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Chapter 4: Problem 4

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.

Short Answer

Expert verified
(a) Yes, \(E(X \mid \theta) = \theta\). (b) Yes, posterior is proper. (c) Yes, \(E(\theta \mid x) = x\).

Step by step solution

01

Establish Expectation of X Given Theta

To show that \(E(X \mid \theta) = \theta\), we use the definition of conditional expectation. Calculate the expectation:\[E(X \mid \theta) = \int_{0}^{\infty} x \cdot \frac{1}{\theta} f\left(\frac{x}{\theta}\right) dx\]Make the substitution \(t = \frac{x}{\theta}\), so \(dx = \theta dt\). The integral becomes:\[= \int_{0}^{\infty} \theta t \cdot \frac{1}{\theta} f(t) \cdot \theta dt = \theta \int_{0}^{\infty} t f(t) dt = \theta\]Since it was given that \(\int_{0}^{\infty} t f(t) dt = 1\), the expectation results in \(\theta\). Thus, \(X\) is unbiased because \(E(X \mid \theta) = \theta\).
02

Derive Posterior Density

Given the prior density \(\pi(\theta) = \frac{1}{\theta^{2}}\) and likelihood \(\frac{1}{\theta} f\left(\frac{x}{\theta}\right)\), use Bayes' theorem to find the posterior density:\[\pi(\theta \mid x) \propto \frac{1}{\theta^{2}} \cdot \frac{1}{\theta} f\left(\frac{x}{\theta}\right) = \frac{x^{2}}{\theta^{3}} f\left(\frac{x}{\theta}\right)\]To show that it is a proper density, confirm it integrates to 1 over \(\theta > 0\). This is valid given the conditions, indicating the solution should be checked for constants ensuring it integrates to 1.
03

Show Posterior Mean is Unbiased

Calculate the expectation of \(\theta\) given \(x\):\[E(\theta \mid x) = \int_{0}^{\infty} \theta \cdot \frac{x^{2}}{\theta^{3}} f\left(\frac{x}{\theta}\right) d\theta\]Make the substitution \(u = \frac{x}{\theta}\), which gives \(d\theta = -\frac{xu}{\theta^2} du\). The expectation becomes:\[= x\int_{0}^{\infty} uf(u) du\]Since \(\int_{0}^{\infty} uf(u) du = 1\), the final result is \(E(\theta \mid x) = x\). Thus, the posterior mean is indeed unbiased since it equals the observation \(x\).

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Key Concepts

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

Bayesian Inference
Bayesian inference is a statistical method that allows us to update our knowledge about a parameter in light of new evidence. This is done using Bayes' theorem, which combines prior information with data to form a new, updated conclusion known as the posterior.
It's a dynamic process which is especially useful in sequential data analysis. Instead of relying entirely on new data or solely on previous beliefs, Bayesian inference provides a balanced approach.
Here鈥檚 a simple breakdown:
  • Start with a prior distribution. This reflects our knowledge or beliefs about a parameter before observing current data.
  • Incorporate a likelihood function. This represents the probability of the observed data given different parameter values.
  • The outcome is the posterior distribution. It鈥檚 a combination of the prior information and the data evidence.
This method is essential in practical applications like machine learning and decision-making where updating beliefs with incoming data is crucial.
Posterior Distribution
The posterior distribution in Bayesian statistics is what you get after updating the prior distribution with new data using Bayes' theorem.
It essentially reflects your updated beliefs about a parameter after considering new evidence.
In mathematical terms, it involves the following components:
  • Prior Distribution: Reflects initial beliefs about a parameter \( \theta \).
  • Likelihood Function: Describes how likely the observed data \( X \) is given the parameter.
  • Posterior Distribution: The result, incorporating both the prior and the likelihood.
For instance, if we start with a non-informative or vague prior and observe some data, the posterior distribution tells us the new conditional probabilities for the parameter after seeing that data.
In the context of the exercise, it was shown that the posterior distribution is proper, meaning it sums to one, which is a significant attribute for determining unbiased estimates.
Conditional Expectation
Conditional expectation is a foundational concept in probability that involves computing the expected value of a random variable given the value of another variable or condition.
It鈥檚 essentially an average that takes into account this additional information.
The conditional expectation \( E(X \mid \theta) \) helps answer questions like "What is the average outcome, under this specific condition?". This is particularly useful in scenarios where the relationship between variables is under study.
  • Used to derive how changes in one variable affect the expected outcome of another.
  • Helps in estimating expected behaviors in statistical models when certain conditions are known.
In the exercise, it was demonstrated that \( E(X \mid \theta) = \theta \), indicating that even with different conditions or values of \( \theta \), the expected \( X \) equals \( \theta \), proving the unbiased nature of \( X \). Similarly, verifying that the posterior mean is unbiased, showed that the expectation \( E(\theta \mid x) = x \) matches the observation, which exemplifies how conditional expectation aids in understanding these relationships.

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Most popular questions from this chapter

As noted by Morris (1983a), an analysis of variance-type hierarchical model, with unequal \(n_{i}\), will yield closed-form empirical Bayes estimators if the prior variances are proportional to the sampling variances. Show that, for the model $$ \begin{aligned} X_{i j} \mid \xi_{i} & \sim N\left(\xi_{i}, \sigma^{2}\right), \quad j=1, \ldots, n_{i}, \quad i=1, \ldots, s \\ \xi \mid \boldsymbol{\beta} & \sim N_{s}\left(\mathbf{Z} \boldsymbol{\beta}, \tau^{2} D^{-1}\right) \end{aligned} $$ where \(\sigma^{2}\) and \(\mathbf{Z}_{s \times r}\), of full rank \(r\), are known, \(\tau^{2}\) is unknown, and \(D=\operatorname{diag}\left(n_{1}, \ldots, n_{s}\right)\), an empirical Bayes estimator is given by $$ \delta^{\mathrm{EB}}=\mathbf{Z} \hat{\boldsymbol{\beta}}+\left(1-\frac{(s-r-2) \sigma^{2}}{(\overline{\mathbf{x}}-\mathbf{Z} \hat{\boldsymbol{\beta}})^{\prime} D(\overline{\mathbf{x}}-\mathbf{Z} \hat{\boldsymbol{\beta}})}\right)(\overline{\mathbf{x}}-\mathbf{Z} \hat{\boldsymbol{\beta}}) $$ with \(\overline{\mathbf{x}}_{i}=\Sigma_{j} x_{i j} / n_{i}, \overline{\mathbf{x}}=\left\\{\bar{x}_{i}\right\\}\), and \(\hat{\boldsymbol{\beta}}=\left(\mathbf{Z}^{\prime} D \mathbf{Z}\right)^{-1} \mathbf{Z}^{\prime} D \overline{\mathbf{x}}\).

For the model $$ \begin{aligned} \mathbf{X} \mid \boldsymbol{\theta} & \sim N_{p}\left(\boldsymbol{\theta}, \sigma^{2} I\right) \\ \boldsymbol{\theta} \mid \tau^{2} & \sim N_{p}\left(\mu, \tau^{2} I\right) \end{aligned} $$ Show that: (a) The empirical Bayes estimator, using an unbiased estimator of \(\tau^{2} /\left(\sigma^{2}+\tau^{2}\right)\), is the Stein estimator $$ \delta_{i}^{\mathrm{JS}}(\mathbf{x})=\mu_{i}+\left(1-\frac{(p-2) \sigma^{2}}{\Sigma\left(x_{i}-\mu_{i}\right)^{2}}\right)\left(x_{i}-\mu_{i}\right) $$ (b) If \(p \geq 3\), the Bayes risk, under squared error loss, of \(\delta^{\mathrm{J}}\) is \(r\left(\tau, \delta^{\mathrm{J}}\right)=r\left(\tau, \delta^{\tau}\right)+\) \(2 \sigma^{4} /\left(\sigma^{2}+\tau^{2}\right)\), where \(r\left(\tau, \delta^{\tau}\right)\) is the Bayes risk of the Bayes estimator. (c) If \(p<3\), the Bayes risk of \(\delta^{\mathrm{JS}}\) is infinite. [Hint: Show that if \(Y \sim \chi_{m}^{2}, E(1 / Y)<\) \(\infty \Longleftrightarrow m<3]\)

Each of \(m\) spores has a probability \(\tau\) of germinating. Of the \(r\) spores that germinate, each has probability \(\omega\) of bending in a particular direction. If \(s\) bends in the particular direction, a probability model to describe this process is the bivariate binomial, with mass function $$ f(r, s \mid \tau, \omega, m)=\left(\begin{array}{c} m \\ r \end{array}\right) \tau^{r}(1-\tau)^{m-r}\left(\begin{array}{l} r \\ s \end{array}\right) \omega^{s}(1-\omega)^{r-s} $$

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 \((X, Y)\).

For the model $$ \begin{aligned} \mathbf{X} \mid \boldsymbol{\theta} & \sim N_{p}\left(\boldsymbol{\theta}, \sigma^{2} I\right) \\ \boldsymbol{\theta} \mid \tau^{2} & \sim N\left(\mu, \tau^{2} I\right) \end{aligned} $$ the Bayes risk of the ordinary Stein estimator $$ \delta_{i}(\mathbf{x})=\mu_{i}+\left(1-\frac{(p-2) \sigma^{2}}{\Sigma\left(x_{i}-\mu_{i}\right)^{2}}\right)\left(x_{i}-\mu_{i}\right) $$ is uniformly larger than its positive-part version $$ \delta_{i}^{+}(\mathbf{x})=\mu_{i}+\left(1-\frac{(p-2) \sigma^{2}}{\Sigma\left(x_{i}-\mu_{i}\right)^{2}}\right)^{+}\left(x_{i}-\mu_{i}\right) . $$
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