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Problem

Let $X \sim N(0,1)$. Suppose that we know \begin{align} Y \; | \; X=x \quad \sim \quad N(x,1). \end{align} Show that the posterior density of $X$ given $Y=y$, $f_{X|Y}(x|y)$, is given by \begin{align} X \; | \; Y=y \quad \sim \quad N\left(\frac{y}{2},\frac{1}{2}\right). \end{align}

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Problem

We can generalize the result of Problem 9.1 using the same method. In particular, assuming \begin{align} X \sim N(\mu,\tau^2) \quad \textrm{and} \quad Y \; | \; X=x \quad \sim \quad N(x,\sigma^2), \end{align} it can be shown that the posterior density of $X$ given $Y=y$ is given by \begin{align} X \; | \; Y=y \quad \sim \quad N \left(\frac{ y / \sigma^2 + \mu / \tau^2 }{1 / \sigma^2 + 1 / \tau^2}, \frac{1}{1 / \sigma^2 + 1 / \tau^2} \right). \end{align} In this problem, you can use the above result. Let $X \sim N(\mu,\tau^2)$ and \begin{align} Y \; | \; X=x \quad \sim \quad N(x,\sigma^2). \end{align} Suppose that we have observed the random sample $Y_1$, $Y_2$, $\cdots$, $Y_n$ such that, given $X=x$, the $Y_i$'s are i.i.d. and have the same distribution as $Y \; | \; X=x$.

1. Show that the posterior density of $X$ given $\overline{Y}$ (the sample mean) is \begin{align} X \; | \; \overline{Y} \quad \sim \quad N \left(\frac{ n \overline{Y} / \sigma^2 + \mu / \tau^2 }{n / \sigma^2 + 1 / \tau^2}, \frac{1}{n / \sigma^2 + 1 / \tau^2} \right). \end{align}
2. Find the MAP and the MMSE estimates of $X$ given $\overline{Y}$.

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Problem

Let $\hat{X}_M$ be the MMSE estimate of $X$ given $Y$. Show that the MSE of this estimator is \begin{align} MSE=E\big[\textrm{Var}(X|Y)\big]. \end{align}

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Problem

Consider two random variables $X$ and $Y$ with the joint PMF given in Table 9.1.

	$Y=0$	$Y=1$
$X=0$	$\frac{1}{5}$	$\frac{2}{5}$
$X=1$	$\frac{2}{5}$	$0$

1. Find the linear MMSE estimator of $X$ given $Y$, ($\hat{X}_L$).
2. Find the MMSE estimator of $X$ given $Y$, ($\hat{X}_M$).
3. Find the MSE of $\hat{X}_M$.

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Problem

Consider Example 9.9 in which $X$ is an unobserved random variable with $EX=0$, $\textrm{Var}(X)=4$. Assume that we have observed $Y_1$ and $Y_2$ given by \begin{align} Y_1&=X+W_1,\\ Y_2&=X+W_2, \end{align} where $EW_1=EW_2=0$, $\textrm{Var}(W_1)=1$, and $\textrm{Var}(W_2)=4$. Assume that $W_1$, $W_2$ , and $X$ are independent random variables. Find the linear MMSE estimator of $X$ given $Y_1$ and $Y_2$ using the vector formula \begin{align} \hat{\mathbf{X}}_L=\mathbf{\textbf{C}_\textbf{XY}} \mathbf{\textbf{C}_\textbf{Y}}^{-1} (\mathbf{Y}-E[\textbf{Y}])+ E[\textbf{X}]. \end{align}

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Problem

Suppose that we need to decide between two opposing hypotheses $H_0$ and $H_1$. Let $C_{ij}$ be the cost of accepting $H_i$ given that $H_j$ is true. That is

$\quad$ $C_{00}$: The cost of choosing $H_0$, given that $H_0$ is true.

$\quad$ $C_{10}$: The cost of choosing $H_1$, given that $H_0$ is true.

$\quad$ $C_{01}$: The cost of choosing $H_0$, given that $H_1$ is true.

$\quad$ $C_{11}$: The cost of choosing $H_1$, given that $H_1$ is true.

It is reasonable to assume that the associated cost to a correct decision is less than the cost of an incorrect decision. That is, $c_{00} \lt c_{10}$ and $c_{11} \lt c_{01}$. The average cost can be written as \begin{align} C =&\sum_{i,j} C_{ij} P( \textrm{choose }H_i | H_j) P(H_j)\\ =& C_{00} P( \textrm{choose }H_0 | H_0) P(H_0)+ C_{01} P( \textrm{choose }H_0 | H_1) P(H_1) \\ & +C_{10} P( \textrm{choose }H_1 | H_0) P(H_0)+ C_{11} P( \textrm{choose }H_1 | H_1) P(H_1). \end{align} Our goal is to find the decision rule such that the average cost is minimized. Show that the decision rule can be stated as follows: Choose $H_0$ if and only if \begin{align} \label{eq:min-cost-test-gen} f_{Y}(y|H_0)P(H_0) (C_{10}-C_{00}) \geq f_{Y}(y|H_1)P(H_1)(C_{01}-C_{11}) \hspace{30pt} (9.8) \end{align}

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Problem

Let \begin{align} X \sim N(0,4) \quad \textrm{and} \quad Y \; | \; X=x \quad \sim \quad N(x,1). \end{align} Suppose that we have observed the random sample $Y_1$, $Y_2$, $\cdots$, $Y_{25}$ such that, given $X=x$, the $Y_i$'s are i.i.d. and have the same distribution as $Y \; | \; X=x$. Find a $95\%$ credible interval for $X$, given that we have observed \begin{align}%\label{} \overline{Y}=\frac{Y_1+Y_2+...+Y_n}{n}=0.56 \end{align} Hint: Use the result of Problem 9.2.

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