text:
\section{Detecting and Locating a Change-in-Mean}
\label{ch3-sec:c-in-mean}

\subsection{Likelihood-Ratio Based Tests}

\label{ch3-subsec:LR}

We start by considering arguably the simplest, but also most widely used, change-in-mean problem. For this we assume data is generated via a signal plus noise model,
\begin{equation} \label{eq:univ_c-in-mean}
X_i = f_i + \varepsilon_i,\quad i = 1, \ldots, n,
\end{equation}
where the noise vector $\bm{\varepsilon} = (\varepsilon_1, \ldots, \varepsilon_n)^T$ consists of independent normal random variables, each with mean 0 and variance $\sigma^2$. We are interested in testing whether the signal is constant, that is $f_1=f_2=\cdots=f_n$, or whether there is a single change-point, $\tau \in \{1,\ldots,n-1\}$, such that
\begin{equation*}
f_1=\cdots=f_{\tau}\neq f_{\tau+1}=\cdots=f_n.
\end{equation*}

Initially assume that the noise variance, $\sigma^2$ is known. If we were interested in testing for a change at a single position, $\tau$, it would be straightforward to use a likelihood-ratio test which compares the maximum of the likelihood for a model with a change at $\tau$ to the maximum of the likelihood for a model with no change:
\begin{eqnarray*}
LR_\tau &=& 2 \left[ \max_{f',g} \frac{-1}{2\sigma^2} \left\{\sum_{i=1}^\tau (X_i-f')^2  +\sum_{i=\tau+1}^n (X_i-g)^2 \right\}  - \max_{f'} \frac{-1}{2\sigma^2}\sum_{i=1}^n (X_i-f')^2
\right]  \\
&=& \frac{1}{\sigma^2} \left[  \min_{f'} \sum_{i=1}^n (X_i-f')^2 -
\min_{f',g} \left\{\sum_{i=1}^\tau (X_i-f')^2  +\sum_{i=\tau+1}^n (X_i-g)^2 \right\}
\right] \\
&=& \frac{1}{\sigma^2} \left[ \sum_{i=1}^n \left(X_i-\bar{X}_{1:n} \right)^2 - \sum_{i=1}^\tau (X_i-\bar{X}_{1:\tau} )^2 - \sum_{i=\tau+1}^n (X_i-\bar{X}_{\tau+1:n} )^2
\right],
\end{eqnarray*}
where we use the notation $\bar{X}_{s:t}$ for $t\geq s$ to be the sample mean of $\bm{X}_{s:t}$:
\begin{equation*}
 \bar{X}_{s:t}=\frac{1}{t-s+1} \sum_{i=s}^t X_i.
\end{equation*}
This likelihood-ratio test statistic can be re-written as $LR_{\tau}=C_{\tau}^2/\sigma^2$, where $C_\tau$ is the so-called CUSUM statistic
\[
 C_\tau = \sqrt{\frac{\tau(n-\tau)}{n}} \left| \bar{X}_{1:\tau} -\bar{X}_{(\tau+1):n}
 \right|.
\]
The CUSUM statistic just compares the sample mean before and after the putative change-point, $\tau$, and re-scales this difference so that $C_{\tau}$ is the absolute value of a random variable that has variance 1.

The likelihood-ratio test would infer a change at $\tau$ if $LR_\tau>c$ for some suitably chosen threshold $c$. The value of $c$ will determine the significance level of the test. The link between the likelihood-ratio test statistic and the CUSUM statistic shows that such a test makes intuitive sense: as the evidence for a change is monotonically increasing in the difference in the sample mean of the data before and after $\tau$. Also, if our model assumptions are correct, then if there is no change, the CUSUM statistic is trivially seen to be the absolute value of a normal random variable with mean 0 and variance $\sigma^2$, and thus the likelihood-ratio statistic has a chi-squared distribution with 1 degree of freedom.

In practice we do not know where the change-point, $\tau$, would be. The natural extension of the likelihood-ratio test to this situation is to use as a test statistic the maximum of $LR_\tau$ as we vary $\tau$:
\[
 LR=\max_{\tau\in\{1,\ldots,n-1\}} LR_\tau.
\]
Again we will detect a change-point if $LR>c$ for some suitably chosen value $c$, and the choice of $c$ will determine the significance level of the test. Such a test is equivalent to one that is based on the maximum of the CUSUM statistics, $C_{\tau}$. If $LR>c$ then we can estimate the location of the change by
\[
 \hat{\tau}=\arg \max_{\tau\in\{1,\ldots,n-1\}} LR_\tau,
\]
and a simple estimate of the size of change is $\bar{X}_{(\hat{\tau}+1):n}-\bar{X}_{1:\hat{\tau}}$.

\subsection{Properties of the Test}
\label{ch3-subsec-theory}

We now turn to properties of this test for a change in mean. Firstly we will consider the distribution of the test statistic under the null hypothesis, and how that helps us determine a suitable threshold for the test. We then turn to its properties when there is change-point, and look at the power of the test and the accuracy with which we can estimate the location and the size of the change.

As has been stated, for a fixed value of $\tau$, $LR_{\tau}$ has a chi-squared distribution with 1 degree of freedom under our null hypothesis. Our test-statistic for a change is $\max_{\tau} LR_{\tau}$, and the challenge with calculating its distribution comes from the dependencies of $LR_{\tau}$ for different values of $\tau$. Furthermore, the standard regularity conditions for likelihood-ratio test statistics do not apply here, as setting size of the actual change in mean to 0 removes the change-point parameter from the model. As a result, even asymptotically, the distribution of  $\max_{\tau} LR_{\tau}$ under the null hypothesis is not chi-squared.

To get the asymptotic distribution under the null we can use the fact that $LR_{\tau}=C_{\tau}^2/\sigma^2$, and that $(C_1,\ldots,C_{n-1})/\sigma$ are the absolute values of a Gaussian process with mean 0 and known covariance. The maximum of a set of Gaussian random variables is known to converge to a Gumbel distribution,  see Theorem 2.1 of \cite{Yao/Davis:1986} \cite[see also][]{gombay1990asymptotic}, this gives
\begin{equation} \label{eq:ch3-gumbel}
\lim_{n \rightarrow \infty} \Pr\left\{  a_n^{-1}(\max_{\tau} C_\tau/\sigma - b_n)\leq u\right\}=\exp\left\{-2\pi^{-1/2}\exp(-u)\right\},
\end{equation}
where $a_n=(2\log \log n)^{-1/2}$ and $b_n=a_n^{-1}+0.5a_n \log \log \log n$. This suggests that the threshold for $LR_{\tau}$ should increase with $n$  like $2\log \log n$.

The rate of convergence of $\max C_{\tau}$ to a Gumbel distribution is slow, and the threshold suggested from this asymptotic distribution can be conservative in practice. A better approximation of the distribution
of $\max LR_{\tau}$ is given by \cite{Yao/Davis:1986}; and \cite{hawkins1977testing} shows how, for any $n$, we can calculate the distribution function for the test statistic numerically. In practice it is often simplest to use Monte Carlo methods to approximate the null distribution of the test statistic, as we show below. Importantly the Monte Carlo approach can be easily applied to more complicated change-point scenarios.

Alternatively,  we can get a bound on the tail probability of the test statistic under the null, and hence a conservative threshold for the test, using a simple Bonferroni argument:
\[
 \Pr( \max_{\tau} LR_{\tau}>c) \leq \sum_{\tau=1}^{n-1} \Pr(LR_\tau>c)= (n-1) \Pr(\chi^2_1>c).
\]
Tail bounds for a $\chi^2_1$ random variable then give that choosing $c=2\log n$ will mean that as $n\rightarrow \infty$ the right-hand side will tend to 0. Thus a threshold of $2 \log n$ for the likelihood ratio statistic will ensure that asymptotically the probability of incorrectly detecting a change-point will tend to 0. Whilst not needed in the single change-point setting, as we have more accurate methods for calculating or bounding the tail probability of the test, this approach will be used frequently in the multiple change-point setting. The bound can be quite loose as it ignores the strong dependence between $LR_{\tau}$ for different values of $\tau$ \cite[the conservative nature of the bound follows from Slepian's lemma, see][]{slepian1962one}. We can see the impact of this by comparing with the thresholds suggested by the asymptotic Gumbel distribution for $\max C_\tau$, which suggested a threshold increasing like $\log \log n$ rather than $\log n$.

To demonstrate the difference in these thresholds, we can compare them to estimates of the quantiles of the null distribution obtained by simulation. For this we need a function that calculates the likelihood ratio statistic. A simple function, for the case where $\sigma=1$, in \texttt{R} is the following

\begin{verbatim}
LR <- function(x){   #input is data vector length >=2
   S <- cumsum(x)   #calculate cummulate sum of data
   n <- length(x)   #number of data points
   tau <- 1:(n-1)   #possible change-point locations to test
   D <- S[tau]/tau-(S[n]-S[tau])/(n-tau)   #difference in means
   LR <- D^2*tau*(n-tau)/n   #LR statistic at locations tau

#return LR statistic and estimate of tau
   return(list(LR=max(LR),tau.hat=which.max(LR)))
}
\end{verbatim}

We would like to emphasise one important computational aspect of calculating the likelihood-ratio statistics. To calculate $LR_{\tau}$ for one specific value of $\tau$ is an $O(n)$ computation as it involves summing the data before and after $\tau$. A naive approach to calculating $LR_{\tau}$ for all $n-1$ possible change-point locations would thus be an $O(n^2)$ computation. However we can calculate $LR_{\tau}$ for all values of $\tau$ in $O(n)$ computation by first storing a set of appropriate summaries of the data. In this case these summaries are just the cumulative sums of the data, $\sum_{i=1}^t X_i$, for $t=1,\ldots,n$. The key point is that once these are calculated and stored, calculating $LR_{\tau}$ for any $\tau$ is just an $O(1)$ calculation. Thus we have an algorithm with just an $O(n)$ cost, for both calculating the cumulative sums and all the $LR_{\tau}$ values, and an $O(n)$ storage. Ideas like this are commonplace in the change-point algorithms we will consider throughout the book, and are important to ensure they scale well with the amount of data.

To find the distribution of  $\max_\tau LR_{\tau}$ under the null hypothesis for any given $n$, we just need to repeatedly simulate IID Gaussian data with mean 0 and variance 1, and use the above function to calculate the  likelihood-ratio statistic for each data set.  If we assume the variance of the data is known, then the test statistic is invariant to the choice of mean or variance, so the distribution of the statistic we obtain is consistent with any data set simulated under our null hypothesis. We can estimate the quantile of the null distribution for a given $n$ by the empirical quantile of the statistic from the set of simulated data. Figure \ref{Fig:ch3-1}(a) compares these quantiles with those suggested by the asymptotic null distribution (\ref{eq:ch3-gumbel}) and with the $2\log n$ threshold suggested by the Bonferroni correction for different values of $n$. There are a couple of important points to draw from this figure. First, the asymptotic null distribution is highly conservative -- with thresholds that are roughly twice what they should be. Second, the true thresholds increase only very slowly with $n$ -- and much more slowly than the $2\log n$ threshold, highlighting the impact of the strong dependence in the series of $LR_{\tau}$ values.

\begin{figure}
\centering
\includegraphics[scale=.65]{figures/ch3-Fig1.pdf}
\caption{ (a) Comparison of thresholds for the likelihood-ratio test for different values of $n$. Solid lines are from the Monte Carlo approximation of the null distribution and dashed lines are from the Gumbel asymptotic distribution; blue dotted line is the $2 \log n$ threshold. For the Monte Carlo and asymptotic distributions, we compare 5\% (black) 1\% (red) and 0.1\% (green) quantiles. (b) Estimated change-point locations (as a proportion of the time series length, $\hat{\tau}/n$) for false positives. Results are shown for data of length $10,000$.}
\label{Fig:ch3-1}
\end{figure}

One further important aspect of the behaviour of the test when there is no change is, if we do detect a change, the distribution of its location is heavily skewed towards estimates near the beginning or end of the data. This can be seen empirically in Figure \ref{Fig:ch3-1}(b). The reason for this comes from the fact that likelihood ratio statistic values for $\tau$ close to the middle of the data are much more heavily dependent than values for $\tau$ near either the beginning or end of the data. The practical implications of this are that, for applications where we know or are willing to assume a minimum segment length, we can substantially improve the accuracy of our test by using that information. For example, if we know that segments have to contain at least $k$ time-points, then our likelihood ratio test-statistic becomes $\max_{k\leq \tau\leq n-k} LR_{\tau}$. Figure \ref{Fig:ch3-2}(a) shows the reduction in threshold for the test that can be obtained as we increase the minimum segment length.

\begin{figure}
\centering
\includegraphics[scale=.65]{figures/ch3-Fig2.pdf}
\caption{ (a) Comparison of thresholds for the likelihood-ratio test for different minimum segment lengths (as proportion of the amount of data). Thresholds calculated by Monte Carlo for a false positive rate of 0.01, results for $n=100$ (black), $n=1000$ (red) and $n=1000$ (green). (b) Lower-bound on the power of a test for a change as a function of the size of change. Results for a test with false positive rate of 0.01, with $n=1000$ and the change in the middle of the data.}
\label{Fig:ch3-2}
\end{figure}

We now turn to the power of the test. We will assume we perform the test so that we detect a change-point if $\max_{\tau} LR_{\tau}>k$.
We will denote the true change-point location as $\tau^0$, and the change in mean to be $\Delta=|f_1-f_n|$. To make clear the dependence of power on sample size, $n$, it is helpful to let $q^0=\tau^0/n$. Using the fact that the likelihood-ratio statistic for a fixed change location is the square of a Gaussian random-variable, it is straightforward to show that the distribution of $LR_{\tau^0}$ is $\chi^2_1(\nu)$ with non-centrality parameter
\begin{equation}
\label{ch3-eq:non-central}
 \nu=\left(\frac{1}{q^0}+\frac{1}{1-q^0}\right)^{-1}n\Delta^2.
\end{equation}
If $\nu>k -1$ then the probability of detecting a change is bounded below by
\begin{equation} \label{ch3-eq:power-simple}
 \Pr(\chi^2_1(\nu)>k) \geq 1-\exp\left\{- \frac{(1+\nu-k)^2}{4+8\nu} \right\},
\end{equation}
where the right-hand side comes from standard tail bounds \cite[e.g Lemma 1 of][]{laurent2000adaptive}. This immediately gives a lower bound on the power of our test, as $\max LR_{\tau} \geq LR_{\tau^0}$.

There are a couple of simple conclusions that can be drawn from this bound on the power of the test. Firstly, if we fix $\Delta$ and $q_0$ then $\nu$ is linear in $n$. As such the right-hand side tends to 1 as $n\rightarrow\infty$ for any choice of $k$ that increases slower than linearly in $n$. So e.g. for $k=2\log n$ the test will have power tending to 1, as well as the probability of a false positive tending to 0. Secondly the power of the test depends primarily on $\nu$, and this is linear in the number of data points but quadratic in the size of change. Thus we would need roughly four times as much data to detect a change that is half as big.
\cite[This is consistent with results, such as Theorem 1 of][which suggest the ease of detecting a change is proportional to the size of the change times the square-root of the segment around it]{chan2013detection}. Also from the definition of $\nu$ we see that the power depends on the location of the change, with changes nearer the boundary of the data harder to detect. This is not surprising, as for such change-points there is much less data to estimate the mean of the data for the shorter segment. Finally, Figure \ref{Fig:ch3-2}(b) shows a lower bound for the power of the likelihood ratio test for a change as a function of the size of the change -- and we can see we move quickly from changes that are hard to detect ($\Delta\leq 0.2$ for a change in the middle of 1000 observations), to ones where the change is easy to detect ($\Delta\geq 0.4$).

The next natural question is, if we detect a change, how accurate will our estimate of the location be. This has been studied for likelihood-ratio tests in general since at least \cite{Hinkley:1970}. The results are particularly simple for the Gaussian model we consider \cite[]{Yao/Davis:1986}.
Define
\[
\mu(q)=\left\{
\begin{array}{cl}
q(1-q^0), & ~~ \mbox{for } q\leq q^0, \\
(1-q)q^0, & ~~ \mbox{for } q>q^0.
\end{array}
\right.
\]
Now given a Brownian bridge, $W_0(t)$; i.e. a Brownian motion on $0\leq t \leq 1$ conditioned so that $W_0(0)=W_0(1)=0$, we can define a continuous-time process for $0<t<1$ as
\begin{equation} \label{eq:ScaledBrownianBridge}
\tilde{W}_0(t)=
\frac{1}{\sqrt{t(1-t)}} \left(\sqrt{n}\Delta\mu(t)+W_0(t)\right)
\end{equation}
It is straightforward to show that the set of CUSUM statistics $\{C_1,\ldots,C_{n-1}\}$ have the same distribution as $\{|\tilde{W}_0(1/n)|,\ldots,|\tilde{W}_0((n-1)/n)| \}$. The estimate of the location is then dependent on the location of the maximum of $|\tilde{W}_0(i/n)|$, conditional on this maximum being above our chosen threshold.

\begin{figure}
\centering
\includegraphics[scale=.65]{figures/ch3-Fig3.pdf}
\caption{ Realisations of the CUSUM statistics and change-point estimates in the presence of a change-point, with $q^0=0.4$, for different values of $n$ and $\Delta$: $n=100$, $\Delta=1$ (top); $n=400$, $\Delta=1$ (middle) and $n=100$, $\Delta=2$ (bottom). Left-hand column plots the CUSUM statistics against re-scaled time, $t/n$; middle-column plots the CUSUM statistics in a region within 20 time-points of the change, $\tau^0=nq^0$; right-hand column shows a histogram of the location of the maximum of the CUSUM statistic, i.e. the estimated change-point location, $\hat{\tau}$ (we use results for all data sets, not just those where the max-CUSUM statistic is above a threshold). For the left and middle columns we show the theoretical mean of the CUSUM statistics in black and 25 realisations in grey.}
\label{Fig:ch3-3}
\end{figure}

We plot realisation of $\{C_1,\ldots,C_{n-1}\}$ and the associated estimated change-point locations for three different choices of $n$ and $\Delta$ in Figure \ref{Fig:ch3-3}. The top two rows are for the same value of $\Delta$ but for differing $n$. If we look at the plot of the CUSUM statistics, $C_t$, against re-scaled time, $t/n$, we can see that increasing $n$ affects the mean of the CUSUM statistics, but does not affect the variability around the mean. We can see that increasing $n$ would increase the power to detect a change, as the maximum value of the CUSUM statistics are substantially greater, and increase the accuracy of the estimate of the change-point location in the re-scaled time, as the mean function is more peaked around $q^0$ but the variability is the same. However if we look at the behaviour in actual time, see middle column of Figure \ref{Fig:ch3-3}, the effect of $n$ on the change-point accuracy is unclear. In fact, empirical results for the distribution of the estimated change-point, $\hat{\tau}$, show similar uncertainty in the location of the change-point on the original time-scale. The bottom row shows results with a larger $\Delta$ but the smaller value of $n$. These have been chosen so that $n\Delta^2$ are identical for the bottom two rows. The plots of the CUSUM statistics against re-scaled time in this case are almost identical -- which is as would be suggested by the distributional result given above, as they have the same value of drift in (\ref{eq:ScaledBrownianBridge}). However this corresponds to much more accurate estimates of the change-point locations when we consider the original time-scale.

The qualitative feature to take from these results, which can be shown by considering the behaviour of the CUSUM statistics under the alternative as given in (\ref{eq:ScaledBrownianBridge}), is that the error in the location of the change-point is $O_p(1)$ if we fix $\Delta$ and $q^0$ and increase $n$; but is $O_p(1/\Delta)$ if we increase $\Delta$. What does this mean in practice? Firstly, the accuracy with which we estimate the location of a change-point depends primarily on the size of the change. By comparison, whilst we have seen that increasing the sample size $n$ improves the power of our test, it has little effect on the accuracy of our estimate of the location of the change-point. This latter point is intuitive, as it is the data closest to the change-point that is most informative about the location. Similarly, the value of $q^0$ impacts the power of our test, with changes in the middle of the data being easier to detect than those near the beginning or the end, but has little impact on the accuracy of the estimate of the location of any detected change-point.

Finally, we turn to the question of estimating the size of change associated with a detected change-point. Given an estimated change-point location, $\hat{\tau}$ a natural estimator of the size of change is to use the difference in empirical mean of the data before and after the data:
\[
\hat{\Delta}_{\hat{\tau}}= \bar{X}_{\hat{\tau}+1:n}-\bar{X}_{1:\hat{\tau}}.
\]
Whilst natural, this naive estimator is biased towards larger absolute values. There are two related reasons for this. The first is that we only estimate the size of change if we have detected a change-point, and the datasets where we detect a change-point are biased to ones where the empirical difference in means is larger; this is a common feature with estimation after a hypothesis test and is often to referred to as the winner's curse \cite[]{zollner2007overcoming}. Second, we have estimated the change-point location based on the difference in the empirical means before and after each possible location, and thus our choice of $\hat{\tau}$ biases us towards values for which $|\hat{\Delta}_{\hat{\tau}}|$ is larger.

The first bias is due to selection of the data-set, and the second is bias due to selection of location within a data-set. The size of each source of bias will depend on the number of data points and the actual location and size of change. For scenarios with greater power to detect a change, the first source of bias will be reduced. For example, in the extreme case where we would detect the change with probability 1, there would be no biasing due to the fact we only estimate the size of change after we have detected a change. Similarly, the more informative the data is about the location of the change, i.e. the larger the actual value of $\Delta$ is, then less bias there will be from estimating its location.

To see this we looked at the bias in estimating the size of change for three different scenarios. These all had a change in the middle of the data, $q^0=0.5$, but had different values of $n$ and $\Delta$: (i) $n=100$ and $\Delta=1/2$; (ii) $n=400$ and $\Delta=1/2$; and (iii) $n=100$ and $\Delta=1$.  Scenario (i) has low power of detecting a change-point of around $50\%$, and substantial uncertainty over the change-point location. The other scenarios had power over $97\%$. Scenario (ii) should have similar uncertainty over the change-point location as (i), with scenario (iii) have substantially less.

\begin{figure}[t]
\centering
\includegraphics[scale=.65]{figures/ch3-Fig4.pdf}
\caption{Histogram of 10,000 realisations of $\hat{\Delta}_{\hat{\tau}}$ for (i) $n=100$ and $\Delta=1/2$ (left); (ii) $n=400$ and $\Delta=1/2$ (middle); and (iii) $n=100$ and $\Delta=1$ (right).}
\label{Fig:ch3-4}
\end{figure}

Histograms of 10,000 realisation of $\hat{\Delta}_{\hat{\tau}}$ are shown in Figure \ref{Fig:ch3-4}. The empirical results suggest an over-estimation of $\Delta$ by 62\%, 7.4\% and 5.7\% for the three scenarios respectively. By comparing (i) with (ii) and (ii) with (iii) we can see that it is the effect of low power that has most impact on bias, rather than the uncertainty over the location of change.  For each scenario the biasing effects mean that the distribution of $\hat{\Delta}_{\hat{\tau}}$ is skewed towards larger absolute values. The deviation from a Gaussian distribution, which is the distribution of $\hat{\Delta}_{t}$ for any time-point $t$ chosen independently of the data, is particularly evident in scenario (i).

We will return to the problem of estimating the size of change, with the related issue of assessing uncertainty regarding estimated change-points, in Chapter 8.

\subsection{Detecting a Single Change when there are Multiple Change-points}
\label{ch3:sec:single-when-multiple}

An important question, which has implications for some methods for detecting multiple change-points that we will look at in the next chapter, is what happens if we perform a test for a single change in mean when there are multiple changes. In particular, should we expect to detect one of the change-points? Or may the signal from the various changes interfere with each other and bias our estimate towards estimating a change location between the change-points?

\begin{figure}[t]
\centering
\includegraphics[scale=.65]{figures/ch3-Fig-multiple.pdf}
\caption{Left plot: example data set (black line) simulated with multiple change-points, with true mean shown (red line). Middle and right plots: log likelihood ratio statistics, $LR_{\tau}$ as we vary $\tau$ (black line), the mean of $LR_{\tau}$ (red line) and change-point locations (blue dashed lines), for analysing the full data (middle) and just the data in the first two segments (right).}
\label{Fig:ch3-mult}
\end{figure}

To give some indication of the answer to this question we will consider an empirical example -- though the ideas we present can be formalised \cite[]{venkatraman1993consistency}. In Figure \ref{Fig:ch3-mult} we display some simulated data from a signal that is piecewise constant with three change-points. We also show the log-likelihood ratio statistic, $LR_{\tau}$, and its expected value given the true piecewise constant mean of the data. The key property to notice from this plot is that all local maxima, and hence the global maximum, of the expected value of $LR_{\tau}$ are at values of $\tau$ that correspond to true change-points. These indicate that the test for a single change-point still has power to detect one of the changes, and that the estimated change-point location should be close to a true change-point. This property of the likelihood ratio test statistic, and equivalently the CUSUM statistic, does not necessarily hold for other test statistics, see for example \cite{venkatraman1993consistency}.

So is anything lost by analysing data with multiple changes using a test for a single change-point? We can see that some information is potentially lost by comparing the middle-plot with the right-hand plot of  Figure \ref{Fig:ch3-mult}. The latter shows $LR_\tau$ and its expected value if we just analyse data from the first two segments. The signal for the presence of the first change-point -- which is the change-point that our test had most power to detect -- is lower when we analyse the full data than if we just analyse the data before the second change. This is not surprising, but it does highlight that we can increase the power for detecting a change-point if we can correctly estimate the location of other change-points -- or equivalently that there may be gains in statistical efficiency if we either try to estimate the location of all change-points jointly, rather than one at a time; or if we use the local nature of the signal for a change-point. We will see both these ideas being used in some of the methods for detecting multiple changes that we describe in the next chapter.

\subsection{Unknown Variance}

If the noise variance, $\sigma^2$, is unknown, we can still use a likelihood-ratio based test, which is constructed in the same way. The only difference comes when maximising the log-likelihood under our models of no change-point or a single change-point. In each case we need to also maximise over $\sigma$. Standard calculations give
\[
 {LR}_\tau=n \log \left\{ \frac{\sum_{i=1}^n (X_i-\bar{X}_{1:n})^2}{\sum_{i=1}^\tau (X_i-\bar{X}_{1:\tau})^2+\sum_{i=\tau+1}^n (X_i-\bar{X}_{\tau+1:n})^2} \right\}
\]
Importantly this test-statistic is still a monotonic function of the CUSUM statistic, as it can be re-written as
\begin{equation} \label{ch3:eq-LR-unknown-var}
 LR_\tau= n \log \left\{ \frac{S^2}{S^2-C_\tau^2} \right\},
\end{equation}
where $S^2=\sum_{i=1}^n (X_i-\bar{X}_{1:n})^2$ is the residual sum of squares under a model with no change, and $C_{\tau}$ is the CUSUM statistic. Thus the likelihood-ratio test for a change is still equivalent to one based on $\max_{\tau} C_{\tau}$, with the only difference being the distribution of the max-CUSUM statistic under the null.

An alternative approach is to estimate $\sigma^2$ and then proceed as per the known $\sigma^2$ approach plugging-in our estimate. As would be expected, for large $n$
such an approach will be very similar to using the likelihood-ratio based test when for the model with $\sigma^2$ unknown. To see this, consider the natural estimate of $\sigma^2$ based on the mean of the residual sum of squares under a model with a change-point at $\tau$, $\hat{\sigma}^2=(S^2-C_\tau^2)/n$. We can re-write  (\ref{ch3:eq-LR-unknown-var}) as
\[
LR_\tau= n \log \left\{ \frac{n\hat{\sigma}^2+C_\tau^2}{n\hat{\sigma}^2} \right\}
= n \log\left\{1+\frac{C_\tau^2}{n\hat{\sigma}^2} \right\},
\]
and a Taylor expansion then gives $LR_\tau=C_\tau^2/\hat{\sigma}^2+O_p(1/n)$. That is the likelihood-ratio test statistic under a model where the common noise variance is unknown is equal to the test statistic where we assume we know the variance and set it equal to our plug-in estimate, $\hat{\sigma}^2$, plus terms that are of order $1/n$. Thus, whilst we would expect using a plug-in estimator to be less accurate, in practice the difference will be negligible if $n$ is large.

In the multiple change-point setting, it is more common to use plug-in estimators of unknown common parameters, such as the variance of the noise in a change-in-mean problem. One approach is to estimate these under a fitted model as we have done above, but this can be computationally expensive given the large number of plausible multiple change-point models for a data set. The other, simpler, approach is to use a simple pre-processing step. Here, we will introduce the simplest and most common approach to estimate the noise variance, and leave further discussion to some of the applications we consider in later chapters.

One approach to estimate $\sigma^2$ in our change-in-mean model is to consider the first difference of the data, $Z_t=X_{t+1}-X_t$, for $t=1,\ldots,n-1$. Under our IID Gaussian model we have that $Z_t$ is Gaussian with variance $2\sigma^2$ and mean 0 provided that there is no change in mean between time $t$ and $t+1$. So we can estimate $2\sigma^2$ using a robust estimate of the variance of $Z_t$. One commonly used robust estimator is the median absolute deviation from the median, or MAD, estimator.

Such an estimator is simple, but can suffer from two issues. First, the use of a robust estimator can lead to some loss of statistical efficiency in estimating the variance. More important is that any approach based on estimating the variance of $Z_t$ will be sensitive to violations of the assumption that the noise is independent. If the noise is auto-correlated, and we define $\rho_1$ to be the lag-1 auto-correlation of $X_{1:n}$, then the variance of $Z_t$ is $2(1-\rho_1)\sigma^2$, which means that any estimate of $\sigma^2$ based on estimating the variance of $Z_{1:n-1}$ will be biased if $\rho_1\neq 0$. Furthermore, if there is positive auto-correlation, $\rho_1>0$, then we will under-estimate $\sigma^2$, which in turn will lead to using a smaller threshold for the CUSUM statistic in our test for a change. This is particularly problematic because, as we will see below, the presence of positive auto-correlated noise means that we want to use a larger threshold for a CUSUM test if we wish to maintain a given significance level for that test.

\subsection{Impact of Model Error}

So far we have developed a test for a change-point in the mean of data where the additive noise is IID Gaussian, and studied its properties when these modelling assumptions hold. It is important to consider what happens when these assumptions are incorrect. In Section \ref{S:ch3-extensions} we will look at how we can construct alternative tests that make different assumptions for the data. Here we will focus on using the likelihood ratio, or equivalently the CUSUM, test that we have derived for data where either the additive noise is not independent or non-Gaussian.

The intuition behind how our test behaves is most clearly seen if we focus on the CUSUM version of the test. First consider the case where the additive noise is stationary, marginally it is Gaussian with mean 0 and variance $\sigma^2$, but it is not independent. We will focus on noise models where the covariance of $\varepsilon_{t}$ and $\varepsilon_{t+h}$ only depends on $h\geq0$, and we will write this as $\sigma^2 \rho_{h}$, and to simplify the exposition assume $\rho_h\geq0$ for all $h$. The main effect of the dependencies in this noise process is that it changes the marginal variance of the CUSUM statistic. In particular
\[
\mbox{Var}(\bar{X}_{t+1:t+m}) = \frac{\sigma^2}{m^2} \sum_{i=1}^m \sum_{j=1}^m \rho_{|i-j|} \leq \frac{\sigma^2}{m} \left(1+2\sum_{h=1}^\infty \rho_h\right).
\]
Furthermore if the dependencies are short-range and $m$ is larger relative to this range, then this upper bound will be a reasonably approximation of the actual variance of the sample mean of data in the segment from $t+1$ to $t+m$.  This suggests that if we ignore the dependencies, the marginal variances of our CUSUM statistics will be inflated by a factor of approximately $1+2\sum_{h=1}^\infty \rho_h$, which will lead to an inflation of the false-error rate of the test. Moreover, we could approximately counteract this effect by inflating the threshold of our test by a factor that is $\sqrt{1+2\sum_{h=1}^\infty \rho_h}$; though the challenge in practice is then to estimate this factor.

\begin{figure}[t]
\centering
\includegraphics[scale=.65]{figures/ch3-Fig5.pdf}
\caption{False positive rate for the CUSUM test in the presence of AR(1) noise. Black line shows the false positive rate using a test threshold that gives a 5\% error rate for IID Gaussian noise. Red dashed lines show false positive rate with inflated threshold as described in the text. Dotted line is the nominal 5\% false error rate.}
\label{Fig:ch3-5}
\end{figure}

To see this in practice we simulated data without a change, but with the noise process being an AR(1) process for differing values of $\rho_1$. Figure \ref{Fig:ch3-5} shows the frequency with which we incorrectly detect change when we apply the test with an appropriate threshold for our test under the assumption of independence, and when we apply the test with the threshold inflated by the true value
\[
\left(1+2\sum_{h=1}^\infty \rho_h\right)^{1/2}=\left( \frac{1+\rho_1}{1-\rho_1} \right)^{1/2}
\]
Ignoring the dependence leads to a substantial increase in the false error rate as we increase $\rho_1$. Inflating the test threshold controls for this, and keeps the empirical false error rate at or below the nominal level. For $\rho_1\approx 1$ we see that the inflation leads to a slightly conservative test. This is because our inflation factor is an upper bound, which is accurate for CUSUM statistics where both segments are sufficiently long relative to the range of the dependencies in the noise process. As $\rho_1$ increases the dependencies are non-negligible over a longer range, and hence our upper bound is less accurate for a slightly higher proportion of the CUSUM statistics.

Whilst this suggests a simple procedure for making our change-in-mean test robust to correlated noise, in terms of controlling the false-positive rate, the fact that our test is designed based on an assumption of independence means that it is likely to lose power, particularly when there is high auto-correlation in the noise. We will investigate this further in Section \ref{S:ch3-extensions} when we will consider how to design tests under different modelling assumptions.

Now consider the case where the noise is non-Gaussian. We will consider the case of IID noise which is heavy tailed, but is still mean 0 and has variance $\sigma^2$. If we consider the signed CUSUM statistics,
\[
C^*_t=\sqrt{\frac{t(n-t)}{n}} (\bar{X}_{1:t}-\bar{X}_{t+1:n} ),
\]
so that $C_t=|C^*_t|$, then as these are linear functions of the data the mean and covariance of $(C^*_1,\ldots,C^*_{n-1})$ will be the same as for Gaussian noise. Furthermore, if $t$ and $n-t$ are ``large enough'' then we would expect, by the central limit theorem, that the distributions of $\bar{X}_{1:t}$ and $\bar{X}_{t+1:n}$ to be approximately Gaussian.

Intuitively this suggests that the impact of heavy-tailed noise will solely be to make the CUSUM statistics heavier tailed, and that this will particularly affect CUSUM statistics for change-points near the beginning or end of the data set. If $n$ is large, the impact on the CUSUM statistics for change-points near  the middle of the data will be negligible. If the noise is not too heavy tailed, this argument can be made rigorous if we impose a minimum segment length, $l_n$ say, and consider an asymptotic regime where $l_n\rightarrow \infty$ as $n\rightarrow \infty$ but $l_n=o(n)$. For this regime, the CUSUM statistics will converge to the modulus of the same Gaussian process as for the Gaussian noise case.

\begin{figure}[t]
\centering
\includegraphics[scale=.65]{figures/ch3-Fig6.pdf}
\caption{Estimated change-points locations for data with no change-point. Histogram for data simulated from a $t_5$ distribution (light-blue) overlaid by Gaussian data (pink). Each histogram shows estimates only for data where a change was detected, but frequency is calculated across all data sets. Left-hand plot is for standard CUSUM test, and right-hand plot for a CUSUM test with a minimum segment length of 25. }
\label{Fig:ch3-6}
\end{figure}

To demonstrate the effect of heavy-tailed noise in practice, we applied the CUSUM test to data with IID student-t distributed noise with 5 degrees of freedom. A comparison of the frequency of false-positive estimates of different values of $\hat{\tau}$ are shown in Figure \ref{Fig:ch3-6}. The key message is that heavy-tailed noise inflates the false rates, and this is primarily through an increase in estimated change-points near the start or end of the data. Introducing a minimum segment length reduces this increase in false-positive rate.

\section{Extensions} \label{S:ch3-extensions}

What if we are interested in detecting a change other than a change in mean? Or we wish to detect a change in mean, but to also impose some additional modelling assumptions on the data? It turns out that the general framework we have used can be applied to most change detection problems where we are willing to specify a parametric model to the data within each segment. In these cases we can first calculate the log-likelihood ratio test statistic conditional on a specific location for the change-point, which we will denote as before by $LR_\tau$, and then use $\max_\tau LR_\tau$ as our test statistic for a change. If we detect a change, then we will estimate its location by $\arg \max LR_\tau$.

We will demonstrate this general procedure through considering a few examples. These attempt to cover some of the more common types of change that we may be required to detect, but also to draw out some of the similarities and differences with the change in mean scenario we have covered so far. But first we make some general informal comments on the properties of $\arg \max LR_\tau$ under the null distribution of no change.

If we are detecting a change in  a single parameter, that is conditional on a known change-point location $\tau$ the alternative for our  test has one additional parameter than the null, then (subject to standard regularity conditions) the distribution of $LR_\tau$ will be approximately chi-squared with one degree of freedom if $\tau$ and $n-\tau$ are reasonably large. This suggests that marginally the null distribution of $LR_\tau$ will be very similar to that for the change in mean problem providing $\tau$ is not close to the start or end of the data.

Thus differences in the null distribution of $\arg \max LR_\tau$ as compared to the change in mean case, will either be due to behaviour of $LR_\tau$ near the start or end of the data; differences in the dependence structure of $LR_\tau$ as we vary $\tau$; or due to detecting changes where more than one parameter changes at a change-point. We will see each of these in the following examples. In practice, in these and other situations, one can use simulation from the null to determine appropriate thresholds for the tests.

\subsection{Change in Mean for Count Data}

For count data with a Poisson model our null hypothesis is that we have IID data
\[
X_i \sim \mbox{Poiss}(\theta_1),~~~ i=1,\ldots,n;
\]
whereas the alternative is that there exists some $1\leq \tau < n$ such that we have independent data with
\[
X_i \sim \mbox{Poiss}(\theta_1),~~~ i=1,\ldots,\tau;~~~X_i \sim \mbox{Poiss}(\theta_2),~~~ i=\tau+1,\ldots,n,
\]
where $\theta_2\neq\theta_1$

The log-likelihood ratio statistic for this model is easily shown to be
\[
LR_\tau = 2\left( \tau\bar{X}_{1:\tau}\log \bar{X}_{1:\tau} + (n-\tau)\bar{X}_{(\tau+1):n}\log \bar{X}_{(\tau+1):n}
-n\bar{X}_{1:n}\log \bar{X}_{1:n}
\right),
\]
where to simplify notation we use the convention that $u\log u=0$ if $u=0$.

An alternative approach to detecting a change in mean in Poisson data, is to use a variance stabilising transformation, for example the Anscombe transform, $\tilde{X}_t=2\sqrt{X_t+3/8}$ and then analyse the transformed data using our earlier test for a change in mean in Gaussian data. These two approaches are closely related, as can be seen by the left-hand plot of Figure \ref{Fig:ch3-Poisson}. Here we show, for one data set simulated under a Poisson model with constant mean of 0.1, the values of $LR_{\tau}$ for the test under the Poisson model, and the test applied to the transformed data under the Gaussian model. For each value of $\tau$ the two test statistics are very similar, and we only see substantial differences for $\tau$ close to either the beginning or end of the data. As can be seen in this example, in general the Gaussian test is more variable in these cases, which means that the threshold for the test statistic $\max_\tau LR_\tau$ under the Gaussian model will be higher than under the Poisson model. The differences we observe will be smaller if we simulate data under a Poisson model with larger mean, and we can also get a closer link between the Gaussian and Poisson tests if we impose a minimum segment length, as this would exclude the values of $\tau$ for which there tend to be the largest differences.

\begin{figure}[t]
\centering
\includegraphics[scale=.65]{figures/ch3-Poisson.pdf}
\caption{ Left plot shows the likelihood-ratio statistic, $LR_{\tau}$ as a function of putative change-point location, $\tau$, for a change in mean in Poisson data (black line) and a change in mean in Gaussian data (red line) for 1000 independent data points drawn from a Poisson distribution with mean 1. Right plot shows  $\max_{\tau}LR_{\tau}$ for the Poisson model against the Gaussian model for 1000 Poisson data sets of size 1000, each simulated with a change in mean from $0.075$ to $0.125$  at time 500. Blue horizontal and vertical lines show threshold for the Gaussian and Poisson tests that empirically give a $5\%$ false error rate for data simulated with no change.}
\label{Fig:ch3-Poisson}
\end{figure}

The right-hand plot of Figure \ref{Fig:ch3-Poisson} compares the value of the test statistic $\max_{\tau}LR_{\tau}$ for the Poisson model against the Gaussian model for 1000 data sets simulated with a change. In this case the Poisson test is more powerful. Performing the test so that we have a $5\%$ false error rate, we detect a change in 50\% of the data sets using the Poisson test and only 20\% of the data sets using the Gaussian test. The higher power is as expected, as the assumptions underlying the Poisson test are correct. The reason for the increased power is primarily due to the different behaviour of the test statistics under the null hypothesis described above, which means we need a substantially higher threshold for the test if we use the Gaussian test statistic. Interestingly, Figure \ref{Fig:ch3-Poisson} shows that for most data sets the value of $\max_{\tau}LR_{\tau}$ is similar for the two tests, but there are a small number of data sets for which $\max_{\tau}LR_{\tau}$  is much larger for the Gaussian test than the Poisson test. Examining these datasets shows that they correspond to data sets where the value of $\tau$ that maximise $LR_{\tau}$ is close to the start or end of the data. Thus these are being driven by the variability in the test statistic rather than the signal of the change, and the estimated change-point locations in these cases would be very inaccurate.

\subsection{Change in Variance}

A natural model for detecting a change in variance is to assume that the data is independent Gaussian with known mean. Without loss of generality we can assume the mean is zero. So our null hypothesis is
\[
X_i=\theta_1 \varepsilon_i~~~ i=1,\ldots,n;
\]
whereas the alternative is that there exists some $1\leq \tau < n$ such that
\[
X_i = \theta_1\varepsilon_i,~~~ i=1,\ldots,\tau;~~~X_i = \theta_2 \varepsilon_i,~~~ i=\tau+1,\ldots,n,
\]
where $\theta_2\neq\theta_1$ and $\varepsilon_1,\ldots,\varepsilon_n$ is a realisation of  IID standard Gaussian random variables.

The log-likelihood ratio statistic for this model is easily shown to be
\[
LR_{\tau}= \left(
n\log S^2_{1:n}
-\tau\log S^2_{1:\tau} - (n-\tau)\log S^2_{(\tau+1):n}
\right),
\]
where for any integers $t\geq s$, $S^2_{s:t}=\sum_{i=s}^t X_i^2/(t-s+1)$ is the estimate of the variance for data $X_s,\ldots,X_t$, using the assumption that the data is mean 0.

\subsection{Change in Mean and Variance}

Again the most natural or common model for situations where one wishes to detect changes in mean and variance is to assume the data is IID Gaussian
\[
X_i=\mu_1+\theta_1 \varepsilon_i~~~ i=1,\ldots,n;
\]
whereas the alternative is that there exists some $1\leq \tau < n$ such that
\[
X_i = \mu_1+\theta_1\varepsilon_i,~~~ i=1,\ldots,\tau;~~~X_i = \mu_2+\theta_2 \varepsilon_i,~~~ i=\tau+1,\ldots,n,
\]
with $\mu_1\neq\mu_2$ and $\theta_1\neq\theta_2$, and
where $\varepsilon_1,\ldots,\varepsilon_n$ is a realisation of  IID standard Gaussian random variables.

The log-likelihood ratio statistic for this model is closely related to that for the change in variance case:
\[
LR_{\tau}= \left(
n\log S^2_{1:n}
-\tau\log S^2_{1:\tau} - (n-\tau)\log S^2_{(\tau+1):n}
\right),
\]
with the difference being that, for any integers $t\geq s$, $S^2_{s:t}=\sum_{i=s}^t (X_i-\bar{X}_{s:t})^2/(t-s+1)$ is the estimate of the variance for data $X_s,\ldots,X_t$. One complication with detecting a change in variance is that the log-likelihood ratio statistic is only finite if $\tau>1$ and $\tau<n-1$. As, for example if $\tau=1$, we have a segment of length 1, the estimate of the variance for that segment will be 0, and $LR_\tau$ has a $\log 0$ term. Thus our test statistic must impose a minimum segment length of 2, i.e. we use $\max_{1<\tau<n-1} LR_{\tau}$.

\begin{figure}[t]
\centering
\includegraphics[scale=.65]{figures/ch3-MeanVar.pdf}
\caption{ Left, a qq-plot for $LR_{\tau}$ for the change in mean and variance model, against a $\chi^2_2$ random variable for data simulated under the null model with $n=1,000$ and $\tau=2$ (black dots) or $\tau=500$ (red dots). Right plot shows the frequency of different values of $\hat{\tau}$ when applying the test for a change in mean and variance to data simulated under the null hypothesis.}
\label{Fig:ch3-MeanVar}
\end{figure}

Compared to the other examples we have considered so far, one of the main differences in terms of the behaviour of $LR_{\tau}$ under the null is that, if $\tau$ and $n-\tau$ are large, the marginal distribution is approximately chi-squared with 2 degrees of freedom, rather than with 1 degree of freedom. Perhaps more importantly, if $\tau$ or $n-\tau$ is small, this approximation is poor, and $LR_{\tau}$ has much heavier tails. This can be seen in the left-hand plot of Figure \ref{Fig:ch3-MeanVar}. One consequence of this is that if one performs a test for a change in mean and variance, then false positives are likely to correspond to estimates of $\tau=2$ or $\tau=n-2$. This is shown in the right-hand plot of Figure \ref{Fig:ch3-MeanVar}, which shows estimates of $\hat{\tau}$ when applying the test to $n=1000$ data points simulated under the null, with the test having a false-positive probability of 5\%. In this case we see that around 60\% of all $\hat{\tau}$ values are correspond to fitting segments with the shortest possible segment length of 2. Again this is a situation where imposing a larger minimum segment length can substantially improve the statistical performance of the test, at least for detecting changes where the segments are larger than this assumed minimum. One way of seeing this is that the threshold for the test of a change in mean and variance for $n=1000$ and a false error probability of 5\% is reduced from 17.3 to 13.5 if we impose a minimum segment length of 10.

\subsection{Change in Slope} \label{ch3:sec-changeinslope}

An alternative to detecting a change in mean is detecting a change in slope. Here we assume the data is of the form signal plus noise
\[
X_i = f_i + \varepsilon_i,\quad i = 1, \ldots, n,
\]
where the noise vector $\bm{\varepsilon} = (\varepsilon_1, \ldots, \varepsilon_n)^T$ consists of independent normal random variables, each with mean 0 and, for simplicity we will assume known variance. As the variance is known, without loss of generality we can assume it is 1. Our null hypothesis is that the signal is linear,
\[
f_i=\theta_0+i\theta_1,\quad i = 1, \ldots, n,
\]
whereas as the alternative is that there is some change-point $\tau\in\{2,\ldots,n-1\}$ where the slope changes
\[
f_i=\theta_0+i\theta_1,\quad i = 1, \ldots, \tau;\quad\quad f_i=\theta_0+i\theta_1+(i-\tau)\theta_2,\quad i = \tau+1, \ldots, n.
\]
The parameters of the model are the initial intercept, $\theta_0$, and slope $\theta_1$, and the change in slope at $\tau$, $\theta_2$.

For this model the log-likelihood ratio statistic can be written as the square of a projection of the data, i.e. $LR_{\tau}=(v_{\tau}^TX_{1:n})^2$, for some contrast column vector $v_{\tau}$  \cite[see][for more details]{bcf16}. This contrast vector is piecewise linear with a change-in-slope at $\tau$, and is such that under the null $v_{1:n}^TX_{1:n}$ has variance 1, and  $v_{\tau}^TX_{1:n}$ is invariant to adding a linear function to the data. Up to an arbitrary sign, these uniquely define $v_{1:n}$, and are intuitive properties. The first property is linked to the test detecting a change in slope at $\tau$, and the other properties mean that under the null $LR_{\tau}$ will have a chi-square distribution with 1 degree of freedom regardless of the value of $\theta_0$ and $\theta_1$.

The fact that we can write $LR_{\tau}$ as the square of a linear projection of the data is a direct analogue to the fact that for the change in mean case $LR_{\tau}$ can be written as the square of the CUSUM statistic. Like the change in mean case, if the noise is IID Gaussian then the marginal distribution of $LR_{\tau}$ under the null is exactly chi-squared with one degree of freedom. However the distribution of $\max_\tau LR_{\tau}$ is different in the change in slope case as the dependence between $LR_{\tau}$ as we vary $\tau$ is different. Whereas the limiting distribution of $\max_\tau LR_{\tau}$  for the change in mean model could be related to the maximum of the square of a scaled version of a Brownian Bridge process (\ref{eq:ScaledBrownianBridge}), it tends to the maximum of the square of a scaled version of an integrated Brownian Bridge process for the change in slope model. Approximations to the tail probabilities of such a process can be obtained using techniques described in \cite{davies1987hypothesis} and \cite{zheng2019consistency}.

\begin{figure}[t]
\centering
\includegraphics[scale=.65]{figures/ch3-Fig-slope1.pdf}
\caption{Comparison of $LR_{\tau}$ for change-in-slope (black) and change-in-mean (red) against $\tau$ for the same IID standard Gaussian data.}
\label{Fig:ch3-slope1}
\end{figure}

Figure \ref{Fig:ch3-slope1} shows the qualitatively different  dependence structure of our test statistic for the change-in-slope as compared to the test statistic for a change-in-mean. We simulated data from a model that is consistent with the null hypothesis for both tests, that is IID Gaussian data with constant mean, and plot $LR_{\tau}$ for both tests. The key thing to note is that for the change-in-slope test the plot of $LR_\tau$ against $\tau$ is much smoother -- indicative of much stronger correlation as we vary $\tau$.

\begin{figure}[t]
\centering
\includegraphics[scale=.65]{figures/ch3-Fig-slope2.pdf}
\caption{Example data (top row) and plots of $\sqrt{LR_{\tau}}=|v_{\tau}^TX_{1:n}|$ (black line) and $|v_{\tau}^T\mbox{E}(X_{1:n})|$ (red line) against $\tau$ (bottom row) under the change-in-slop model; change-point locations are shown by the vertical blue dashed lines. The left hand column shows an example where the mean (red line in plots in top row) has a single change in slope, and the right hand column shows an example where there are two changes.  }
\label{Fig:ch3-slope2}
\end{figure}

This extra correlation has an impact on the accuracy with which we can estimate the change-point locations. We can see this qualitatively in the left-hand plots of Figure \ref{Fig:ch3-slope2}. In particular if we compare the plot if $|v_{\tau}^TX_{1:n}|$ or $|v_{\tau}^T\mbox{E}(X_{1:n})|$, with the corresponding plots for the equivalent CUSUM statistic in Figure \ref{Fig:ch3-3} we can see that these are much flatter near the change-point location, which corresponds to more uncertainty about the location. The impact of this can be seen on asymptotic results for the accuracy of estimating the location of a change in mean \cite[]{chen2020jump}.

The right-hand plots of Figure \ref{Fig:ch3-slope2} show an example where we try and detect a single change-in-slope for data where there are two changes. Looking the $|v_{\tau}^T\mbox{E}(X_{1:n})|$ shows that this function attains its maximum at a value of $\tau$ that is part way between the two change-points. This is saying that if we observed the mean function without error, we would detect a change in slope but it would be estimated in the wrong place -- in this case at time 500, when the actual change-points are at 400 and 600. This property will be important when we consider approaches to detect multiple change in slopes, and will be revisited in Chapter 5.

\subsection{Change in Mean in Correlated Noise}
\label{S:ch3-correlated}

Our final example revisits the change in mean model (\ref{eq:univ_c-in-mean}) but assumes that the noise process is auto-correlated, and considers how we can incorporate knowledge of the auto-correlation in the test for a change. For simplicity we will focus on the case where it is a stationary Gaussian auto-regressive process of lag 1. We will assume the marginal variance is known, and without loss of generality set it to 1, and that the lag 1 autocorrelation, $\phi$ say, is also known.

Again the log-likelihood ratio statistic can be written as the square of a projection of the data, i.e. $LR_{\tau}=(v_{1:n}^TX_{1:n})^2$, for some contrast column vector $v_{1:n}$. The contrast vector, and its properties, naturally differ from the change in slope case. The contrast vector is such that $v_{1:n}^TX_{1:n}$ is invariant to adding a constant to the data, and $v_{1:n}^T\varepsilon_{1:n}$ has variance 1 under our modelling assumptions -- this again means that under, if in addition the null holds, then the test statistic is chi-squared with 1 degree of freedom regardless of the mean of the data. The contrast vector has the final property that under all vectors that satisfy these two properties, it is the one for which $|\sum_{i=1}^\tau v_i|$ is maximum. This final property means that this is the optimal linear projection, satisfying our two properties, in terms of power for detecting a change in mean at $\tau$ -- see \cite{Romano:2020} for more details.

\begin{figure}[t]
\centering
\includegraphics[scale=.65]{figures/ch3-Fig-ARcomp.pdf}
\caption{ Power (left-hand plot) and accuracy, defined as mean absolute error of the estimated change location, (right-hand plot) for detecting a change-in-mean in AR(1) noise as we vary the lag-1 autocorrelation, $\rho$. Results are for a likelihood-ratio test under the correct model (black full line) and for a test based on the CUSUM test statistic that ignores the correlation (red dashed line). For both method we chose the threshold for the test statistic using Monte Carlo so that it has a 1\% false positive rate under the true model. For each value of $\rho$, results are based on 2000 simulated data sets of length $n=80$ with a change at $\tau=20$ of size 2. }
\label{Fig:ch3-slope3}
\end{figure}

Figure \ref{Fig:ch3-slope3} show results where we compare the likelihood-ratio test for the true model with the CUSUM, or equivalently likelihood-ratio test, under a model which ignores the autocorrelation. For the latter test we inflate the threshold for the test so that it has the required significance level under the actual model used to simulate the data. As we vary $\rho$ we fix the noise to have the same marginal variance, of 1,  the true change-in-mean is of size 2 at time point 20 out of a data set of length 80.

   There are two key messages from these results. Firstly, looking at the performance of the ideal test-statistic we see that the difficulty of detecting and locating the change is hardest for intermediate values of $\rho$, but becomes easier for $\rho$ close to 0 and close to 1 \cite[the latter property is discussed in][]{Romano:2020}. More importantly, by comparing the performance to the performance of the standard CUSUM test, that ignores correlation, we see the two tests perform almost identically unless $\rho$ is very large -- around $0.7$ or higher. For higher values of $\rho$, ignoring correlation leads to a substantial loss of power and accuracy. However in many applications, whilst there will be auto-correlation in the noise it will be at the sort of level for which there is little or no power lost if it is ignored and the standard CUSUM test is used; though care is needed in such situations to choose the threshold for the CUSUM test appropriately to account for the level of correlation that is present.

\section{Bibliographical notes}

Whilst the sequential, or online, detection of  change-points dates back to at least \cite{page1954continuous}, the offline estimation of a change-point seems to only be studied more recently. The first paper that we are aware of that considers this problem is \cite[]{chernoff1964estimating}, and in that paper the main interest is in estimating the post-change mean rather than the location of any change. However, partly due to a series of papers by David Hinkley \cite[]{Hinkley:1970,hinkley1970inference,hinkley1971inference} there was a sustained interest in this problem from 1970 onwards.

This early work on a range of approaches to test whether there is change either based on, or closely related to the, likelihood ratio test we have presented, and on calculating the distribution of the test statistic if there is no change. Early methods to calculate the null distribution of the likelihood-ratio test include \cite{sen1975tests},  \cite{hawkins1977testing} and \cite{worsley1979likelihood}, with the latter correcting an error in \cite{hawkins1977testing}. \cite{james1987tests} gives an overview and comparison of some of the early methods for detecting a single change.

\section{Summary}

Intuition from the behaviour of tests for a single change will be valuable in understanding properties of tests for multiple changes. For example, many approaches to testing for multiple changes involve repeatedly performing a test for a single change.

The following are some of the key messages from this chapter that will be relevant as we consider detecting multiple change-points in the remainder of the book.

\begin{itemize}
 \item For change in mean in Gaussian data, the likelihood-test and the CUSUM test are equivalent.
 \item Sample size affects power to detect, but has little impact on the accuracy of estimates of the location of a change.
 \item False positives are most likely to occur at the boundary of the region.
 \item Whilst CUSUM methods are primarily designed for detecting changes in mean, the likelihood-ratio test is a general approach that can be used to test for general types of change in a range of data types.
 \item change-point methods have some robustness to model error. Often model error will primarily impact on the null distribution of the test, and thus thresholds for detecting a change will need to be adapted to allow for model error. However estimates of the location of a change are more robust. We will see in the next chapter a related phenomena: that estimating the number of changes is more difficult than estimating their locations.
 \item In some situations, such as the change in slope model, estimating a single change when there are multiple changes, can lead to an estimator that is not consistent with any of the actual changes.
\end{itemize}

\bibliographystyle{royal}

redpajama_set_name: RedPajamaArXiv


text:
Comments (2,847 comments)
Limbaugh: Obama is 'frightened of me'
CNN Associate Political Editor Rebecca Sinderbrand
Conservative radio talk show host Rush Limbaugh responded Monday to recent reported comments by President Obama.
(CNN) - Radio host Rush Limbaugh said Monday that President Obama is "frightened of me."
"He's obviously more frightened of me than he is [Senate Minority Leader] Mitch McConnell," Limbaugh told listeners. "He's more frightened of me, then he is of say, [House Minority Leader] John Boehner, which doesn't say much about our party."
Limbaugh's comments followed reports Obama warned GOP congressional leaders last week that they should stop listening to the conservative talker, who had said on air he wanted the new president to fail.
"Now this is the great unifier," he told listeners Monday. "This is the man who's going to unify everybody and usher in a new era of bipartisanship and love."
White House Press Secretary Robert Gibbs gave reporters the administration's latest message to Limbaugh at Monday's press briefing: "Tell him I said, 'Hi,'" he joked.
Filed under: Popular Posts • President Obama • Rush Limbaugh
Podcast: Saving the economy - and the planet
CNN's Sarah Parker
CNN=Politics Daily is The Best Political Podcast from The Best Political Team.
(CNN) - It was a one-day unemployment bloodbath as tens of thousands of people lost their jobs at some of America's biggest and best-known corporations. Critics say President Barack Obama's actions could make economic matters worse by hitting one industry particularly hard, in an effort to make the U.S. more energy independent. In the latest installment of CNN=Politics Daily, CNN White House correspondent Dan Lothian takes a look at the two big issues on the president's plate – saving the economy and saving the planet.
Also: Vice President Joe Biden makes a deal with President Obama - he won't be a "deputy president" like those before him. So how will Biden's VP role differ from former Vice President Dick Cheney's? CNN's Brian Todd explains.
Meanwhile: President Obama launches his first peace mission. CNN's foreign affairs correspondent Jill Dougherty takes a look at his new Middle East envoy.
Plus: Embattled Gov. Rod Blagojevich is boycotting his impeachment and going on a media blitz. CNN's Susan Roesgen takes a look at Blago's recent media frenzy.
Finally: Former rivals face off, again. Former Republican presidential candidate John McCain is not on board with the president's economic stimulus plan, and he's letting everyone know. CNN's Brianna Keilar reports on how the Arizona senator blasted President Obama's trillion dollar stimulus package.
Click here to subscribe to CNN=Politics Daily.
Filed under: CNN=Politics Daily • Podcasts
Geithner sworn in as Treasury Secretary
The Senate voted Monday to confirm Timothy Geithner.
WASHINGTON (CNN) - Vice President Joe Biden swore in Timothy Geithner as the nation's next Treasury Secretary as President Obama looked on Monday evening.
Click here to read more on CNNMoney.com
Updated 7:55 p.m.
Filed under: Tim Geithner
Conyers again subpoenas Rove
(CNN) - House Judiciary Committee Chairman John Conyers issued a new subpoena for former Bush adviser Karl Rove Monday.
Rove was subpoenaed by Conyers last year, but refused to testify before Congress, pointing to executive privilege although he was no longer part of the administration.
"I have said many times that I will carry this investigation forward to its conclusion, whether in Congress or in court, and today's action is an important step along the way," Conyers said in a statement released Monday. "Change has come to Washington, and I hope Karl Rove is ready for it. After two years of stonewalling, it's time for him to talk."
The Michigan congressman is again looking to question Rove over whether the prosecution of former Democratic Gov. Don Siegelman, and the firing of the U.S. attorneys, were politically motivated. Rove has denied playing a part in either affair.
Filed under: John Conyers • Karl Rove
Borger: Obama, GOP on their first date
CNN Senior Political Analyst Gloria Borger
President Barack Obama has pledged to work with Republicans on his economic stimulus package.
WASHINGTON (CNN) - If there was one message that resonated loud and clear from the Obama campaign, it was this: Let's get past all of the partisanship and work together to get something done.
Turns out, President Barack Obama actually meant it. He's been meeting with congressional Republicans (and will trek to the Hill again Tuesday).
He's been listening to their ideas, incorporating some tax-cut ideas. And, yes, he's also been turning some down. After all, as he told Republicans last week, he was the one who won the election.
So what we're watching now as the stimulus package takes shape is the legislative equivalent of the first date: Republicans meeting with the new president, sizing him up. Obama deciding whether the GOP is serious about working with him.
"He [Obama] came to a conclusion early on in his political career that partisan polarization is a large part of the problem," says a senior Obama adviser. "Both in reality and in the minds of the American people."
Ah, but it's never that simple.
Filed under: GOP • President Obama
Virginia Republican promises to get high-tech
CNN Political Producer Peter Hamby
WASHINGTON (CNN) The GOP - looking for a glimmer of hope in the wake of November's drubbing - has trained their eyes on Virginia, where Republican gubernatorial candidate Bob McDonnell is pitching himself as a solutions-oriented leader who can reverse the commonwealth's leftward drift.
And McDonnell, running in a state The Weekly Standard has dubbed "Ground Zero for the GOP," is working overtime to avoid mistakes that have vexed Republicans in recent election cycles.
The commonwealth's current attorney general is promising that unlike other recent Republican campaigns, his will aggressively target younger voters using new technology. On Monday, the same day Democratic candidate Terry McAuliffe used text messaging to announce his first TV spot, McDonnell posted a promise on the popular conservative blog RedState that Republicans, not Democrats, will be the tech-wizards of the 2009 governor's race.
Filed under: Bob McDonnell • Virginia
What's next for Guantanamo Bay detainees?
President Barack Obama recently ordered the closing of Guantanamo Bay detention facility.
(CNN) - With President Barack Obama limiting how and where detainees at Guantanamo Bay can be interrogated, some analysts are asking if intelligence agencies will be able to get the information they need to keep the country safe - and where the prisoners will eventually end up.
Fresh off his inauguration, Obama issued executive orders relating to Guantanamo, including one requiring that the detention facility be closed within a year.
During a signing ceremony at the White House on January 22, Obama reaffirmed his inauguration pledge that the United States does not have "to continue with a false choice between our safety and our ideals."
The president said he was issuing the order to close the facility in order to "restore the standards of due process and the core constitutional values that have made this country great even in the midst of war, even in dealing with terrorism."
At its peak, Gitmo held 770 people the U.S. government believed may have been involved in terrorist activity or military action against the nation. The facility drew sharp criticism, including from Obama as he campaigned for the presidency.
Filed under: Guantanamo Bay • President Obama
Winfrey shocked over 'Senator Oprah' news
CNN's Denise Horn
Illinois Gov. Rod Blagojevich said Monday that he considered Oprah Winfrey to replace President Barack Obama in the U.S. Senate.
(CNN) - Surprised to hear Illinois might have had a Senator Oprah? So was Winfrey herself.
The talk show host said Monday she'd had no clue Illinois Gov. Rod Blagojevich had considered tapping her for the state's open Senate slot before he confirmed the news in a Monday morning TV interview.
Watch: Ill. governor on considering Winfrey
"If I had been watching [the ABC interview], as I watch from the treadmill, I probably would have fallen off the treadmill," Winfrey told friend Gayle King in a radio interview after hearing the news.
Blagojevich admitted Monday he once considered asking the media mogul to fill the senate seat vacated by President Barack Obama, before deciding to appoint former state attorney general Roland Burris.
"She seemed to be someone who had helped Barack Obama in a significant way to become President," he told ABC, adding that Winfrey "had a much broader bully pulpit than a lot of senators."
The embattled governor - currently facing impeachment proceedings for allegedly plotting to sell the vacant position to the highest bidder - will take calls from viewers tonight on CNN's Larry King Live at 9 pm ET.
Related: Blagojevich takes his case to TV
Filed under: Oprah Winfrey • Popular Posts • Rod Blagojevich
White House press office hit by e-mail outage
Staff for the new, wired president found themselves without e-mail Monday.
WASHINGTON (CNN) - President Obama has his blackberry — but his tech-savvy press operation has been without e-mail for most of the day, and it's not yet clear when they'll get it back.
Obama press staffers had just made the leap to their official White House e-mail addresses Monday morning when the Outlook server went down. They weren't the only ones affected by the crash, which also hit the first lady's office and other White House offices.
An apologetic Press Secretary Robert Gibbs told reporters at 1:30 pm press briefing that the server meltdown was responsible for the delay in responding to their e-mailed questions. But late Monday, nearly six hours since the server first went haywire, staffers still can't use their official e-mail addresses — and a press aide could not tell CNN when the press office might be able to start sending and receiving messages again.
Related: Access to 'BarackBerry' limited to small, select group
Filed under: White House
Poll: Caroline Kennedy to blame for Clinton replacement process
New York Gov. David Paterson announced his pick of Rep. Kirsten Gillibrand to replace Hillary Clinton in the Senate.
(CNN) - More than three times as many New Yorkers in a new poll blame Caroline Kennedy and her team for the messy process surrounding the search for Hillary Clinton's Senate replacement than fault the state's governor, David Paterson - although, on balance, his final selection meets with their approval.
Forty-nine percent of voters surveyed in a Quinnipiac University poll released Monday said Kennedy and her advisers were to blame, to 15 percent who pointed to Paterson. Twelve percent blame both, and 24 percent are undecided.
Overall, the state's voters approve of Paterson's selection of Rep. Kirsten Gillibrand, backing the pick 46 to 30 percent, with 24 percent undecided. That margin is higher upstate, where the choice of the Albany-born Gillibrand draws the approval of 55 percent of the region's voters to 25 percent who disapprove. In New York City, that margin is far smaller: there, the conservative Democrat draws the approval of 41 percent to 34 percent who disapprove. But in the state's suburbs, her edge falls within the survey's 3 point margin of error: 35 percent approve, 32 percent do not.
Filed under: Caroline Kennedy • David Paterson • Kirsten Gillibrand • Poll • Popular Posts

redpajama_set_name: RedPajamaCommonCrawl


Ok, opposite words, but what's the opposite word of a circus? Anyways, here's the Dead Letter Circus at the Key Club with Intronaut, Animals As Leaders, Last Chance to Reason and I think Cynic was on this bill. My memory is a bit foggy, or was it Intronaut headlining? Hmm, I'll have to chase down the archives to find this answer... I believe this was my first viewing of DLC live and was thoroughly impressed by them and became a fan. Super cool guys and had a flyer signed by all of them. The Catalyst Fire is a great album too. Need to get the newer stuff... enjoy!


Jerid has lived and worked in the Capital Region his entire life. Born in the small town of Averill Park, NY, he grew to cherish small town ways and values. Being honest, up-front, and helping families became second nature while working for a local auction house for most of his adult life. He built a solid foundation from donating to special needs organizations to participating in many charitable events. He has a love for the outdoors, fitness, as well as spending time with his family and dog. So if you're a first time home buyer or selling and looking for the right tools to make your transition easier, Jerid's dedication to values and family will help you achieve your goals. Contact Jerid at Berkshire Hathaway HomeServices Blake, Realtors® for all your real estate needs today!


text:
Обмін () — російський короткометражний фільм  Романа Качанова. Прем'єра фільму відбулася 22 листопада 2017 року кінотеатрі «Ілюзіон» в Москві.

Сюжет
Сімейна пара Федя і Тетяна приїжджають в заміський будинок до друзів Костянтину і Олені.  Увечері, коли всі вже добряче випили, Костянтин вирубується.  Випивають ще. Далі Федір нічого не пам'ятає…  Рано вранці відбувається серйозне з'ясування стосунків.

У ролях
 Олександр Ільїн — Федір
 Ольга Арнтгольц — Тетяна
 Михайло Владимиров — Костянтин
 Валерія Шкірандо — Олена

Знімальна група
 Режисер:  Роман Качанов
 Автор сценарію:  Роман Качанов
 Продюсер:  Роман Качанов
 Оператор:  Михайло Мукасей
 Саундтрек: План Ломоносова
 Художник: Микола Поспєлов
 Костюми: Марія Рау
 Монтаж:  Маргарита Смирнова
 Звук: Девід Шуфутинський
 Лінійний продюсер: Катерина Дубровська
 Виконавчий продюсер: Олег Бабицький

Зйомки і прем'єра
Влітку 2017 року культовий режисер Роман Качанов повідомив, що хоче зняти короткометражне кіно «без будь-якої цензури, підневільної редактури і іншої корозії». Через мережу Качанов оголосив, що збирає гроші на новий фільм. Відгукнулося понад 250 осіб. За два тижні було сформовано бюджет фільму. Можливо, вперше в історії російського кіно все, що брали участь в краудфандінге були включені в фінальні титри фільму.  Зйомки пройшли в Підмосков'ї. 22 листопада 2017 роки фільм був представлений глядачам. Відразу після першого показу фільму в кінотеатрі була проведена інтернет-прем'єра фільму — прямо в кінозалі був відкритий вільний доступ до фільму в інтернеті.

Примітки

Посилання
 «Известия»
 «Прес-реліз.ру»
 Aspmedia24.ru
 «Першоджерело»
 «Білоруські новини»
 Сайт Молодіжного центру Спілки кінематографістів РФ
 Punxlove.ru
 Фільм «Обмін» в  мережі

Кінокомедії Росії
Кінокомедії 2017
Панк
Фільми Романа Романовича Качанова

redpajama_set_name: RedPajamaWikipedia


text:
Principles of Unity
Parallel Development
SDCEDSV Board of Directors & Supporting Organizational Programs
Sacred Turtle Woman
Rerighting Herstory
Crisis Lines
DV/SV info
Is it Domestic Violence, Abuse, Conflict, Violence or Battering?
What Is Sexual Violence?
SAAM Campaign Flyers
What is Sex Trafficking?
Prevention Materials
Mental Health Consequences of Crime
Sisseton Wahpeton Oyate Passes Landmark Sexual-Abuse Statute
Issues Confronting Indigenous Women in Rural USA
Stalking Is a Problem Everywhere, Including Indian Country
Web site for information on social networking.
The Real Violence Against Women Act
A Feminist Perspective on the Ethics of Communication
Advocacy on Behalf of Battered Women
A Look at the Indian Health Service Policy of Sterilization, 1972-1976
MAKING SOCIAL CHANGE: THE DYNAMICS OF EDUCATION, ACTION AND REFLECTION
Exit this siteA new tab will open and this page will redirect to Google.
Bruises on Mind, the Body, and the Spirit
Bruises on the Mind, Body, and Spirit. A handbook for women who have been battered and are seeking safety, support and healing. 38 page booklet.
Funding for this project was made possible by Grant #2017-MU-AX-0020 awarded by the Office on Violence Against Women, U.S. Department of Justice. The views expressed in written conference materials or publications and by speakers and moderators do not necessarily reflect the official policies of the Department of Justice nor does mention of trade names, commercial practices, or organizations imply endorsement by the U.S. Government.
© 2019 South Dakota Coalition Ending Domestic & Sexual Violence | All rights reserved.


text:
Delsea grapplers hold off Kingsway
JFRIEDMAN@THEDAILYJOURNAL.COM
WOOLWICH TWP. – The marquee matchup of Friday night's Delsea-Kingsway high school wrestling match lived up to the billing, but it was the following act that made the biggest difference on the scoreboard.
The crucial performance was delivered by Delsea senior Tom Carney, as he rallied from a first-period deficit to pin Brian Maiden at 138 pounds as the Crusaders took down Kingsway, 37-22.
"It means a lot," Carney said. "They're a good team and it shows how good we are and how good we can be if we wrestle the way we can."
Prior to Carney's bout, the Dragons had pulled within 25-20 as Kingsway's Stanley Atkinson edged Delsea's Billy Janzer 1-0 in a matchup of two of the best 132-pounders in South Jersey.
Atkinson, No. 1 in The Mat Pack's Region 7 and 8 rankings, got a second-period escape to go up 1-0, but Janzer, No. 2 in the rankings, decided to start the third neutral. Janzer had several close takedowns, including one in the final five seconds, but was unable to secure the two points and the win.
"It was a good match, it's not the outcome we wanted, but we know what kind of caliber of wrestlers they both are," Delsea coach Greg Sawyer said. "They both are state-level wrestlers. We kind of saw that today. It was a 1-0 match but it wasn't a 1-0, no-action match. There was a lot of scrambling, and Janzer was in on quite a few shots and I thought it was good match."
It wasn't looking good for Carney when he fell behind 4-0 in the opening two minutes of his match. But Carney escaped to open the second, closing the gap to 4-1, but it was his takedown at the buzzer that swung momentum, as he entered the third down 4-3.
He controlled the final 97 seconds before earning the fall in 5:37 to push the Crusaders' lead out to 31-20.
"I knew I just had to stay calm and do what I do best, and just wrestle and not worry about if I was losing or winning and just go out there and wrestle my best," Carney said.
Then, during the 145-pound match, Kingsway was docked a team point for unsportsmanlike conduct before Delsea's Kevin Huck earned a 7-6 win to clinch the Crusaders' victory.
Kingsway earned the first win of the night as Rory Bushby, the No. 4 160-pounder in The Mat Pack rankings, bumped up to 170 and topped Delsea's Brett Hughes 5-1, giving his team a 3-0 lead.
The Crusaders (8-1) bounced back quick though, as Vince DeGeorge controlled his 182-pound bout with Sean Calhoun to earn a 16-4 decision, which gave his team a 4-3 advantage.
Then Jake Hallowell (195), Jake Hughes (220) and Joe Fabrizio (heavyweight) recorded back-to-back-to-back falls, with each coming quicker than the next to give Delsea a commanding 22-3 edge.
But it wasn't just Delsea's victories that proved vital to its victory. At 106, Chris Recinto worked his tail off to stay off his back against Kingsway's Quinn Kinner, The Mat Pack's top-ranked wrestler in his weight class, and lost by tech fall, 17-1 (5:59), to save his team a point.
John Hassett had a similarly strong performance in defeat, as he fell to Kingsway's Trace Kinner at 120, 5-2. Kinner is the No. 3 ranked 113-pounder by The Mat Pack.
"If you're not going to win you might as well wrestle tough, and that's the contributions they make to the team in those matches," Sawyer said. "When you're getting a little outmatched, it's a matter of being able to wrestle tough and save points from being put on the board."


COMEDIANS ON TV! - Clue Search Puzzles - Combining Trivia, Crosswords, and Word Search Puzzles for Classroom Use, Homeschooling and Brain Exercises!
Wayne Brady was a regular on Whose Line Is It __?
Brad Garrett and Ray Romano are brothers on "Everybody Loves __"
His Weekend Update ended with "That's the news, and I am outta here!"


Feel free to submit letter of interest, resume and copies of cert(s) to Mrs. Allison Angermeyer M.A. Chief School Administrator/Superintendent, Prospect Park School District, 290 North 8th St, Prospect Park, NJ 07508. (973) 720-1981 *Fax: (973) 720-1992.
* SPED Teacher ABA/Autism Experience.
* Physical Education/Health Half Time, Full Time possible with K-6 Certification/HQ.
EOE A place to play, learn and grow together.


text:
also known as Bryan Lee Cranston
Actor, VoiceCast
Born: Mar 07, 1956, California, U.S.A.
The eponymous lead of Breaking Bad (2008-2013), one of the most successful TV shows of all time, Bryan Cranston is an American actor, director, screenwriter, and producer. His transpiring performance as Walter While in AMC's Breaking Bad won him four Primetime Emmy Awards and he also twice won the Outstanding Drama Series Award as a producer of the show. He was also nominated for his role as Hal on the Fox comedy series Malcolm in the Middle. In 2015, he gave a compelling performance in Jay Roach's biographical dramaTrumbo and received an Academy Award nomination for Best Actor.
Cranston was born to Joseph Louis "Joe" Cranston, who was a former amateur boxer and actor, and Audrey Peggy, who was a radio actress, in Hollywood, USA. He has two siblings and he grew up living on a poultry farm in Canoga Park, Los Angeles, USA. Cranston's father skipped many jobs before he took up acting as a career but unfortunately his acting career never fetched him enough to provide support for his family. Eventually, the actor's father walked out on the family when he was just 11 years old. It was only 11 years later that Cranston reunited with his father after his brother tracked him down.
The actor completed his studies from the Canoga Park High School and later, in 1976, earned an associate degree in police science from the Los Angeles Valley College.
Cranston married to writer Mickey Middleton in 1977 and the couple separated in 1982. The actor was 33 when he married his second wife Robin Deardon. They both met on the sets of the TV show Airwolf in 1984. Their daughter, Taylor Deardon Cranston, was born in 1993 and has also appeared in one of the episodes of Breaking Bad, which was directed by Cranston.
Theater Career
Cranston first started performing in local and regional theaters, before he became associated with the Granada Theater in California. Years later he appeared in plays such as The God of Hell, The Taming of the Shrew, and Wrestlers. The actor also won a Drama-Logue award for his performance in the play The Steven Weed Show.
TV Career
Cranston began working on TV shows in the '80s: mostly appearing in minor roles. He first appeared in an uncredited role in the TV film To Race the Wind. Throughout the '80s and the '90s, the actor starred in numerous TV shows but none of his roles were substantial barring a few roles. He appeared in three episodes of the TV show Murder, She Wrote and also essayed the character of Russel in the TV series Raising Miranda; he starred in nine episodes. He went on to guest star on TV shows such as Baywatch (1989)The Flash (1991), and Viper (1994). He then appeared as Dr. Tim Whatley on the TV show Seinfeld (1994).
Although in the next six years Cranston appeared on several popular shows such as The X-Files (1998), Chicago Hope (1998), and The King of Queens (1999). It was only in 2000 that the actor was signed for the leading role of Hal in the comedy series Malcolm in the Middle; a role that later went on to become one of the most substantial roles of the actor. Cranston appeared on the show for seven seasons for 151 episodes and even was nominated for the Primetime Emmy Award thrice. The show lasted for six years from 2000 to 2006, and during the time he appeared on a few TV films viz. 'Twas the Night (2001), The Santa Clause Brothers (2001), and National Lampoon's Thanksgiving Family Reunion (2003). In 2006, he also starred in the popular sitcom How I Met Your Mother as Hasmond Druthers, Ted Mosby's (Josh Radnor) former boss.
A couple of years later Cranston was cast as Walter White in Vince Gilligan's Breaking Bad. The actor portrayed the character of a high school chemistry teacher diagnosed with lung cancer, who teams up with his former student Jesse Pinkman (Aaron Paul) and manufactures & sells methamphetamine in order to ensure the well-being of his family. The role that not only defined Cranston's career but also went on to become one of the most acclaimed and celebrated roles in the history of TV industry. The actor went on to win three consecutive Primetime Emmy Awards for Outstanding Lead Actor in Drama series from 2008 to 2010 and won the award again in 2014.
Following the success of Breaking Bad, Cranston went on to guest appear in TV shows such as The Simpsons and The Cleveland Show. In 2015, he was cast in the stop-motion animated comedy television series SuperMansion to voice the titular character, Titanium Rex.
As a Director
Cranston has directed several episodes of the TV shows Malcolm in the Middle and Breaking Bad. In 2012, he also directed two episodes of the TV series Modern Family.
Movie Career
As an Actor
Cranston started appearing in movies in the late '80s and made his screen debut in a minor role in the 1987 movie Amazon Women on the Moon. He followed this up with several other minor roles in movies such as Fletch Lives (1989), Clean Slate (1994), That Thing You Do (1996), and Strategic Command which was released in 1997. The following year he played the role of a War Department Colonel in Steven Speilberg's Saving Private Ryan - he appears as the colonel who insists that Ryan must be saved.
In 2005, Cranston gave the voice to the character of Buzz Aldrin in the IMAX 3D documentaryMagnificent Desolation: Walking on the Moon 3D. The following year he starred in the Academy Award-winning movie Little Miss Sunshine portraying the role Stan Grossman. The following years saw him taking up several supporting roles is successful movies such as The Lincoln Lawyer (2001) wherein he starred as detective Lankford, Drive (2011), wherein he essayed the role of Shannon, and Contagion (2012), wherein he starred in the role of Haggerty.
In 2012, the actor appeared in six films of which two were commercial and critical hits vizMadagascar 3: Europe's Most Wanted and Academy Award-winning Argo directed by Ben Affleck. In Madagascar 3Cranston voiced the character of Vitaly and in Argo, he played a major role - Jack O'Donnel. However, one of his most substantial and acclaimed roles came with the movie 2015 movie Trumbo a biographical drama on the American screenwriter Dalton Trumbo. Essaying the role of Trumbo, the actor received rave reviews for his performance with many critiques praising him. In the following year, Cranston voiced the character of Li, Po's (voiced by Jack Black) biological father and the uncle of Dim and Sum, in the animated movie Kung Fu Panda 3 directed by Jennifer Yuh Nelson and Alessandro Carloni. In 2016, he also starred in the movies Get a Job, The Infiltrator, In Dubious Battle, Wakefield, and Why Him?. His 2017 releases include the movies Power Rangers, The Masterpiece, and Last Flag Flying.
Cranston also tried his hand at directing in 2000 and made his directorial debut with the movie Last Chance.
During his struggling days as an actor, Bryan Cranston performed weddings for $150; he was ordained as a minister by the Universal Life Church.
Bryan Cranston is only the second actor to win three consecutive Primetime Emmy Awards for Outstanding Lead Actor in a Drama Series. The first actor to do so was Bill Cosby, who won three consecutive Emmys from 1966 to 1968 for his performance in the TV series I Spy.
While shooting the final episode of Breaking Bad, Bryan Cranston and Aaron Paul both got tattoos done - Cranston's tattoo consists of the show's logo on one of his fingers.
Cranston at some point in his career had decided to become a police officer.
Cranston co-owns the independent theater - Cinemas Palme d'Or, Palm Desert, California.
If you're a person who complains about everything all the time, then you're just the boy who cried 'wolf.' But if you do it on occasion and about the right reasons, then people listen
When you're an actor in grade school, high school, college, whatever, you start to realize what you're really good at, what you're kinda good at, what you're okay at, and you start to compartmentalize. But if you know yourself and what you're capable of, it's just a matter of opportunity.
I learned long ago to focus on things you can control and don't even pay attention to things you don't.
I can tell you for sure: people who are at their peak right now will not sustain that. You can't. It's against the law of nature.
If you have a level of expectation in your life that you have to be a quote-unquote star, whatever that means, you might be setting yourself up for failure.
Luck is a component that a lot of people in the arts sometimes fail to recognise: that you can have talent, perseverance, patience, but without luck you will not have a successful career.
If you like vanilla, you're not going to like 'Breaking Bad' - you need to like a specific flavor that is unusual, that is different, that takes risks
Love between two adults is always conditional. You can fall out of love because you are able to fall in love.
I don't have spare time.
Love is not as important as good health. You cannot be in love if you're not healthy. You can't appreciate it.
Born: Apr 08, 1963
Born: Oct 22, 1962
Born: Jul 26, 1959
Born: Mar 20, 1958
Robin Dearden
Born: Dec 04, 1953
Mickey Middleton


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Tessa Traeger
Tessa Traeger is a British photographer born in 1938, who trained at the Guildford School of Fine Art and Photography. She has worked in London since the 1960s, and also spent time in the Haut-Vivarais in South-Western France in the 1990s, photographing farmers from the region. Her work mainly consists of still-life and food photography, and is represented in the National Portrait Gallery, the Victoria and Albert Museum, Tate Britain and the Bibliothèque Nationale de France.
'La Couronne, Ardèche' (1994) is a black-and-white photograph of a woman holding a big round loaf of bread on her lap. It is part of the book, Voices of the Vivarais (2010), which tells the story of a collection of individuals living a very traditional lifestyle in a remote mountainous region. Of this series of pictures, Traeger has said that she was looking at a 19th-century culture, and capturing it with 19th-century tools.
In this portrait, the focus on the bread conveys a certain sense of pride and tradition. The title, which translates from French as 'The Crown' is used to describe this specific shape of loaf, but is also a symbol of royalty. The photographs' slightly faded colours, which have come about as a result of polaroid negatives being developed in a sodium sulphide solution, give the piece a sense of nostalgia and melancholia.
Title: La Curonne, Ardeche
Medium: Photography


A streamlined version of the classic Raven®. Perform all your favorite vanishes of coins and small objects from a table or a spectator's hand!
One effect: Fold a dollar and place it on your palm. Wave your other hand over it and it vanishes. This is one of many effects possible.
Includes Raven® Mini, accessories and basic instructions, plus a download code for additional tips and routines.


Shannon ( @shasherslife ) and I met The Lorax while we were at Blissdom this past week - photo courtesy of Shannon's iPhone :).
Blissdom was amazing - full of inspiration and fun and so much more. I promise to tell you all more about my Blissdom experience as soon as I can gather my thoughts enough to get them all down in a post. I can't wait for Blissdom Canada this fall!!


This ambitious project is at the groundbreaking Launchpad development centre of training excellence, offering a wide range of diplomas recognised in commercial kitchens for 14 to 19 year olds. The centre needed a truly modern catering training facility on a par with the best commercial kitchens, with commercial ovens, hobs and food service equipment to match.
Durability and practicality had to stand side by side with flexibility. When not being put through its paces by students, the kitchen has to turn out all the food required for the college café and refectory really putting all the food service equipment to the test.
Space created a real power-house of a commercial kitchen with numerous green features to take advantage of the latest cooking methods, save energy and be kinder on the environment. Advanced i-hob technology, widely predicted as the future cooking method of choice in commercial kitchens, is evident throughout.
Two and four ring flush induction hob cookers, with electric commercial oven and grill beneath, use clever sensor technology to switch on the hob when a pan is placed on top, and automatically turn it off when it's removed. It's a more efficient way to cook – 90% of the energy goes into the pan compared to 40% with gas – costs less and is safer – no hot surfaces and no risk of leaving the food service equipment on.
The smart new kitchen features a central island of eleven induction hobs neatly suited together, equipped with ambient storage underneath. Five more training modules with commercial ovens are fitted around the perimeter of the kitchen including one designed to be easily lowered for wheelchair users.
Everything from the work surfaces to the hobs and commercial ovens gleams with crisp stainless steel finishes for added durability and lifespan – a far cry from the old days when domestic style cabinets were the norm in food tech classrooms – and totally in keeping with the emphasis on professional catering qualifications.


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Home/Encryption Keys/Biss/ATV Türkiye Biss Key on Turksat
ATV Türkiye Biss Key on Turksat
ATV (also stylized aTV) is a nationwide TV channel in Turkey, began broadcasting in September 1993. ATV is an acronym of Actual Television (Aktüel Televizyonu). As of August 2013, ATV is Turkey's 5th popular channel with a market share of 6.71%.
ATV was founded by Dinç Bilgin, who is the original owner of the channel. The channel changed its ownership to Ciner Medya Grubu in 2002, and Tasarruf Mevduat? Sigorta Fonu in 2007. ATV was owned by Çal?k Holding's Turkuvaz Media Group, but in 2013, the ownership changed again to Kalyon Group.
Satellite technical details with Biss key.
Türksat 3A / 4A 42°E
ID : FEED
Biss CW: 81 18 AB 44 00 18 CC E4
Turksat is the name of a series of Turkish communications satellites operated or projected by the Türksat A.?..
Satellite telecommunication is provided by the Gölba?? Ground Station in Ankara.
List of channels using Biss encryption on SES 6 at 40.5°W
List of channels using Biss encryption on Intelsat 905 & Alcomsat 1 at 24.5°W
MCA Nigeria package biss
BISS KEY OF ETV SPORT ON NSS 12


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Daisy Ridley responds to Spider-Woman Rumors!
Star Wars star Daisy Ridley responds to rumors of her being cast as 'Spider-Woman,' saying if the role comes to her she will gladly accept it.
Celebrities, Movies
After exploding into the limelight in Star Wars sequels, Daisy Ridley is now being linked to many other projects. Now, Ridley has addressed the rumors of her dawning the Spidey-suit and becoming Spider-Woman!
Related: Daisy Ridley responds to Labeled 'Aggressive' on Star Wars set
Daisy Ridley is thrilled to become Spider-Woman
Disney ©
While talking to author Sariah Wilson, Daisy Ridley was asked if she had heard rumors about her playing Spider-Woman. To that, Ridley said:
OMG, I would love to play Spider-Woman. But is it more of like an extended Spiderverse thing?
What so many of you have been waiting for: I asked Daisy Ridley I'd she'd heard the Spider-Woman rumors. She said no and asked what they were. I explained it to her and she said, "OMG, I would love to play Spider-Woman. But is it more of like an extended Spiderverse thing?"
— Sariah Wilson (@sariahwilson) February 20, 2021
This shows that Ridley is extremely eager to play the female version of the web-slinger, Jessica Drew. Furthermore, Ridley told Sariah that she absolutely loved the animated feature Spider-Man: Into the Spider-Verse. She has also been following Marvel's latest show WandaVision on Disney+.
Later on, Daisy Ridley clarified while appearing on The Graham Norton Show that her comments were from an interview taken out of context. She said:
It's funny, I did this chat with a lovely woman and she asked me if it was true – it was the first time I'd heard of it and then the next day people kept texting me about it and it keeps being brought up. There are a million other people who could do it, but perhaps I can will it into existence!
Spider-Woman first appeared in Marvel Spotlight in 1977 and has been a regular in Marvel comics ever since. In some comics, even Mary Jane has worn her costume. But, the most notable version of this hero has always been Jessica Drew.
Furthermore, Daisy Ridley will not be the first Star Wars actor to toy with the MCU. Oscar Isaac will soon don the role of Moon Knight, and will soon get a series of his own on Disney+.
Related: Daisy Ridley fires back at Ted Cruz for defending Gina Carano
# Daisy Ridley# Spider-Woman
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