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Chapter 4: Problem 9

To understand the convergence of the Gibbs sampler, let \((X, Y) \sim f(x, y)\), and define $$ k\left(x, x^{\prime}\right)=\int f_{X \mid Y}(x \mid y) f_{Y \mid X}\left(y \mid x^{\prime}\right) d y $$ (a) Show that the function \(h^{*}(\cdot)\) that solves \(h^{*}(x)=\int k\left(x, x^{\prime}\right) h^{*}\left(x^{\prime}\right) d x^{\prime}\) is \(h^{*}(x)=\) \(f_{X}(x)\), the marginal distribution of \(X\). (b) Write down the analogous integral equation that is solved by \(f_{Y}(y)\). (c) Define a sequence of functions recursively by \(h_{i+1}(x)=\int k\left(x, x^{\prime}\right) h_{i}\left(x^{\prime}\right) d x_{I}^{\prime}\) where \(h_{0}(x)\) is arbitrary but satisfies \(\sup _{x}\left|\frac{h_{0}(x)}{h^{*}(x)}\right|<\infty\). Show that $$ \int\left|h_{i+1}(x)-h^{*}(x)\right| d x<\int\left|h_{i}(x)-h^{*}(x)\right| d x $$ and, hence, \(h_{i}(x)\) converges to \(h^{*}(x)\). [The method of part (c) is called successive substitution. When there are two variables in the Gibbs sampler, it is equivalent to data augmentation (Tanner and Wong 1987). Even if the variables are vector-valued, the above results establish convergence. If the original vector of variables contains more than two variables, then a more general version of this argument is needed (Gelfand and Smith 1990).]

Short Answer

Expert verified
(a) \( h^*(x) = f_X(x) \). (b) \( h'(y) = f_Y(y) \). (c) \( h_i(x) \) converges to \( h^*(x) \).

Step by step solution

01

Define the function

To solve part (a), we need the function \( h^*(x) \) that satisfies the given integral equation: \[ h^*(x) = \int k\left(x, x'\right) h^*\left(x'\right) \, dx' \].The kernel \( k(x, x') \) is already defined as\[ k(x, x') = \int f_{X|Y}(x|y) f_{Y|X}(y|x') \, dy \].Our aim is to show that \( h^*(x) = f_X(x) \), the marginal distribution of \( X \), satisfies this equation.
02

Marginalize Joint Distribution

Recall that the joint distribution \( f(x, y) \) equals \( f_{X|Y}(x|y)f_Y(y) \) and equivalently \( f_{Y|X}(y|x')f_X(x') \). Thus, the marginal distribution of \( X \) is obtained by integrating out \( Y \):\[ f_X(x) = \int f(x, y) \, dy = \int f_{X|Y}(x|y)f_Y(y) \, dy \].
03

Substitute and Simplify

Substitute \( f_{X|Y}(x|y)f_Y(y) = f(x, y) \) into the right-hand side of the integral equation,\[ \int k(x, x') f_X(x') \, dx' = \int \left ( \int f_{X|Y}(x|y)f_{Y|X}(y|x') \, dy \right )f_X(x') \, dx' \].By Fubini鈥檚 theorem, change the order of integration,\[ \int \int f_{X|Y}(x|y)f_{Y|X}(y|x') f_X(x') \, dx' \, dy = \int f_{X|Y}(x|y) \left( \int f_{Y|X}(y|x')f_X(x') \, dx' \right) \, dy \].
04

Verify the Marginalization

The term \( \int f_{Y|X}(y|x')f_X(x') \, dx' \) simplifies to \( f_Y(y) \). Therefore the integral becomes\[ \int f_{X|Y}(x|y)f_Y(y) \, dy = f_X(x) \], which confirms \( h^*(x) = f_X(x) \). Thus, the equation is satisfied with \( h^*(x) = f_X(x) \).
05

Formulate Integral Equation for f_Y(y)

For part (b), the analogous equation for \( f_Y(y) \) can be written by switching \( X \) and \( Y \):\[ k'(y, y') = \int f_{Y|X}(y|x) f_{X|Y}(x|y') \, dx \],\[ h'(y) = \int k'(y, y') f_Y(y') \, dy', \]which is solved by \( h'(y) = f_Y(y) \).
06

Define Successive Substitution

For part (c), define a sequence \( h_{i+1}(x) = \int k(x, x') h_i(x') \, dx' \). We need to show\[ \int \left|h_{i+1}(x) - h^*(x)\right| \, dx < \int \left|h_i(x) - h^*(x)\right| \, dx \].
07

Establish Convergence

Assume \( h_0(x) \) satisfies \( \sup_x \left| \frac{h_0(x)}{h^*(x)} \right| < \infty \). By defining the error term \( e_i(x) = h_i(x) - h^*(x) \), expand into the error integral:\[ h_{i+1}(x) - h^*(x) = \int k(x, x')[h_i(x') - h^*(x')] \, dx' = \int k(x, x')e_i(x') \, dx' \].Applying integral properties,\[ \int \left| \, h_{i+1}(x) - h^*(x) \, \right| \, dx \] should decrease as it integrates over kernel \( k \), signifying convergence.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convergence
Convergence in the context of a Gibbs sampler is essential because it tells us how well the algorithm approximates the target distribution after several iterations. When we say that a sequence of functions converges, it means that each successive term gets closer and closer to a specified function or value. In our case, the sequence of functions \( h_i(x) \) converges to \( h^*(x) = f_X(x) \), the marginal distribution of \( X \). This is proven by demonstrating that the integral \(\int |h_{i+1}(x) - h^*(x)| \, dx < \int |h_i(x) - h^*(x)| \, dx\) consistently decreases with each iteration.
  • Starts with an arbitrary function \( h_0(x) \) that bounds the ratio \( \frac{h_0(x)}{h^*(x)} \).
  • Each iteration refines the function \ h_i(x) \ closer to \ h^*(x) \.
  • Eventually, the difference between \( h_i(x) \) and \( h^*(x) \) becomes negligible, indicating convergence.
Marginal Distribution
The marginal distribution is a fundamental concept in probability and statistics. It refers to the probability distribution of a subset of variables within a joint distribution. In simpler terms, it's the distribution of one variable without considering the others. For the exercise, the marginal distribution of \( X \), denoted as \( f_X(x) \) is derived by integrating the joint distribution \( f(x, y) \) over all possible values of \( y \).\[ f_X(x) = \int f(x, y) \, dy \]This expression indicates that the marginal distribution is obtained by summing up all probabilities along the \( y \) axis. Essentially, this provides insight into the behavior of \( X \) without factoring in \( Y \). The Gibbs sampler relies on understanding these marginal distributions to refine estimates with each sampling iteration.
Successive Substitution
Successive substitution is a technique used to solve complex equations by iteratively substituting approximations into an equation to get closer to a solution. In the context of the Gibbs sampler, the method starts with an initial guess, \( h_0(x) \), and iteratively refines it through the relation:\[ h_{i+1}(x) = \int k(x, x') h_i(x') \, dx' \]Here, \( k(x, x') \) is the kernel function that maps the old distribution to the new one. Each application of this integral gradually adjusts \( h_i(x) \) until it closely approximates \( h^*(x) = f_X(x) \).
  • Initial function \( h_0(x) \) doesn't need to be specific, but it must be bounded.
  • With each iteration, the adjustment term reduces error.
  • Successive substitution mirrors how Gibbs sampling approaches target distributions iteratively.
Integral Equations
Integral equations involve an unknown function under an integral sign, providing a relationship between this unknown function and given data. In the Gibbs sampler, understanding integral equations is crucial because they express specific constraints linking distributions. In the exercise, we need to show that \( h^*(x) \) satisfies:\[ h^*(x) = \int k(x, x') h^*(x') \, dx' \]This integral equation encodes the connection between the marginal distribution \( f_X(x) \) and the conditional distributions \( f_{X|Y}(x|y) \) and \( f_{Y|X}(y|x') \).
  • Connects marginal distributions to underlying joint probabilities.
  • Critical to proving the convergence of the Gibbs sampler.
  • Mathematical manipulation often involves Fubini鈥檚 theorem for changing integration order.

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Most popular questions from this chapter

An entertaining (and unjustifiable) result which abuses a hierarchical Bayes calculation yields the following derivation of the James-Stein estimator. Let \(X \sim N_{p}(\theta, I)\) and \(\boldsymbol{\theta} \mid \tau^{2} \sim N_{p}\left(0, \tau^{2} I\right)\) (a) Verify that conditional on \(\tau^{2}\), the posterior and marginal distributions are given by $$ \begin{aligned} \pi\left(\boldsymbol{\theta} \mid \mathbf{x}, \tau^{2}\right) &=N_{p}\left(\frac{\tau^{2}}{\tau^{2}+1} \mathbf{x}, \frac{\tau^{2}}{\tau^{2}+1} I\right) \\ m\left(\mathbf{x} \mid \tau^{2}\right) &=N_{p}\left[0,\left(\tau^{2}+1\right) I\right] \end{aligned} $$ (b) Show that, taking \(\pi\left(\tau^{2}\right)=1,-1<\tau^{2}<\infty\), we have $$ \begin{aligned} &\iint_{\Re p} \boldsymbol{\theta} \pi\left(\boldsymbol{\theta} \mid \mathbf{x}, \tau^{2}\right) m\left(\mathbf{x} \mid \tau^{2}\right) d \boldsymbol{\theta} d \tau^{2} \\ &=\frac{\mathbf{x}}{(2 \pi)^{p / 2}\left(|\mathbf{x}|^{2}\right)^{p / 2-1}}\left[\Gamma\left(\frac{p-2}{2}\right) 2^{(p-2) / 2}-\frac{\Gamma(p / 2) 2^{p / 2}}{|\mathbf{x}|^{2}}\right] \end{aligned} $$ and $$ \begin{aligned} &\iint_{\Re P P} \pi\left(\boldsymbol{\theta} \mid \mathbf{x}, \tau^{2}\right) m\left(\mathbf{x} \mid \tau^{2}\right) d \theta d \tau^{2} \\ &=\frac{1}{(2 \pi)^{p / 2}\left(|\mathbf{x}|^{2}\right)^{p / 2-1}} \Gamma\left(\frac{p-2}{2}\right) 2^{(p-2) / 2} \end{aligned} $$ and hence $$ E(\boldsymbol{\theta} \mid \mathbf{x})=\left(1-\frac{p-2}{|\mathbf{x}|^{2}}\right) \mathbf{x} $$ (c) Explain some implications of the result in part (b) and, why it cannot be true. [Try to reconcile it with (3.3.12).] (d) Why are the calculations in part (b) unjustified?

Bickel and Mallows (1988) further investigate the relationship between unbiasedness and Bayes, specifying conditions under which these properties cannot hold simultaneously. In addition, they show that if a prior distribution is improper, then a posterior mean can be unbiased. Let \(X \sim \frac{1}{\theta} f(x / \theta), x>0\), where \(\int_{0}^{\infty} t f(t) d t=1\), and let \(\pi(\theta)=\frac{1}{\theta^{2}} d \theta\) \(\theta>0\) (a) Show that \(E(X \mid \theta)=\theta\), so \(X\) is unbiased. (b) Show that \(\pi(\theta \mid x)=\frac{x^{2}}{\theta^{3}} f(x / \theta)\) is a proper density. (c) Show that \(E(\theta \mid x)=x\), and hence the posterior mean, is unbiased.

Let \(X\) and \(Y\) be independently distributed according to distributions \(P_{\xi}\) and \(Q_{\eta}\), respectively. Suppose that \(\xi\) and \(\eta\) are real-valued and independent according to some prior distributions \(\Lambda\) and \(\Lambda^{\prime} .\) If, with squared error loss, \(\delta_{\Lambda}\) is the Bayes estimator of \(\xi\) on the basis of \(X\), and \(\delta_{\Lambda^{\prime}}^{\prime}\) is that of \(\eta\) on the basis of \(Y\), (a) show that \(\delta_{\Lambda^{\prime}}^{\prime}-\delta_{\Lambda}\) is the Bayes estimator of \(\eta-\xi\) on the basis of \((X, Y)\); (b) if \(\eta>0\) and \(\delta_{\Lambda^{\prime}}^{*}\) is the Bayes estimator of \(1 / \eta\) on the basis of \(Y\), show that \(\delta_{\Lambda} \cdot \delta_{\Lambda^{\prime}}^{*}\) is the Bayes estimator of \(\xi / \eta\) on the basis of \((X, Y)\).

Let \(\mathcal{F}=\\{f(x \mid \theta) ; \theta \in \Omega\\}\) be a family of probability densities. The Kullback-Leibler information for discrimination between two densities in \(\mathcal{F}\) can be written $$ \psi\left(\theta_{1}, \theta_{2}\right)=\int f\left(x \mid \theta_{1}\right) \log \left[\frac{f\left(x \mid \theta_{1}\right)}{f\left(x \mid \theta_{2}\right)}\right] d x $$ Recall that the gradient of \(\psi\) is \(\nabla \psi=\left\\{\left(\partial / \partial \theta_{i}\right) \psi\right\\}\) and the Hessian is \(\nabla \nabla \psi=\) \(\left\\{\left(\partial^{2} / \partial \theta_{i} \partial \theta_{j}\right) \psi\right\\}\) (a) If integration and differentiation can be interchanged, show that $$ \nabla \psi(\theta, \theta)=0 \quad \text { and } \quad \operatorname{det}[\nabla \nabla \psi(\theta, \theta)]=I(\theta) $$ where \(I(\theta)\) is the Fisher information of \(f(x \mid \theta)\). (b) George and McCulloch (1993) argue that choosing \(\pi(\theta)=(\operatorname{det}[\nabla \nabla \psi(\theta, \theta)])^{1 / 2}\) is an appealing least informative choice of priors. What justification can you give for this?

The Taylor series approximation to the estimator ( \(5.5 .8\) ) is carried out in a number of steps. Show that: (a) Using a first-order Taylor expansion around the point \(\bar{x}\), we have $$ \begin{aligned} \frac{1}{\left(1+\theta^{2} / v\right)^{(v+1) / 2}}=& \frac{1}{\left(1+\bar{x}^{2} / \nu\right)^{(v+1) / 2}} \\ &-\frac{v+1}{v} \frac{\bar{x}}{\left(1+\bar{x}^{2} / v\right)^{(v+3) / 2}}(\theta-\bar{x})+R(\theta-\bar{x}) \end{aligned} $$ where the remainder, \(R(\theta-\bar{x})\), satisfies \(R(\theta-\bar{x}) /(\theta-\bar{x})^{2} \rightarrow 0\) as \(\theta \rightarrow \bar{x}\). (b) The remainder in part (a) also satisfies $$ \int_{-\infty}^{\infty} R(\theta-\bar{x}) e^{-\frac{p}{2 \sigma^{2}}(\theta-\bar{x})^{2}} d \theta=O\left(1 / p^{3 / 2}\right) $$ (c) The numerator and denominator of \((5.5 .8)\) can be written $$ \int_{-\infty}^{\infty} \frac{1}{\left(1+\theta^{2} / \nu\right)^{(\nu+1) / 2}} e^{-\frac{p}{2 \sigma^{2}}(\theta-\bar{x})^{2}} d \theta=\frac{\sqrt{2 \pi \sigma^{2} / p}}{\left(1+\bar{x}^{2} / \nu\right)^{(\nu+1) / 2}}+O\left(\frac{1}{p^{3 / 2}}\right) $$ and $$ \begin{aligned} \int_{-\infty}^{\infty} & \frac{\theta}{\left(1+\theta^{2} / \nu\right)^{(v+1) / 2}} e^{-\frac{p}{2 \sigma^{2}}(\theta-\bar{x})^{2}} d \theta \\ &=\frac{\sqrt{2 \pi \sigma^{2} / p}}{\left(1+\bar{x}^{2} / \nu\right)^{(v+1) / 2}}\left[1-\frac{(v+1) / v}{\left(1+\bar{x}^{2} / v\right)}\right] \bar{x}+O\left(\frac{1}{p^{3 / 2}}\right) \end{aligned} $$ which yields \((5.6 .32)\).
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