi}\right)^n}\int_{\mathbb{R}^n}e^{-\mathbf{y^TDy}-\mathbf{(S^Tb)^Ty}}\mathrm{d}\mathbf{y}\\
	&=&\frac{1}{\sqrt{2^n\det \mathbf{D}}}e^{\frac{\mathbf{(S^Tb)^TD^{-1}S^Tb}}{4}}
	=\frac{1}{\sqrt{2^n\det\mathbf{D}}}e^{\frac{\mathbf{b^TSD^{-1}S^Tb}}{4}}\\
	&=&\frac{1}{\sqrt{2^n\det\mathbf{A}}}e^{\frac{\mathbf{b^TA^{-1}b}}{4}},
\end{eqnarray*}
since $\det\mathbf{D}=\det\mathbf{A}$ and $\mathbf{A^{-1}}=\mathbf{SD^{-1}S^T}$.
\end{proof}

Now we can compute the expression in \eqref{integralformofexp} using Lemma \ref{integral3lemma}:
\begin{equation}\label{matrixformofexp}
\mathbb{E}\left[e^{-Q_T(\phi,X_0,\mathbf{\varepsilon})}|X_0=z\right]
=\frac{1}{\sqrt{2^T\det \mathbf{A}_T}}e^{\frac{\mathbf{b^T}(\phi,z)\mathbf{A}_T\mathbf{^{-1}b}(\phi,z)}{4}-c(\phi,z)}.
\end{equation}

We proceed to examining the determinant of $\mathbf{A}_T$
to prove that $\mathbf{A}_T$ is positive definite (as \eqref{matrixformofexp} holds only in this case)
and we will need to compute one element of the inverse matrix, $\left(\mathbf{A^{-1}}\right)_{1,1}$.

 First we present a lemma which will be very useful later.
 \begin{lemma} \label{simplelemma}
 For $\theta_t^T$ (defined in \autoref{maintheorem}) and for all $m\leq n$
 \begin{equation}
 \sum_{i=m}^n\theta_i^T\alpha^i=(T+1-m)\alpha^m-(T-n)\alpha^{n+1}.
 \end{equation}
\end{lemma}

\begin{proof}
\begin{eqnarray*}
\sum_{i=m}^n\theta_i^T\alpha^i&=&\sum_{i=m}^n(T+1-i-(T-i)\alpha)\alpha^i=\sum_{i=m}^n(T+1-i)\alpha^i-\sum_{i=m}^n(T-i)\alpha^{i+1} \\
		&=&\sum_{i=m}^n(T+1-i)\alpha^i-\sum_{i=m+1}^{n+1}(T+1-i)\alpha^i=(T+1-m)\alpha^m-(T-n)\alpha^{n+1}
	\end{eqnarray*}
\end{proof}

\begin{lemma}\label{sumofrowsstatement}
 For $\mathbf{A}_T=[a_{ij}]$ we have
 \begin{equation}\label{sumofrows}
 a_{1,n}-\beta\sum_{i=2}^Ta_{in}=0\quad \textrm{for all }2\leq n\leq T. \end{equation}
 \end{lemma}

\begin{proof} First we consider the case $n=T$,
 \begin{eqnarray*}
 		a_{1,T}-\beta\sum_{i=2}^Ta_{i,T}&=&\frac{\beta\alpha^{T-2}}{2}
 			-\beta\left(\frac{1}{2}+\sum_{i=2}^{T-1}\frac{\beta\alpha^{T-i-1}}{2}\right)=
 			\alpha^{T-2}-1-\beta\sum_{i=2}^{T-1}\alpha^{T-i-1} \\
 		&=&\alpha^{T-2}-1-\sum_{i=2}^{T-1}\alpha^{T-i}+\sum_{i=2}^{T-1}\alpha^{T-i-1}=
 		\alpha^{T-2}-1-\sum_{i=2}^{T-1}\alpha^{T-i}+\sum_{i=3}^{T}\alpha^{T-i} \\
 		&=&\alpha^{T-2}-1-\alpha^{T-2}+1=0.
 		\end{eqnarray*}
 	Then we consider the case $n\neq T$,
 	\begin{equation}\label{sumofrows2}
 	a_{1,n}-\beta\sum_{i=2}^Ta_{i,n}=a_{1,n}-\beta\sum_{i=2}^{n-1}a_{i,n}-\beta a_{n,n}-\beta\sum_{i=n+1}^{T-1}a_{i,n}-\beta a_{T,n}.
 	\end{equation}
 	We compute the sums separately.
 	\begin{eqnarray*}
 	\beta\sum_{i=2}^{n-1}a_{i,n}&=&\sum_{i=2}^{n-1}\left(\frac{\beta^2\theta_n^T\alpha^{n-i-1}}{2}
 		+\beta^3\sum_{j=n}^{T-1}\theta_{j+1}^T\alpha^{2j-n-i}\right) \\
 	&=&\frac{\beta^2\theta_n^T}{2}\sum_{i=2}^{n-1}\alpha^{n-i-1}+\beta^3\sum_{j=n}^{T-1}\theta_{j+1}^T\sum_{i=2}^{n-1}\alpha^{2j-n-i} \\
 	&=&\frac{\beta\theta_n^T}{2}\left(\sum_{i=2}^{n-1}\alpha^{n-i}-\sum_{i=2}^{n-1}\alpha^{n-i-1}\right)
 		+ \beta^2\sum_{j=n}^{T-1}\theta_{j+1}^T\left(\sum_{i=2}^{n-1}\alpha^{2j-n-i+1}-\sum_{i=2}^{n-1}\alpha^{2j-n-i}\right) \\
 	&=&\frac{\beta\theta_n^T}{2}\left(\alpha^{n-2}-1\right)
 		+\beta^2\sum_{j=n}^{T-1}\theta_{j+1}^T \left(\alpha^{2j-n-1}-\alpha^{2j-2n+1}\right) \\
 	&=&\frac{\beta\theta_n^T\alpha^{n-2}}{2}-\frac{\beta\theta_n^T}{2}+\beta^2\sum_{j=n}^{T-1}\theta_{j+1}^T\alpha^{2j-n-1}
 		-\beta^2\sum_{j=n}^{T-1}\theta_{j+1}^T\alpha^{2j-2n+1} \\
\beta\sum_{i=n+1}^{T-1}a_{in}&=&\sum_{i=n+1}^{T-1}\left(\frac{\beta^2\theta_i^T\alpha^{i-n-1}}{2}
		+\beta^3\sum_{j=i}^{T-1}\theta_{j+1}^T\alpha^{2j-i-n}\right)\\
 	&=&\frac{\beta^2\alpha^{-n-1}}{2}\sum_{i=n+1}^{T-1}\theta_i^T\alpha^i
 		+ \beta^3\sum_{j=n+1}^{T-1}\sum_{i=n+1}^{j}\theta_{j+1}^T\alpha^{2j-i-n} \\
 	&=&\frac{\beta^2\alpha^{-n-1}}{2}\sum_{i=n+1}^{T-1}\theta_i^T\alpha^i
 	+\beta^2\sum_{j=n+1}^{T-1}\theta_{j+1}^T\left(\sum_{i=n+1}^{j}\alpha^{2j-i-n+1}
 		-\sum_{i=n+1}^{j}\alpha^{2j-i-n}\right) \\
 	&=&\frac{\beta^2\alpha^{-n-1}}{2}\left((T-n)\alpha^{n+1}-\alpha^T\right)
 		+\beta^2\sum_{j=n+1}^{T-1}\theta_{j+1}^T\left(\alpha^{2j-2n}-\alpha^{j-n}\right) \textrm{ by Lemma \ref{simplelemma}},\\
 	&=&\frac{\beta^2(T-n)}{2}-\frac{\beta^2\alpha^{T-n-1}}{2}
 		+ \beta^2\sum_{j=n+1}^{T-1}\theta_{j+1}^T\alpha^{2j-2n}-\beta^2\sum_{j=n+2}^{T}\theta_{j}^T\alpha^{j-n-1} \\
 	&=&\frac{\beta^2(T-n)}{2}-\frac{\beta^2\alpha^{T-n-1}}{2}
 		+ \beta^2\sum_{j=n+1}^{T-1}\theta_{j+1}^T\alpha^{2j-2n}-\beta^2\alpha(T-n-1)\textrm{ by Lemma \ref{simplelemma}}.
 	\end{eqnarray*}

 		The other terms in \eqref{sumofrows2} are
 		\begin{eqnarray*}
 		a_{1,n}&=&\frac{\beta\theta_n^T\alpha^{n-2}}{2}+\beta^2\sum_{j=n}^{T-1}\theta_{j+1}^T\alpha^{2j-n-1} \\
 		\beta a_{nn}&=&\frac{\beta}{2}+\beta^3\sum_{j=n}^{T-1}\theta_{j+1}^T\alpha^{2j-2n} \\
 		\beta a_{Tn}&=&\frac{\beta^2\alpha^{T-n-1}}{2}.
 		\end{eqnarray*}

Substituting these into \eqref{sumofrows2}:
\begin{eqnarray*}
 	a_{1,n}-\beta\sum_{i=2}^Ta_{i,n}&=&\frac{\beta\theta_n^T\alpha^{n-2}}{2}+\beta^2\sum_{j=n}^{T-1}\theta_{j+1}^T\alpha^{2j-n-1}
 		-\frac{\beta\theta_n^T\alpha^{n-2}}{2}+\frac{\beta\theta_n^T}{2}-\beta^2\sum_{j=n}^{T-1}\theta_{j+1}^T\alpha^{2j-n-1}\\
 	&\,&+\beta^2\sum_{j=n}^{T-1}\theta_{j+1}^T\alpha^{2j-2n+1}
 		 -\frac{\beta}{2}-\beta^3\sum_{j=n}^{T-1}\theta_{j+1}^T\alpha^{2j-2n}-\frac{\beta^2(T-n)}{2}+\frac{\beta^2\alpha^{T-n-1}}{2} \\
 	&\,&-\beta^2\sum_{j=n+1}^{T-1}\theta_{j+1}^T\alpha^{2j-2n}+\beta^2\alpha(T-n-1)-\frac{\beta^2\alpha^{T-n-1}}{2} \\
 	&=&\frac{\beta-\beta^2(T-n)}{2}+\beta^2\sum_{j=n}^{T-1}\theta_{j+1}^T\alpha^{2j-2n}-\frac{\beta}{2}-\frac{\beta^2(T-n)}{2} \\
  &\,& -\beta^2\sum_{j=n+1}^{T-1}\theta_{j+1}^T\alpha^{2j-2n}+\beta^2\alpha(T-n-1)\\
 	&=&\beta^2\theta_{n+1}^T-\beta^2(T-n)+ \beta^2\alpha(T-n-1)\\
 	&=&\beta^2(\theta_{n+1}^T-(T-n-\alpha(T-n-1)))=0.
\end{eqnarray*}
\end{proof}

\begin{definition} For a matrix $\mathbf{A}\in\mathbb{R}^{n\times n}$
let $\mathbf{A}(i,j)\in\mathbb{R}^{(n-1)\times (n-1)}$ denote the appropriate minor matrix of $\mathbf{A}$,
i.e. the matrix obtained by omitting the \emph{i}th row and the \emph{j}th column of $\mathbf{A}$.
\end{definition}

\begin{lemma}\label{lemmaformatrices}
We have
 		 \begin{equation} \mathbf{A}_T(1,1)=\mathbf{A}_{T-1}. \end{equation}
\end{lemma}

\begin{proof} We denote the elements of $\mathbf{A}_T(1,1)$ and $\mathbf{A}_{T-1}$ by $u_{i,k}$ and $v_{i,k}$, respectively.
 		 \begin{eqnarray*}
 		 u_{i,i}&=&a_{i+1,i+1}=\frac{1}{2}\beta^2\sum_{j=i+1}^{T-1}\theta_{j+1}^T\alpha^{2j-2(i+1)}
 		 	= \frac{1}{2}\beta^2\sum_{j=i}^{T-2}\theta_{j+2}^T\alpha^{2j-2i}\\
 		 &=&\frac{1}{2}\beta^2\sum_{j=i}^{T-2}\theta_{j+1}^{T-1}\alpha^{2j-2i}=v_{i,i} \qquad 1\leq i\leq T-2,\\
 		 u_{i,k}&=&a_{i+1,k+1}=\frac{\beta\theta_{i+1}^T\alpha^{i-k-1}}{2}+\beta^2\sum_{j=i+1}^{T-1}\theta_{j+1}^T\alpha^{2j-i-k-2}\\
 		 	&=&\frac{\beta\theta_{i}^{T-1}\alpha^{i-k-1}}{2}+\beta^2\sum_{j=i}^{T-1}\theta_{j+2}^T\alpha^{2j-i-k}\\
 		 	&=&\frac{\beta\theta_{i}^{T-1}\alpha^{i-k-1}}{2}+\beta^2\sum_{j=i}^{T-1}\theta_{j+1}^{T-1}\alpha^{2j-i-k}=v_{i,k} \qquad 1\leq k<i\leq T-2, \\
 		u_{T-1,k}&=&a_{T,k+1}=\frac{\beta\alpha^{T-(k+1)-1}}{2}=\frac{\beta\alpha^{(T-1)-k-1}}{2}=v_{T-1,k} \qquad 1\leq k\leq T-2\\
 		u_{T-1,T-1}&=&a_{TT}=\frac{1}{2}=v_{T-1,T-1}
 	\end{eqnarray*}
 	\end{proof}

\begin{lemma}\label{determinant1statement}
 		For the determinant of $\mathbf{A}_T$, for all $T\geq 2$, we have
 		\begin{equation}\label{determinant1}
 		\det \mathbf{A}_T=\frac{1+\beta^2(T-1)}{2}\det \mathbf{A}_{T-1}.
 		\end{equation}
\end{lemma}

\begin{proof} We construct a matrix $\mathbf{B}_T$ in such a way that we subtract the rows of $\mathbf{A}_T$
multiplied by $\beta$ from the first row. Then, according to Lemma \ref{sumofrowsstatement},
in the first row of $\mathbf{B}_T$ all elements expect the first one ($b_{1,1}$) are zero.
Hence, using Lemma \ref{lemmaformatrices}

\[
\det \mathbf{B}_T=b_{1,1}\det \mathbf{B}_T(1,1)
=\left(a_{1,1}-\beta\sum_{i=2}^Ta_{i,1}\right)\det \mathbf{A}_T(1,1)
=\left(a_{1,1}-\beta\sum_{i=2}^Ta_{i,1}\right)\det \mathbf{A}_{T-1}.
\]

We need to check that
\begin{equation}\label{firstelementofB}
a_{1,1}-\beta\sum_{i=2}^Ta_{i,1}=\frac{1+\beta^2(T-1)}{2}.
\end{equation}
Indeed,
 \begin{eqnarray*} \beta\sum_{i=2}^Ta_{i,1}&=&\beta\sum_{i=2}^{T-1}\left(\frac{\beta\theta_i^T\alpha^{i-2}}{2}
 +\beta^2\sum_{j=i}^{T-1}\theta_{j+1}^T\alpha^{2j-i-1} \right)+\frac{\beta^2\alpha^{T-2}}{2} \\
 &=&\frac{\beta^2\alpha^{-2}}{2}\sum_{i=2}^{T-1}\theta_i^T\alpha^i
 +\beta^3\sum_{j=2}^{T-1}\sum_{i=2}^j\theta_{j+1}^T\alpha^{2j-i-1}+ \frac{\beta^2\alpha^{T-2}}{2} \\
 &=&\frac{\beta^2\alpha^{-2}}{2}\left((T-1)\alpha^2-\alpha^T\right)+\beta^2\sum_{j=2}^{T-1}
 \theta_{j+1}^T\alpha^{2j-1} (\alpha-1)\sum_{i=2}^j\alpha^{-i}+ \frac{\beta^2\alpha^{T-2}}{2} \textrm{ by Lemma \ref{simplelemma}},\\
 &=&\frac{\beta^2(T-1)}{2}+\beta^2\sum_{j=2}^{T-1}\theta_{j+1}^T\alpha^{2j-1}\left(\sum_{i=2}^j\alpha^{-i+1}
 -\sum_{i=2}^j\alpha^{-i}\right)\\
 &=&\frac{\beta^2(T-1)}{2}+\beta^2\sum_{j=2}^{T-1}\theta_{j+1}^T\alpha^{2j-1}\left(\alpha^{-1}-\alpha^{-j}\right)\\
 &=&\frac{\beta^2(T-1)}{2}+\beta^2\sum_{j=2}^{T-1}\theta_{j+1}^T\alpha^{2j-2}-\beta^2\sum_{j=2}^{T-1}\theta_{j+1}^T\alpha^{j-1} \\
 &=&\frac{\beta^2(T-1)}{2}+\beta^2\sum_{j=2}^{T-1}\theta_{j+1}^T\alpha^{2j-2}-\beta^2\sum_{j=3}^{T}\theta_j^T\alpha^{j-2} \\
 &=&\frac{\beta^2(T-1)}{2}+\beta^2\sum_{j=2}^{T-1}\theta_{j+1}^T\alpha^{2j-2}-\beta^2(T-2)\alpha
 \textrm{ by Lemma \ref{simplelemma}}.
 \end{eqnarray*}
 We substitute this into \eqref{firstelementofB}:

 \begin{eqnarray*}
 a_{1,1}-\beta\sum_{i=2}^Ta_{i,1}&=&\frac{1}{2}+\beta^2\sum_{j=1}^{T-1}\theta_{j+1}^T\alpha^{2j-2}- \frac{\beta^2(T-1)}{2}-\beta^2\sum_{j=2}^{T-1}\theta_{j+1}^T\alpha^{2j-2}+\beta^2(T-2)\alpha \\
 &=&\frac{1}{2}+\beta^2\theta_{2}^T-\frac{\beta^2(T-1)}{2}+\beta^2(T-2)\alpha \\
 &=&\frac{1}{2}+\beta^2(T-1-(T-2)\alpha)-\frac{\beta^2(T-1)}{2}+\beta^2(T-2)\alpha=\frac{1+\beta^2(T-1)}{2}.
 \end{eqnarray*}
 \end{proof}

 \begin{lemma}\label{determinant2statement}
 $\mathbf{A}_T$ is positive definite and its determinant is
 \begin{equation}\label{determinant2}
 \det \mathbf{A}_T=\frac{1}{2^T}\prod_{i=0}^{T-1}\left(1+\beta^2i\right)\qquad \textrm{for all }T\geq 1.
 \end{equation}
 \end{lemma}

\begin{proof} For $T=1$, \eqref{determinant2} gives $1/2$. During the computation of $\hat{\phi}_1^1$ we
saw that the coefficient of the quadratic term was indeed $1/2$. Let's assume that \eqref{determinant2} holds for $T-1$, namely
 \begin{equation}\label{assumption}
 \det \mathbf{A}_{T-1}=\frac{1}{2^{T-1}}\prod_{i=0}^{T-2}\left(1+\beta^2i\right).
 \end{equation}
 Then
 \begin{eqnarray*}\det \mathbf{A}_T&=&\frac{1+\beta^2(T-1)}{2}\det \mathbf{A}_{T-1},
 \quad\textrm{by Lemma \ref{determinant1statement},}\\
 		&=&\frac{1+\beta^2(T-1)}{2}\frac{1}{2^{T-1}}\prod_{i=0}^{T-2}\left(1+\beta^2i\right),
 		\quad\textrm{by \eqref{assumption},}\\
 		&=&\frac{1}{2^T}\prod_{i=0}^{T-1}\left(1+\beta^2i\right).
 \end{eqnarray*}

 	Since $\det \mathbf{A}_{T}>0$ for all $T\geq 1$, Lemma \ref{lemmaformatrices} applies and
 	the determinants of the matrices $[a_{ij}]_{i=n,\ldots,T;j=n,\ldots,T}$ are positive for all
 	$1\leq n\leq T$,
 	therefore $\mathbf{A}_{T}$ is positive definite for all $T\geq 1$.
 \end{proof}

Obviously, we can express $\det \mathbf{A}_T$ with the well-known $\Gamma$ function.
\begin{lemma}\label{determinantwithgamma}
We have $\det\mathbf{A}_T=\gamma_\beta(T)/2^T$, where $\gamma_\beta$ is the function defined in \ref{gamma}.
\hfill$\Box$\end{lemma}

 Later on we will need the value of $p:=\left(\mathbf{A}_{T}^{-1}\right)_{1,1}$.
 Now we compute it using Lemmas \ref{lemmaformatrices} and \ref{determinant1statement}:

 \begin{equation}\label{inverse}
 p=\frac{\det \mathbf{A}_T(1,1)}{\det \mathbf{A}_T}=\frac{\det \mathbf{A}_{T-1}}{\det \mathbf{A}_T}=\frac{2}{1+\beta^2(T-1)}.
 \end{equation}

We need to  compute the minimiser of \eqref{matrixformofexp}. Note that in \eqref{matrixformofexp} only the exponent depends on $\phi$, so we can focus on this. Let
\begin{equation}\label{f}
f(\phi, z):=\frac{\mathbf{b^T}(\phi,z)\mathbf{A}_T^{-1}\mathbf{b}(\phi,z)}{4}-c(\phi,z).
\end{equation}

Then, we need to solve
\begin{equation}\label{diffoff}
\partial_{\phi}f(\phi,z)=\frac{\partial_{\phi}\mathbf{b^T}(\phi,z)\mathbf{A}_T^{-1}\mathbf{b}(\phi,z)}{2}-\partial_{\phi}c(\phi,z)=0.
\end{equation}

From the definition a $\mathbf{b}(\phi,z)$ and $c(\phi,z)$
\begin{eqnarray}
\partial_{\phi}\mathbf{b^T}(\phi,z)&=&\left(\sigma,0,\dots,0\right), \label{diffofb}\\
\partial_{\phi}c(\phi,z)&=&\beta z. \label{diffofc}
\end{eqnarray}

 Note that we can write $\mathbf{b}(\phi,z)$ as

 \begin{equation}\label{b}
 \mathbf{b}(\phi,z)=\left(\sigma\phi-\frac{\alpha z}{\sigma}\right)\mathbf{e_1}+\frac{2\alpha z}{\sigma}\mathbf{A}_T(:,1),
 \end{equation}

 where $\mathbf{e_1}=\left(1,0\dots,0\right)^{\mathbf{T}}$,
 and $\mathbf{A}_T(:,1)$ is the first column of $\mathbf{A}_T$.
 Let's substitute \eqref{diffofb}, \eqref{diffofc} and \eqref{b} into \eqref{diffoff}:

\begin{eqnarray*}
	(\sigma, 0,\dots,0)\mathbf{A}_T^{-1}\left[ \left(\sigma\phi-\frac{\alpha z}{\sigma}\right)\mathbf{e_1}+
		\frac{2\alpha z}{\sigma}\mathbf{A}_T(:,1)\right]-2\beta z&=&0 \\
	\sigma\mathbf{A}_T^{-1}(1,:)\left[\left(\sigma\phi-\frac{\alpha z}{\sigma}\right)\mathbf{e_1}+
		\frac{2\alpha z}{\sigma}\mathbf{A}_T(:,1)\right]-2\beta s_0&=&0 \\
	\left(\sigma^2\phi-2\alpha z\right)p+2\alpha z-2\beta z&=&0 \\
	\sigma^2\phi-\alpha z&=&-\frac{2z}{p} \\
	\sigma^2\phi-\alpha z&=&-\left(1+\beta^2(T-1)\right)z\quad\textrm{by \eqref{inverse}, hence} \\
	\phi&=&\frac{\beta z}{\sigma^2}\left(1-(T-1)\beta\right) \\
	\phi&=&\frac{\beta z}{\sigma^2}\theta_1^T.
\end{eqnarray*}

	We can see from the above calculation that this $\phi$ is a global minimiser of $f$ for a given $z$.
	Hence the minimiser $\phi$ for \eqref{phi1def} is
	\[\phi=\hat{\phi}_1^T(z)=\frac{\beta z}{\sigma^2}\theta_1^T,\]
	and we have proved the first part \autoref{maintheorem} in the case of using past information.
	As we have found explicit optimal strategies for the expected utility problem, we can now turn to \eqref{condexp}
	and \eqref{exp}.

First we compute the maximal conditional expected utility which is
\begin{equation}\label{almostdone}
\mathbb{E}\left[U\left(L_T^{{\phi}^*}\right)|X_0=z\right]=-\frac{1}{\sqrt{\gamma_\beta(T)}}e^{f\left(\hat{\phi}_1^T(z),z\right)}
\quad\textrm{by \eqref{matrixformofexp},  Lemma \ref{determinantwithgamma} and \eqref{f}}.
\end{equation}

Let
$\hat{f}(z):=f\left(\hat{\phi}_1^T(z),z\right)$,
$\mathbf{\hat{b}}(z):=\mathbf{b}\left(\hat{\phi}_1^T(z),z\right)$ and
$\hat{c}(z):=c\left(\hat{\phi}_1^T(z),z\right)$.

First we compute $\mathbf{\hat{b}^T}(z)\mathbf{A}_T^{-1}\mathbf{\hat{b}}(z)$:

\begin{eqnarray*}
	\mathbf{\hat{b}^T}(z)\mathbf{A}_T^{-1}\mathbf{\hat{b}}(z)&=& \left[\frac{2\alpha z}{\sigma}\mathbf{A}_T(:,1)
	-\frac{z}{\sigma}\left(1+(T-1)\beta^2\right)\mathbf{e_1}\right]^{\mathbf{T}}\mathbf{A}_T^{-1}\\
	&\,&\cdot\left[\frac{2\alpha z}{\sigma}\mathbf{A}_T(:,1)-\frac{z}{\sigma}\left(1+(T-1)\beta^2\right)\mathbf{e_1}\right]
	 \quad\textrm{by \eqref{b},}\\
&=&\left[\frac{2\alpha z}{\sigma}\left(1,0,\dots,0\right)-\frac{z}{\sigma}\left(1+(T-1)\beta^2\right)\mathbf{A}_T^{-1}(1,:)\right]
\cdot\left[\frac{2\alpha z}{\sigma}\mathbf{A}_T(:,1)-\frac{z}{\sigma}\left(1+(T-1)\beta^2\right)\mathbf{e_1}\right],\\
&=&\frac{z^2}{\sigma^2}\left(\left(1+\beta^2(T-1)\right)^2p-4\alpha\left(1+\beta^2(T-1)\right)+4\alpha^2a_{1,1}\right)\\
&=&\frac{z^2}{\sigma^2}\left(2\left(1+\beta^2(T-1)\right)-4\alpha\left(1+\beta^2(T-1)\right)+4\alpha^2a_{1,1}\right)
\quad\textrm{by \eqref{inverse}.}
\end{eqnarray*}

 For $\hat{f}(z)$
 \begin{eqnarray*}
 \hat{f}(z)&=&\frac{\mathbf{\hat{b}^T}(z)\mathbf{A}_T^{-1}\mathbf{\hat{b}}(z)}{4}-\hat{c}(z)\\
  &=&\frac{z^2}{\sigma^2}\left(\frac{\left(1+\beta^2(T-1)\right)}{2}-\alpha\left(1+\beta^2(T-1)\right)
  +\alpha^2a_{1,1}- \beta(1-(T-1)\beta)-\beta^2\sum_{j=2}^T\theta_j^T\alpha^{2j-2}\right)\\
  &=&\frac{z^2}{\sigma^2}\left(\frac{\left(1+\beta^2(T-1)\right)}{2}-\alpha\left(1+\beta^2(T-1)\right)
  +\frac{\alpha^2}{2}+\beta^2\sum_{j=1}^{T-1}\theta_{j+1}^T\alpha^{2j}
  -\beta(1-(T-1)\beta)-\beta^2\sum_{j=2}^T\theta_j^T\alpha^{2j-2}\right)\\
  &=&-\frac{\beta^2 z^2}{2\sigma^2}T.
 \end{eqnarray*}

 Hence the maximal achievable conditional expected utility is
 \begin{equation}
 \mathbb{E}\left[U\left(L_T^{{\phi}^*}\right)|X_0=z\right]=-\frac{1}{\sqrt{\gamma_\beta(T)}}e^{-\frac{\beta^2z^2}{2\sigma^2}T},
 \end{equation}
 and we have proved \eqref{condexp}. Now we prove \eqref{exp}. For stable processes, in case of $\mathrm{var}(X_t)=1$
 for all $t\geq 0$

 \begin{equation} 1=\mathrm{var}(X_t)=\mathrm{var}\left(\sigma\sum_{j=0}^{+\infty}\alpha^j\varepsilon_{t-j}\right)=\sigma^2\sum_{j=0}^{+\infty}\alpha^{2j}=\frac{\sigma^2}{1-\alpha^2},\nonumber
 \end{equation}
 therefore $\sigma^2=1-\alpha^2$. As $X_0$ is $N(0,1)$, the maximal expected utility can be found using
 \eqref{integral1}:
 \begin{eqnarray*}
 \mathbb{E}\left[U\left(L_T^{{\phi^*}}\right)\right]&=&
 \mathbb{E}\left[\mathbb{E}\left[U\left(L_T^{{\phi}^*}\right)|X_0 \right]\right]
  =\mathbb{E}\left[-\frac{1}{\sqrt{\gamma_\beta(T)}}e^{-\frac{\beta^2X_0^2}{2\sigma^2}T}\right]\\
	&=&-\frac{1}{\sqrt{2\pi \gamma_\beta(T)}}\int_{\mathbb{R}}e^{-\frac{\beta^2x^2}{2\sigma^2}T-\frac{x^2}{2}}\mathrm{d}x
	=-\frac{1}{\sqrt{\gamma_\beta(T)\left(\frac{\beta^2T}{1-\alpha^2}+1\right)}} \\
	&=&-\sqrt{\frac{\beta+2}{\left(2-(T-1)\beta\right)\gamma_\beta(T)}},
	\end{eqnarray*}
so we have proved \eqref{exp}.

Now we focus on the case where the strategies depend only on the initial value $X_0$
of the autoregressive process. In this case using the strategy $\eta=(\eta_1,\dots,\eta_T)$  we get
\begin{eqnarray*}
L_T^\eta&=&\sum_{j=1}^T\eta_j\left(X_j-X_{j-1}\right) \\
		&=&\sum_{j=1}^T\eta_j\left(\alpha^{j-1}\beta X_0+\sigma\varepsilon_j+\sigma\beta\sum_{k=1}^{j-1}\alpha^{j-1-k}\varepsilon_k\right) \\
		&=&\beta X_0\sum_{j=1}^T\eta_j\alpha_{j-1}+\sigma\sum_{j=1}^T\eta_j\varepsilon_j+\beta\sigma\sum_{j=1}^T\sum_{k=1}^{j-1}\eta_j \alpha^{j-k-1}\varepsilon_k \\
		&=&\beta X_0\sum_{j=1}^T\eta_j\alpha_{j-1}+\sigma\sum_{j=1}^T\eta_j\varepsilon_j +\beta\sigma\sum_{k=1}^T\varepsilon_k\sum_{j=k+1}^T\eta_j\alpha^{j-k-1}.
	\end{eqnarray*}

	Let $c: \mathbb{R}^T\rightarrow\mathbb{R}$, and $\mathbf{b}:\mathbb{R}^T\rightarrow\mathbb{R}^T$, where
	\[c(\eta)=\beta X_0\sum_{j=1}^T\eta_j\alpha^{j-1},\quad
	\textrm{and}\quad b_k(\eta)=\sigma\varepsilon_k+\beta\sigma\sum_{j=k+1}^T\eta_j\alpha^{j-k-1},\quad 1\leq k\leq T.\]
Using the notation $L_T^\eta=c(\eta)+\mathbf{b^T}(\eta)\mathbf{\varepsilon}$ we get from
Lemma \ref{integral2lemma} that
\begin{equation}\label{almostdonenomem}
\mathbb{E}\left[U\left(L_T^\eta\right)|X_0\right]=-e^{g(\eta)}, \textrm{ where }
g(\eta)=-c(\eta)+\frac{\mathbf{b^T}(\eta)\mathbf{b}(\eta)}{2}.
\end{equation}

We need to solve the system of equations $\nabla g(\eta)=\mathbf{0}$.
We denote these equations by $(E_k)$, where $1\leq k\leq T$:
\[0=\frac{\partial g}{\partial \eta_k}(\eta)=-X_0\beta\alpha^{k-1}+\mathbf{b^T}(\eta)\frac{\partial\mathbf{b}}{\partial\eta_k}(\eta).
\quad (E_k)\]
The partial derivatives of $\mathbf{b}$ are:
\[\left(\frac{\partial\mathbf{b}}{\partial\eta_k}(\eta)\right)_l=\left\{\begin{array}{cl} \sigma\beta\alpha^{k-l-1} & \textrm{if }l<k \\
																																										\sigma & \textrm{if } l=k \\
																																										0 & \textrm{if } l>k. \end{array} \right. \]
Hence we have
\begin{eqnarray*}
\mathbf{b^T}(\eta)\frac{\partial\mathbf{b}}{\partial\eta_k}(\eta)&=&\sigma\beta\sum_{l=1}^{k-1}\alpha^{k-l-1}\left(\sigma\eta_l +\sigma\beta\sum_{j=l+1}^T\eta_j\alpha^{j-l-1}\right)+\sigma^2\eta_k+\sigma^2\beta\sum_{j=k+1}^T\eta_j\alpha^{j-k-1} \\
	&=&\sigma^2\beta\sum_{l=1}^{k-1}\eta_l\alpha^{k-l-1}+\sigma^2\beta^2\sum_{l=1}^{k-1}\sum_{j=l+1}^T\eta_j\alpha^{j+k-2l-2} +\sigma^2\eta_k+\sigma^2\beta\sum_{j=k+1}^T\eta_j\alpha^{j-k-1}.
	\end{eqnarray*}

Therefore the equations $E_k$ take the form
\[\beta\sum_{l=1}^{k-1}\eta_l\alpha^{k-l-1}+\beta^2\sum_{l=1}^{k-1}\sum_{j=l+1}^T\eta_j\alpha^{j+k-2l-2}
+\eta_k+\beta\sum_{j=k+1}^T\eta_j\alpha^{j-k-1}=\frac{X_0}{\sigma^2}\beta\alpha^{k-1}. \]

Let $k\in\{1, 2, \dots ,T-1\}$. Then for $E_{k+1}$ we have
\[\beta\sum_{l=1}^{k}\eta_l\alpha^{k-l}+\beta^2\sum_{l=1}^{k}\sum_{j=l+1}^T\eta_j\alpha^{j+k-2l-1} +\eta_{k+1}+\beta\sum_{j=k+2}^T\eta_j\alpha^{j-k-2}=\frac{X_0}{\sigma^2}\beta\alpha^{k}. \]
We define equation $F_k$ for $1\leq k\leq T-1$ by substract equation $E_k$ multiplied by $\alpha$ from equation $E_{k+1}$, so we get
\begin{eqnarray*}
0&=&\beta\eta_k+\beta^2\sum_{j=k+1}^T\eta_j\alpha^{j-k-1}+\eta_{k+1}-\alpha\eta_k
+\beta\sum_{j=k+2}^T\eta_j\alpha^{j-k-2} -\beta\sum_{j=k+1}^T\eta_k\alpha^{j-k}, \\
0&=&\eta_{k+1}-\eta_k+\beta\left(\sum_{j=k+1}^T\eta_j\alpha^{j-k}-\sum_{j=k+1}^T\eta_j\alpha^{j-k-1}\right)
+\beta\sum_{j=k+2}^T\eta_j\alpha^{j-k-2} -\beta\sum_{j=k+1}^T\eta_k\alpha^{j-k}, \\
0&=&\eta_{k+1}-\eta_k-\beta\sum_{j=k+1}^T\eta_j\alpha^{j-k-1}+\beta\sum_{j=k+2}^T\eta_j\alpha^{j-k-2}, \\
0&=&\eta_{k+1}-\eta_k-\beta\eta_{k+1}+\beta\sum_{j=k+2}^T\eta_j\alpha^{j-k-2}(1-\alpha), \\
0&=&(1-\beta)\eta_{k+1}-\eta_k-\beta^2\sum_{j=k+2}^T\eta_j\alpha^{j-k-2}. \\
\end{eqnarray*}

\begin{lemma}\label{eqnstatement}
For the solutions of the system $F_k$, $k=1,\ldots,T-1$,
\begin{equation}\label{eqn}
 \eta_k=\theta_k^T\eta_T=(1-(T-k)\beta)\eta_T
\end{equation}
hold, for all $k=1,\ldots, T-1$.
\end{lemma}

\begin{proof} First we consider the equation $F_{t-1}$,
\begin{eqnarray*}
0&=&(1-\beta)\eta_T-\eta_{T-1} \\
\eta_{T-1}&=&\theta_{T-1}^T\eta_T
\end{eqnarray*}

Let's assume that \eqref{eqn} holds for $l=k+1, \dots ,T-1$.
Considering the equation $F_k$, using Lemma \ref{simplelemma} we get
\begin{eqnarray*}
0&=&(1-\beta)\eta_{k+1}-\eta_k-\beta^2\sum_{j=k+2}^T\eta_j\alpha^{j-k-2}, \\
0&=&(1-\beta)(1-(T-k-1)\beta)\eta_T-\eta_k-\beta^2\sum_{j=k+2}^T\theta_j^T\eta_T\alpha^{j-k-2}, \\
0&=&(1-\beta)(1-(T-k-1)\beta)\eta_T-\eta_k-\beta^2(T-k-1)\eta_T, \\
\eta_k&=&(1-(T-k-1)\beta-\beta)\eta_T ,\\
\eta_k&=&\theta_k^T\eta_T.
\end{eqnarray*}
\end{proof}

Because of $\theta_T^T=1$, $\eta_T=\theta_T^T\eta_T$ also holds. To prove the first part of
\autoref{maintheorem} for the case without using past information, we only need to show that $\eta_T=\beta\frac{X_0}{\sigma^2}$.

Substituting \eqref{eqn} into $E_T$,
\begin{equation}
\beta\sum_{l=1}^{T-1}\theta_l\eta_T\alpha^{T-l-1}+\beta^2\sum_{l=1}^{T-1}\sum_{j=l+1}^T\theta_j^T\eta_T\alpha^{j+T-2l-2}
+\eta_T=\frac{X_0}{\sigma^2}\beta\alpha^{T-1} \\
\end{equation}
We compute the two sums separately,
\begin{eqnarray*}
\sum_{l=1}^{T-1}\theta_l^T\alpha^{T-l-1}&=&\sum_{l=1}^{T-1}\left(T+1-l-(T-l)\alpha\right)\alpha^{T-l-1}
	=\sum_{l=1}^{T-1}(T+1-l)\alpha^{T-l-1}-\sum_{l=1}^{T-1}(T-l)\alpha^{T-l} \\
	&=&\sum_{l=2}^{T}(T+2-l)\alpha^{T-l}-\sum_{l=1}^{T-1}(T-l)\alpha{T-l}
	=\sum_{l=2}^T2\alpha^{T-l}-(T-1)\alpha^{T-1}; \\
\beta\sum_{l=1}^{T-1}\sum_{j=l+1}^T\theta_j^T\alpha^{j+T-2l-2}&=&\beta\sum_{l=1}^{T-1}\alpha^{T-2l-2}\sum_{j=l+1}^T\theta_j^T\alpha^j
	=\beta\sum_{l=1}^{T-1}(T-l)\alpha^{T-l-1} \\
	&=&\sum_{l=1}^{T-1}(T-l)\alpha^{T-l}-\sum_{l=1}^{T-1}(T-l)\alpha^{T-l-1}
	=\sum_{l=1}^{T-1}(T-l)\alpha^{T-l}-\sum_{l=2}^{T}(T-l+1)\alpha^{T-l}\\
	&=&-\sum_{l=2}^T\alpha^{T-l}+(T-1)\alpha^{T-1}.
\end{eqnarray*}

Hence, from $E_T$:
\begin{eqnarray*}
\left(1+\beta\sum_{l=2}^T\alpha^{T-l}\right)\eta_T&=&\alpha^{T-l}\beta\frac{X_0}{\sigma^2} \\
\left(1+\sum_{l=2}^T\alpha^{T-l+1}-\sum_{l=2}^T\alpha^{T-l}\right)\eta_T&=&\alpha^{T-l}\beta\frac{X_0}{\sigma^2} \\
\alpha^{T-1}\eta_T&=&\alpha^{T-l}\beta\frac{X_0}{\sigma^2} \\
\eta_T&=&\beta\frac{X_0}{\sigma^2}.
\end{eqnarray*}
So we have proved the firs part of \autoref{maintheorem}. Now we prove \eqref{condexpnomem}. First we compute
the element of  $\hat{\mathbf{b}}:=\mathbf{b}(\hat{\eta})$, using Lemma \ref{simplelemma},
\begin{equation}
\hat{b}_k=\frac{\beta X_0}{\sigma}\theta_k+\frac{\beta^2 X_0}{\sigma}\sum_{j=k+1}^T\theta_j\alpha^{j-k-1}
	=\frac{\beta X_0}{\sigma}(1-(T-k)\beta)+\frac{\beta^2 X_0}{\sigma}(T-k)
	=\frac{\beta X_0}{\sigma}.
\end{equation}
Therefore
$\mathbf{\hat{b}^T\hat{b}}=\frac{\beta^2 X_0^2}{\sigma^2}T$, and
$\hat{c}:=c(\hat{\eta})=\frac{\beta^2X_0^2}{\sigma^2}\sum_{j=1}^T\theta_j\alpha^{j-1}=\frac{\beta^2X_0^2}{\sigma^2}T$, by
Lemma \ref{simplelemma}.
Hence, using \eqref{almostdonenomem}, we get
$\mathbb{E}\left[U\left(L_T^{\hat{\eta}}\right)|X_0\right]
=e^{-\frac{\beta^2X_0^2}{2\sigma^2}T}$, which proves \eqref{condexpnomem}.
Based on the same calculation that we did to get \eqref{exp} from \eqref{condexp},
we get \eqref{expnomem} from \eqref{condexpnomem}.

The only statment of \autoref{maintheorem} left to prove is the forth part. For $\beta=0$, the statement is trivial. For the other case, using $y(T):=T-1+\frac{1}{\beta^2}$, we get
\begin{equation}
\frac{\gamma_\beta(T)}{h_\beta(T)}=\frac{\beta^{2T}\Gamma(y(T)+1)}
{\left(\beta^2\right)^{1-\frac{1}{\beta^2}}\sqrt{2\pi y(T)}\left(\frac{\beta^2y(T)}{e}\right)^{T-1+\frac{1}{\beta^2}}}
=\frac{\Gamma\left(y(T)+1\right)}{\sqrt{2\pi y(T)}\left(\frac{y(T)}{e}\right)^{y(T)}},
\end{equation}
and it is well-known  that this expression tends to $1$ if $T$ (and hence also $y(T)$) tend to infinity.

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text:
Q: Jmeter- how to extract data from redirection I have any API. When i hit this API by jmeter, ir redirects to server A(do some processing), then redirects to B, and finally redirects to C( which is a HTML page).
In view result tree, i only get he final HTML response. Is there some way, by which i can see the request response to server A & B

A: Just uncheck "Follow redirects" option in HTTP Request

See:

*

*http://jmeter.apache.org/usermanual/component_reference.html#HTTP_Request
You will then need to emit yourselft the requests by Using A Regexp Extractor to extract from Header the URL and then call it.

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Spot gold was up 0.2 percent at $1,306.88 an ounce at 10:27 a.m EDT (1427 GMT), its highest since March 28. U.S. GCv1 were also up 0.2 percent at $1,310.40.
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TaskRabbit hops into five new markets
TaskRabbit is scampering into five new towns today and announced that it's far on track to gain 1.25 million new customers this year. The peer-to-peer marketplace for outsourcing obligations is now to be had in Atlanta, Dallas, Houston, Philadelphia, and Washington, D.C. Spokesperson Johnny Brackett stated this expands its "geographical footprint," employing 50 percent. Existing markets include Boston, Chicago, Los Angeles, New York City, Seattle, the S.F. Bay location, Austin, Texas, and Portland, Ore.
The 5 new cities have over 2,000 new TaskRabbits, and TaskRabbit's army is now 15,000 strong. These TaskRabbits carry out responsibilities and run errands for users, including constructing Ikea furnishings or choosing up your dry cleaning. The marketplace allows users to accomplish duties that they don't have the time, motivation, knowledge, or assets to do on their very own and provide TaskRabbits with an additional circulate of profits.
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Rachel Botsman wrote an e-book known as What's Mine Is Yours: The Rise of Collaborative Consumption. She estimates that the purchaser peer-to-peer rental market becomes a $26 billion region and that the sharing financial system is a $one hundred ten billion marketplace. It is growing speedy as Internet penetration will increase around the arena, and those end up extra at ease with the idea of "disownership." TaskRabbit won its initial traction in early adopter markets like San Francisco and New York but is now making it's way deeper into the mainstream. The corporation has also made a massive push on cellular and is currently working on a revamp of the iOS app, a local Android app, and an app for TaskRabbits by using the give up of the 12 months.
How Silicon Valley's Obsession With Narrative Changed TaskRabbit
The gig economy is a double-edged sword for migrant workers

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Entertainment December 6, 2021
7 Facts About Julia Child That May Surprise You
By Tess Bonn
What we learned from the new documentary about the culinary icon.
There's no question that famed chef and personality Julia Child continues to have an impact on how Americans cook and eat. You'll still find Mastering the Art of French Cooking, the book that first put her on the map decades ago, in countless households across the U.S.
But she was anything but your typical chef — she towered over most people standing at 6-feet, 2-inches, and then there was her distinctive, bird-like voice. What you may not know is that Child also had some convictions that were considered unconventional during her time. For instance, she became a strong advocate for abortion rights at a time when it was highly scrutinized and spoke out for the LGBT community.
A new documentary, Julia, is capturing a more complete portrait of her — literally, the film includes an intimate shot of the infamous chef posing nude. Finally, Child gets the credit she deserves as not only a trailblazer in the professional cooking world but also for women in television.
"Women on television had really been young, petite, pretty, and supportive, not the main focus of much programming and then Julia Child comes on, and she's tall, a little gangly, and middle-aged with a loud voice and she's telling people what to do, and people are eating it up," co-director Betsy West tells us.
Here are seven things we learned about Child from the new film.
Julia Child didn't grow up in a household with a love of cooking.
Though she would go on to inspire foodie culture, food was not the centermost focus in her home growing up in Pasadena, California. In fact, Child's mother rarely cooked, and the family largely relied on chefs for their meals, which consisted of "very sensible New England-type food" like roasts, mashed potatoes, and fresh peas. "Nobody discussed food a great deal because it just wasn't done," Child says in the film.
In fact, it was her husband Paul Child who first introduced her to what would become her passion and legacy. "She meets Paul — a guy who's 10 years older and is actually quite worldly — and he has a great love of culture and food and so he introduced her and really stoked her love of food," says co-director Julie Cohen. Child went on to enroll at the infamous French cooking school Le Cordon Bleu in Paris when she was 37 years old and she didn't publish her first cookbook until she was 49 years old because it took her nine years to complete.
Julia Child worked for what we know today as the CIA.
Like many women of her time, when World War II broke out, Child wanted to do her part. So, she enlisted and eventually got into the Office of Strategic Services [OSS], which was a precursor to the CIA or special intelligence. Though she never became a spy herself, she worked with spies and managed top secret files as a clerk typist. She would later volunteer to go to Ceylon (now known as Sri Lanka), which is where she met Paul, who was working as a graphics artist at the time.
"She had lived a fairly privileged, but sheltered life. She was looking for adventure but she didn't really know that much about the world," Cohen says.
Julia Child and Paul Child's lifelong romance wasn't love at first sight.
Child wasn't initially attracted to Paul. As revealed in letters she sent back home, she first described him as "not at all nice looking," and said he had "an unbecoming mustache and a long nose."
But, with time, their love grew and Paul would become one of — if not the — biggest champions of Child's career. "It really is a feminist romance and it was another thing that we just loved about Julia's story," Cohen says.
She helped make public television what it is today.
Are you a fan of instructional cooking shows like the ones you see on the Food Network? Well, you can thank Child for that — she practically invented them with her 1963 show The French Chef.
The idea for the show first came about when Child made an appearance on a local Boston television station while promoting her debut cookbook, Mastering the Art of French Cooking. After she demonstrated how to make a proper omelet, fans practically fell in love with her right off the bat. West credits Child's immediate popularity to her exuberance and authenticity.
"She was authentic — she wasn't hiding if she made a mistake, she acknowledged it and then showed you what to do when you make a mistake in the kitchen," she says. "She saw it as an opportunity."
Julia Child beat breast cancer.
Child was diagnosed with breast cancer in 1968 at the age of 55 after finding a lump in her breast. But she was relatively private about her battle, and even in her own journal, simply wrote, "Left breast off."
But the life-threatening experience did have an impact on her, even taking a toll on her self-esteem, though Paul would promptly put that to rest. "'Paul said, 'I didn't marry you for your breasts, I married you for your legs,' and she never gave it another thought and that was that," according to her personal assistant Stephanie Hersh, who made appearances throughout the film.
Julia Child was a champion of abortion rights.
Child became a passionate supporter of Planned Parenthood. She started out fundraising and hosting benefits before becoming a board member. In 1982, she wrote a letter for the nonprofit to donors, saying, "Few politicians will take the risk of publicly supporting either contraception or abortion — and who is 'for abortion' anyway?," according to The New Yorker, "We are concerned with freedom of choice."
Even though it was the 1980s, it was still considered a risky move at the time, and her support for abortion drew backlash. Some of her fans picketed her events but that didn't deter her from remaining an outspoken advocate. "Julia was a mainstream cultural icon, not someone who just the elites were following — she had a big fan base," says West. "Some of them objected to her association with Planned Parenthood, but she just shrugged that off; it was something she really truly believed in."
Julia Child became a vocal supporter of the LGBT community.
As revealed by even some of her own friends in the documentary, Child was known for making homophobic remarks, but her mindset changed when her attorney and friend Bob Johnson died of AIDS in 1986. In wake of his death, she poured herself into hosting benefits and raising money to fight the disease.
"What she did was actually really important — it was a complete 180-degree turn," says Cohen. "And we just thought it was interesting from the story perspective and also tells you something about Julia's character."
Julia is playing in select theaters across the country and will be available to stream at the end of December.
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text:
Businesses aren't telling anyone when they've been hit with ransomware
Half of American businesses have been hit with ransomware this year but many stay silent and spend money instead of informing customers or law enforcement of the breach.
There may be more businesses dealing with ransomware than the private sector community cares to talk about.
Half of American businesses suffered a ransomware attack in the last year, according to new research from the security firm SentinelOne. The standard business response to the attacks —hackers encrypt a target's data and demand payment to get it back — is to spend more money, lose faith and feel helpless on cybersecurity, the survey shows. Then, in many cases, they try to keep the entire incident under wraps as best they can.
While laws and regulations vary across states and countries, staying silent can be illegal. The specifics depend on a variety of factors including what kind of data was exposed, how it happened and where the breach took place. What stays the same is the perception that disclosure may be more trouble than it's worth.
"Ransomware is an incredibly simple business model, highly lucrative, and often leverages the current malware infrastructure with minimal new investment required," Jeremiah Grossman, Chief of Security Strategy at SentinelOne security firm told CyberScoop. In October, one particular ransomware variant, Locky, could be found in over 14 million malicious emails in a single week.
The new survey was conducted by research firm Vanson Bourn and commissioned by SentinelOne, a security firm that specializes in endpoint security. The figures match up with other research, as well as information from law enforcement.
Ransomware is now a major focus for the FBI. Director James Comey has urged businesses to report breaches. In many cases, hacked businesses prefer to keep the whole incident under wraps.
The survey found that just 61 percent of cybersecurity professionals notified the CEO or board of their company when they suffered an attack. Just 54 percent did or would notify law enforcement and lawyers. If you're out of that exclusive circle, the numbers plummet further. Only 38 percent of cybersecurity professionals say they did or would notify customers if they suffered a ransomware attack.
Security teams often keep malware infections quiet from company leadership unless it's highly disruptive. They won't go to law enforcement unless they need help or want to capture the extortionist.
"If the cyber-extortionist willingly takes a 'reasonable' ransom demand and turns over the encryption key to allow access to the data, the business will simply recover as best they can and quietly move on," Grossman said. Customers often won't be notified unless regulations require them to do so or unless customers noticed a problem. If not, "they'll quietly take care of the matter and move on."
That silence doesn't leave the feds too pleased.
"We have discovered that the majority of our private partners do not turn to law enforcement when they face an intrusion. And that is a very big problem," Comey said in August. "It is fine to turn to one of the many excellent private sector entities that will help with attribution and with remediation—that's good. But we have to get to a place where it is routine for people who are victimized to turn to us for assistance."
Eighty-five percent of American businesses surveyed said they'd been hit three or more times with ransomware in the last year.
"The situation is likely to get far worse, as some of the ill-gotten gains will be invested into research and development designed to improve encryption strength and utilize new delivery methods, as witnessed with Locky," Grossman explained.
"These criminals have evolved over time and now bypass the need for an individual to click on a link," FBI Cyber Division Assistant Director James Trainor said earlier this year. "They do this by seeding legitimate websites with malicious code, taking advantage of unpatched software on end-user computers."
The vast majority of infections are still phishing via email and social media. But malicious websites and unpatched software account have been seen at least once by most businesses surveyed.
After the attack, the money starts flowing into preventative and reactive measures. Patching software, disabling macros, and reviewing anti-malware becomes a renewed focus.
"For reactive, it's all about backups," Grossman said. "Having backups that are both immutable and stored offline can save you big time."
Millions of dollars are being poured into the fight from all sides—the defenders, the hackers, and the victims who pay them off—but it's a war that's been kept mostly quiet so far.
Jeremiah Grossman
Locky
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It's never too early to make predictions.
Hey, if we can do a 2021 mock draft the day after the 2020 draft, we can definitely do some 2020 Eagles predictions in mid-May, right?
All of these are contingent on the players staying healthy, which around these parts hasn't been real common lately. But, hey, new training staff, new medical staff. Maybe things will really change.
So here are our never-too-early-to-make Eagles predictions for the coming season. If I'm close, I'll remind everyone in December. If I'm way off, hopefully nobody will remember them!
1. The surprise leading WR: Greg Ward began last year on the practice squad and didn't play on offense until Week 11. But he caught 31 passes the last seven games, which was only 12 off the team lead by a WR for the entire season. Despite him not playing until Week 11. Ward was really the Eagles' only reliable WR the last month and a half of the season, and that won't be the case this coming season. But he runs such good routes out of the slot and has such good hands and knows how to get open, and Carson Wentz clearly likes throwing to him.
Prediction: Ward leads all Eagles WRs with 62 catches.
2. Another record for Wentz: If Wentz could throw 27 TD passes with the lowly group of receivers he had last year, there's no reason he can't throw a dozen or so more with an upgraded cast. Wentz set the franchise record of 33 TD passes in just 13 games in 2017. That's about 2 1/2 per game. Project that over 16 games and you've got 40. I really think that's doable this year.
Prediction: Wentz extends his Eagles record and throws 40 TDs in 2020.
3. A full season of Sanders: Miles Sanders had such a promising rookie season it's easy to forget that he didn't really become a consistent weapon until Week 8 in Buffalo. If you just increase his rookie average of 83 yards per game by 20 percent, you're looking at about 1,600 yards. Sanders is that good.
Prediction: Sanders becomes the fifth running back in franchise history with a 1,600-yard season. Let's go with 1,055 rushing yards and 616 receiving yards for 1,671 scrimmage yards in his second season.
4. Will the rookie produce?: Only four Eagles WRs have had as many as 30 catches as a rookie in the last 25 years (Reggie Brown, DeSean Jackson, Jeremy Maclin and Jordan Matthews). Jalen Reagor has to beat that. Has to.
Prediction: Let's go 52 catches for 676 yards and five TDs for Reagor.
5. Holding it down at corner: The Eagles haven't had a cornerback with five INTs in a season since Brandon Boykin in 2013, but Darius Slay ends that drought in his first season with the Eagles.
Prediction: Slay finishes with six interceptions.
6. What should we expect from Boston Scott?: Scott had only 23 touches until the last month of the season, when he averaged 80 scrimmage yards over the last five games. It's not a huge body of work, but Scott showed that, used the right way, he can be a weapon as a receiver and situational runner. We'll see if the Eagles bring in a veteran back and how the roles will be defined, but I'd be surprised if Scott doesn't give the Eagles six or seven touches per game.
Prediction: I'm giving Scott 688 scrimmage yards. That's 43 per game. That's doable.
7. What about Jalen Hurts?: I'm not sure he'll justify where he was drafted, but I do expect Hurts to make an impact, mainly as a short-yardage and goal-line runner.
Prediction: Hurts becomes the first Eagles rookie with five rushing TDs in a season since Don "Heartbeat" Johnson in 1953.
8. Who's getting the sacks?: Brandon Graham is 32 now, but he's had his two best seasons in 2017 and 2019, with 9 ½ sacks during the Super Bowl season and 8 1/2 last year. Graham is one of only six active players with 50 or more career sacks that's never made a Pro Bowl. He takes great care of himself, he's consistent and he's durable, having missed just one game since opening day 2012.
Prediction: In his 11th season, Graham records a career-high 11 1/2 sacks and makes his first Pro Bowl.
9. How about the tight ends?: Zach Ertz goes into the eighth season of his remarkable NFL career with 525 catches. He needs 23 to move into the all-time NFL top-10 for tight ends and 65 to pass Hall of Famer Harold Carmichael and become the all-time leading Eagles receiver.
Prediction: Ertz finishes with 81 catches, becomes the third TE In NFL history with six straight 70-catch seasons and passes Carmichael in Week 13 against the Seahawks.
10 more quickie predictions
Nelson Agholor doesn't make it to opening day with the Raiders … Dallas Goedert leads the Eagles with nine TD catches … Derek Barnett has a career-high 7 ½ sacks … Fletcher Cox joins Reggie White and Pete Pihos as the third Eagle to make six straight Pro Bowls … Undrafted rookie Raequan Williams out of Michigan State makes the team as the fourth defensive tackle … Jason Peters signs with the Eagles … DeSean Jackson plays 11 games and catches 37 passes for 592 yards and four TDs … JJ Arcega-Whiteside is a little better but only catches 29-for-493 … Jalen Mills winds up back at corner and rookie K'Von Wallace starts the last seven games of the season … The Eagles sign and release Jordan Matthews at least twice.
I like this roster. I like this team. I expect the Eagles to be better on both sides of the football, and I like Doug Pederson coaching in a division with three new head coaches. If the new medical and training staff can help the Eagles avoid another injury-plagued season, this will be another playoff season in Philly.
Prediction: The Eagles go 11-5 and become the first team since the 2003 and 2004 Eagles to win the NFC East in consecutive seasons.
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text:
Human Spaceflights
Private Spaceship Factory Opens for Business in Calif. Desert
By Leonard David
The WhiteKnightTwo/SpaceShipTwo launch system is showcased during opening day ceremonies of the Final Assembly, Integration and Test Hangar. The Spaceship Company is ramping up its ability to produce both suborbital rocket ships and flying launch platforms. (Image credit: Virgin Galactic/Griffin Communications Group)
In a grand and ceremonious style, a factory site that will crank out private spaceships has opened its hangar doors.
The $8 million hangar was specifically designed to support the final stages of assembly and integration for prime customer Virgin Galactic's fleet of passenger-carrying suborbital SpaceShipTwos and the mothership launch craft, WhiteKnightTwos.
Called the Final Assembly, Integration and Test Hangar, or FAITH, the special building was unveiled Sept. 19 at the Mojave Air and Space Port in California. That's the home port for Scaled Composites, builder of the SpaceShipTwo/WhiteKnightTwo launch system. [Photos: SpaceShipTwo's First Glide Test Flight]
Important step
"The opening of the new facility is an important step on a journey that will culminate in commercial operations at Spaceport America" in New Mexico, said George Whitesides, CEO and president of Virgin Galactic, which will launch paying customers to suborbital space aboard SpaceShipTwo.
"The modern plant is energy-efficient and will provide ample space for future growth of The Spaceship Company. It can hold two WhiteKnightTwo carrier aircraft and several SpaceshipTwos at the same time," Whitesides told SPACE.com.
The Spaceship Company (TSC) was established to manufacture additional vehicle sets beyond the first pair from Scaled Composites.
"We believe there is tremendous possibility for growth in the future," Whitesides added.
FAITH is a 68,000-square-foot (6,317 square meters) structure. It's big enough to accommodate the first-ever side-by-side public viewing of WhiteKnightTwo/SpaceShipTwo and the WhiteKnightOne/SpaceShipOne system, which happened at the grand opening ceremony.
In 2004, SpaceShipOne bagged the $10 million Ansari X Prize as the world's first privately developed piloted spacecraft.
Onlookers take part in Sept. 19, 2011 ceremonies at the Mojave Air and Space Port. The Final Assembly, Integration and Test Hangar is designed to put together spaceships for Virgin Galactic. (Image credit: Courtesy: Virgin Galactic/Griffin Communications Group)
TSC is the aerospace production joint venture of Sir Richard Branson's Virgin Galactic and Scaled Composites, which teamed up to build the world's first fleet of commercial spaceships and flying launch pads.
Virgin Galactic's SpaceShipTwo is being built to carry six customers and two pilots on suborbital jaunts. That trek to the edge of space and back provides passengers several minutes of out-of-the-seat, zero-gravity experience. [Video: SpaceShipTwo's First Crewed Flight]
During the flight, space travelers are promised astounding views of the planet from the black sky of space for about 1,000 miles (1,609 kilometers) in every direction.
In the space reservation department, Virgin Galactic has signed up more than 450 individuals who have made deposits between $20,000 and the total per-seat cost of $200,000.
Total deposits received are over $57 million. Over 85,000 people from 125 countries have registered their interest in becoming a Virgin Galactic astronaut, according to the company.
Kick-in-the-pants velocity
The first fully built SpaceShipTwo is dubbed VSS Enterprise, and the WhiteKnightTwo is called VMS Eve. Both have already undergone a step-by-step test flight program.
Still ahead, however, are critical flights involving SpaceShipTwo's hybrid rocket motor. That engine is central to giving the craft a kick-in-the-pants velocity to attain a desired suborbital trajectory.
The progression of shake-out flights is necessary before Virgin Galactic can start safe and sound commercial operations, which will be based at Spaceport America in New Mexico.
FAITH was completed within 10 months, as scheduled and on budget by Bakersfield-based Wallace & Smith General Contractors. It can support the production of two WhiteKnightTwos and at least two SpaceShipTwo vehicles in parallel. Also, the facility can support major return-to-base maintenance for the rocket plane and carrier mothership fleet once in operation, officials said.
Virgin Galactic's SpaceShipTwo private spaceliner flies over San Francisco's Bay Area Bridge with the city as a backdrop while en route to San Francisco International Airport with the Virgin America aircraft "My Other Ride is a Spaceship" to open a new terminal at San Francisco International Airport on April 6, 2011. (Image credit: Mark Greenberg)
Workforce wants
FAITH is one of two facilities that TSC will use to produce commercial spacecraft. The other is a 48,000-square-foot (4,459 square meters) existing building at the Mojave Air and Space Port that TSC recently upgraded to serve as the company's fabrication and vehicle sub-assembly facility.
TSC has secured options to expand the size of the FAITH facility and build an adjacent flight test hangar, as the customer base grows.
The opening of FAITH has been also billed as a means to boost local economies in California and New Mexico. TSC currently employs more than 80 people and is looking to double its work force within the next year, with numerous high-tech and engineering positions available in the next few months, according to TSC officials.
Others in the aerospace field see the opening of the new hangar as a good sign for the nascent space tourism industry.
"Not only are we welcoming a new neighbor at the Mojave Air and Space Port…we're ushering in another phase in the development of commercial space travel," said Doug Shane, president of Scaled Composites. "It's exciting to see the vision becoming a reality."
For more information on The Spaceship Company, visit: http://www.thespaceshipcompany.com
Leonard David has been reporting on the space industry for more than five decades. He is a winner of this year's National Space Club Press Award and a past editor-in-chief of the National Space Society's Ad Astra and Space World magazines. He has written for SPACE.com since 1999.
Leonard David
Space Insider Columnist
Leonard David is an award-winning space journalist who has been reporting on space activities for more than 50 years. Currently writing as Space.com's Space Insider Columnist among his other projects, Leonard has authored numerous books on space exploration, Mars missions and more, with his latest being "Moon Rush: The New Space Race" published in 2019 by National Geographic. He also wrote "Mars: Our Future on the Red Planet" released in 2016 by National Geographic. Leonard has served as a correspondent for SpaceNews, Scientific American and Aerospace America for the AIAA. He was received many awards, including the first Ordway Award for Sustained Excellence in Spaceflight History in 2015 at the AAS Wernher von Braun Memorial Symposium. You can find out Leonard's latest project at his website and on Twitter.
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On January 13, 2017, Jody HiGHROLLER, aka, Peach Panther, aka, RiFF RaFF, dropped by Full Sail for an hour long interview. RiFF RaFF claims he got his start in the music business because of his entrepreneurial childhood, work ethic, business mind-set, and limitless, let's do it now, mindset.
Because of RiFF RaFF's limitless mind-set, he has delved into many projects. These projects include: Freestyle mixtapes, studio quality recordings, reality shows, being a guest on a comedy show, directing and creating his own music video, creating a movie that will go to theaters, creating 100 YouTube videos in one year, and modeling fashion lines.


Description en-US: Plan and conduct an investigation to provide evidence of the effects of balanced and unbalanced forces on the motion of an object.
Comment en-US: Clarification Statement: Examples could include an unbalanced force on one side of a ball can make it start moving; and, balanced forces pushing on a box from both sides will not produce any motion at all.
Comment en-US: Assessment Boundary: Assessment is limited to one variable at a time: number, size, or direction of forces. Assessment does not include quantitative force size, only qualitative and relative. Assessment is limited to gravity being addressed as a force that pulls objects down.