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\begin{document}

<<setup, cache=FALSE, echo=FALSE>>=
library(knitr)
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opts_chunk$set(cache=TRUE, autodep=TRUE, fig.align='center')
options(scipen=999)
options(repos=c(CRAN="https://cloud.r-project.org/"))
inline_hook <- function (x) {
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      sprintf("%.3f", x),
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\begin{center}
\textsclarge{MATH 559: Bayesian Theory and Methods}

\vspace{0.1 in} \textsc{Decision Theory: Two Special Cases}
\end{center}

The Bayesian decision for loss function $\Loss(\d(\by),\theta)$ and posterior distribution $\pi_n(\theta)$ is obtained as
\[
\min_d \int \Loss(\d(\by),\theta) \pi_n(\theta) \ d \theta.
\] 
We examine the solution to two specific decision problems.

\begin{enumerate}[label=\arabic*.]
  \item \textbf{Credible intervals:}  Suppose $0 < \kappa < 1$, and consider the loss function for $\theta \in \R$ defined as
  	\[
	\Loss(\d,\theta)  = \kappa(\d-\theta )\mathbb{1}(\theta < \d) + (1-\kappa)(\theta - \d)\mathbb{1}(\theta \ge \d) = \Loss_{1-\kappa}(\d,\theta)
	\] 
    say. The expected posterior loss for possible decision $t$ is then
    \begin{align*}
      \int \Loss(t,\theta) \pi_n(\theta) \ d \theta & =  \int \kappa(t-\theta )\mathbb{1}(\theta < t) \pi_n(\theta) \ d \theta + \int (1-\kappa)(\theta - t)\mathbb{1}(\theta \ge t) \pi_n(\theta) \ d \theta \\[6pt]
      & = \kappa \int_{-\infty}^{t} (t - \theta) \pi_n(\theta) \ d \theta + (1-\kappa) \int_{t}^\infty (\theta-t) \pi_n(\theta) \ d \theta
    \end{align*}
    Differentiating with respect to $t$ we obtain by the chain rule and fundamental rule of calculus
    \begin{itemize}
      \item First term:
      \[
      \kappa \left[ \int_{-\infty}^t \pi_n(\theta) \ d \theta + t \pi_n(t) - t \pi_n(t) \right] = \kappa \int_{-\infty}^t \pi_n(\theta) \ d \theta
      \]
      \item Second term:
      \[
      (1-\kappa) \left[ -t \pi_n(t)  - \int_{t}^\infty \pi_n(\theta) \ d \theta + t \pi_n(t) \right] = -(1-\kappa) \int_{t}^\infty \pi_n(\theta) \ d \theta
      \]
      \end{itemize}
      and so the minimizing value of $t$ solves
      \[
      \kappa \int_{-\infty}^t \pi_n(\theta) \ d \theta = (1-\kappa) \int_{t}^\infty \pi_n(\theta) \ d \theta
      \]
      so that
      \[
      \kappa = \int_{t}^\infty \pi_n(\theta) \ d \theta.
      \]
      Thus the solution to the loss minimization problem is the $1-\kappa$ quantile of $\pi_n(\theta)$. Now consider the decision $\d = (\d_1,\d_2)$ that comprises two components, with loss function
      \[
      \Loss((\d_1,\d_2),\theta) = \Loss_{1-\kappa}(\d_1,\theta) + \Loss_{\kappa}(\d_2,\theta).
      \]
      By the above calculation, we obtain the solution
      \[
      (\d_1(\by), \d_2(\by)) 
      \]
      as the $\kappa$ and $1-\kappa$ posterior quantiles respectively, so that the interval between the two values contains posterior probability $1-2 \kappa$, and is the $100 (1-2 \kappa)$\% equal-tailed credible interval.
      
      \medskip
      The other versions of the Bayesian credible interval can also be obtained as loss minimization decisions.

  
  \item \textbf{Predictive Distribution:} Suppose our task is to estimate the unknown conditional density $f_Y(y;\theta_0)$.  We aim to construct a Bayesian estimate of $f_Y(y;\theta_0)$ using the Kullback-Leibler loss, that is, we aim to minimize 
      \[
      \int \left\{ \int \log \frac{f_Y(y;\theta)}{f(y)}  f_Y(y;\theta) \ d y \right\} \pi_n(\theta) \ d \theta
      \]
      with respect to the function $f(\cdot)$, that is, to maximize
      \[
      \iint \log f(y)  f_Y(y;\theta) \pi_n(\theta) \ d \theta \ d y.
      \]
      Re-writing the double integral
      \[
      \int \log f(y) \left\{\int f_Y(y;\theta) \pi_n(\theta) \ d \theta  \right\} \ dy = \int \log f(y) \ p_n(y) \ dy
      \]
      and maximizing this quantity is the same as minimizing
      \[
      \int \log \frac{p_n(y)}{f(y)} p_n(y) \ d y.
      \]
      Thus by properties of the KL divergence, the optimal $f$ is $p_n(y)$
      \[
      p_n(y) = \int f_Y(y;\theta) \pi_n(\theta) \ d \theta .
      \]
      that is, the posterior predictive distribution. 
\end{enumerate}



\end{document}