9.1.9 Bayesian Interval Estimation
Given the observation $Y=y$, the interval $[a,b]$ is said to be a $(1-\alpha)100\%$ credible interval for $X$, if the posterior probability of $X$ being in $[a,b]$ is equal to $1-\alpha$. In other words, \begin{align} P(a \leq X \leq b |Y=y)=1-\alpha. \end{align}
Example
Let $X$ and $Y$ be jointly normal and $X \sim N(0,1)$, $Y \sim N(1,4)$, and $\rho(X,Y)=\frac{1}{2}$. Find a $95\%$ credible interval for $X$, given $Y=2$ is observed.
- Solution
- As we have seen before, if $X$ and $Y$ are jointly normal random variables with parameters $\mu_X$, $\sigma^2_X$, $\mu_Y$, $\sigma^2_Y$, and $\rho$, then, given $Y=y$, $X$ is normally distributed with \begin{align}%\label{} \nonumber &E[X|Y=y]=\mu_X+ \rho \sigma_X \frac{y-\mu_Y}{\sigma_Y},\\ \nonumber &\textrm{Var}(X|Y=y)=(1-\rho^2)\sigma^2_X. \end{align} Therefore, $X|Y=2$ is normal with \begin{align}%\label{} \nonumber &E[X|Y=y]=0+ \frac{1}{2} \cdot \frac{2-1}{2}=\frac{1}{4}, \end{align} \begin{align} \nonumber \textrm{Var}(X|Y=y)=\left(1-\frac{1}{4}\right)\cdot 1=\frac{3}{4}. \end{align} Here $\alpha=0.05$, so we need an interval $[a,b]$ for which \begin{align} P(a \leq X \leq b |Y=2)=0.95 \end{align} We usually choose a symmetric interval around the expected value $E[X|Y=y]=\frac{1}{4}$. That is, we choose the interval in the form of \begin{align} \left[\frac{1}{4}-c, \frac{1}{4}+c\right]. \end{align} Thus, we need to have \begin{align} P\left(\frac{1}{4}-c \leq X \leq \frac{1}{4}+c |Y=2\right) &=\Phi \left( \frac{c}{\sqrt{3 /4} }\right)-\Phi \left( \frac{-c}{\sqrt{3 /4} }\right)\\ &=2 \Phi \left( \frac{c}{\sqrt{3 /4} }\right)-1=0.95 \end{align} Solving for $c$, we obtain \begin{align} c=\sqrt{3 /4} \Phi^{-1}(0.975) \approx 1.70 \end{align} Therefore, the $95\%$ credible interval for $X$ is \begin{align} \left[\frac{1}{4}-c, \frac{1}{4}+c\right] \approx [-1.45,1.95]. \end{align}
