/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Noninformative prior densities: ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Noninformative prior densities: (a) For the binomial likelihood, \(y \sim \operatorname{Bin}(n, \theta)\), show that \(p(\theta) \propto \theta^{-1}(1-\) \(\theta)^{-1}\) is the uniform prior distribution for the natural parameter of the exponential family. (b) Show that if \(y=0\) or \(n\), the resulting posterior distribution is improper.

Short Answer

Expert verified
For (a), the noninformative prior is \( p(\theta) \propto \theta^{-1}(1-\theta)^{-1} \). For (b), the posterior is improper if \( y = 0 \) or \( y = n \).

Step by step solution

01

Identify the Exponential Family Form

The binomial likelihood can be written as: \[ L( heta) = \binom{n}{y} \theta^y (1-\theta)^{n-y} \]This is part of the exponential family as it can be expressed in the form: \[ L(\theta) = \exp \left( y \ln \left( \frac{\theta}{1-\theta} \right) + n \ln (1-\theta) + \ln \binom{n}{y} \right) \]where \( \theta \) transforms as \( \eta = \ln \left( \frac{\theta}{1-\theta} \right) \), identifying \( \eta \) as the natural parameter.
02

Derive Noninformative Prior for Natural Parameter

The Jacobian of the transformation between \( \theta \) and \( \eta \) is \( \frac{d\theta}{d\eta} = \theta(1-\theta) \). The noninformative prior for the natural parameter is uniform over the range of \( \eta \), leading to:\[ p(\theta) \propto |\theta(1-\theta)|^{-1} = \theta^{-1}(1-\theta)^{-1} \]This confirms that \( p(\theta) \propto \theta^{-1}(1-\theta)^{-1} \) is the uniform prior for \( \eta \).
03

Analyze Posterior for Extreme Observations

The posterior distribution given observations \( y = 0 \) or \( y = n \) and prior \( p(\theta) \propto \theta^{-1}(1-\theta)^{-1} \) can be expressed as:\[ p(\theta | y) \propto \theta^y(1-\theta)^{n-y} \times \theta^{-1}(1-\theta)^{-1} \]For \( y = 0 \):\[ p(\theta | y = 0) \propto (1-\theta)^{n-1} \times \theta^{-1} \]For \( y = n \):\[ p(\theta | y = n) \propto \theta^{n-1} \times (1-\theta)^{-1} \]Both forms are improper because they integrate to infinity over the interval \( (0,1) \).
04

Conclusion about Improper Posterior

The posterior is improper for \( y = 0 \) and \( y = n \) since:- For \( y = 0 \), the integral of \( (1-\theta)^{n-1}/\theta \) diverges as \( \theta \to 0 \).- For \( y = n \), the integral of \( \theta^{n-1}/(1-\theta) \) diverges as \( \theta \to 1 \).Thus, the posterior distributions are improper in both cases.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Exponential Family
The exponential family of probability distributions is a broad class of distributions that can be neatly characterized using a specific mathematical form. Many commonly used distributions like normal, binomial, and Poisson fall into this category. A distribution belongs to the exponential family if its probability density function (PDF) or probability mass function (PMF) can be expressed as follows:
\[ P(y | \theta) = h(y) \exp \left( \eta T(y) - A(\eta) \right) \]
where \(\theta\) is the parameter, \(\eta\) represents the natural parameter, \(T(y)\) is a sufficient statistic for the parameter, and \(A(\eta)\) is the log-partition function ensuring the PDF/PMF integrates to 1 over its domain.

In the case of the binomial distribution, with likelihood \(y \sim \operatorname{Bin}(n, \theta)\), it can be rewritten in exponential family form:
  • \(T(y) = y\)
  • \(\eta = \ln \left( \frac{\theta}{1-\theta} \right)\) – the transformation from \(\theta\) to the natural parameter.
This transformation is key in simplifying analyses in statistics, including finding priors, like the noninformative prior.
Binomial Likelihood
In statistics, the likelihood function represents the probability of observed data given a set of parameter values. For the binomial distribution, we consider the likelihood of observing \(y\) successes in \(n\) trials often expressed as:
\[ L(\theta) = \binom{n}{y} \theta^y (1-\theta)^{n-y} \]
This expression takes form under a binomial distribution assumption, showcasing the probability of observing exactly \(y\) successes out of \(n\) independent trials each with success probability \(\theta\).

In the context of the exponential family, this likelihood can be rewritten using natural parameters, providing a neat mathematical format for further analyses like determining the noninformative prior. One crucial feature for this transformation is the Jacobian, \(\frac{d\theta}{d\eta}\), showing the rate of change from \(\theta\) to its natural counterpart. This aids in finding a uniform prior over \(\eta\), confirming \(p(\theta) \propto \theta^{-1}(1-\theta)^{-1}\) for a noninformative prior.
Improper Posterior
In Bayesian statistics, the posterior distribution represents the updated beliefs about a parameter \(\theta\) after observing data. It combines prior beliefs with the likelihood, but not all resulting posteriors are proper. Proper posteriors integrate to one, but improper ones do not, potentially causing issues in statistical inference.

For a binomial likelihood with noninformative prior \(p(\theta) \propto \theta^{-1}(1-\theta)^{-1}\), if we observe data with extreme outcomes like \(y = 0\) or \(y = n\), we face an improper posterior. For \(y = 0\), the posterior becomes:
\[ p(\theta | y = 0) \propto (1-\theta)^{n-1} \times \theta^{-1} \]
And for \(y = n\):
\[ p(\theta | y = n) \propto \theta^{n-1} \times (1-\theta)^{-1} \]
In both scenarios, integration over \((0,1)\) leads to an undefined behavior signaling the posterior's impropriety: it diverges. This is critical to note, as bayesian inference requires proper posteriors for meaningful statistical insights, implying special care or alternative techniques are needed when encountering these situations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Beta-binomial distribution and Bayes' prior distribution: suppose \(y\) has a binomial distribution for given \(n\) and unknown parameter \(\theta\), where the prior distribution of \(\theta\) is \(\operatorname{Beta}(\alpha, \beta)\). (a) Find \(p(y)\), the marginal distribution of \(y\), for \(y=0, \ldots, n\) (unconditional on \(\theta\) ). This discrete distribution is known as the beta- binomial, for obvious reasons. (b) Show that if the beta-binomial probability is constant in \(y\), then the prior distribution has to have \(\alpha=\beta=1\).

Computing with a nonconjugate single-parameter model: suppose \(y_{1}, \ldots, y_{5}\) are independent samples from a Cauchy distribution with unknown center \(\theta\) and known scale 1: \(p\left(y_{i} \mid \theta\right) \propto 1 /\left(1+\left(y_{i}-\theta\right)^{2}\right) .\) Assume, for simplicity, that the prior distribution for \(\theta\) is uniform on \([0,1]\). Given the observations \(\left(y_{1}, \ldots, y_{5}\right)=(-2,-1,0,1.5,2.5)\) (a) Compute the unnormalized posterior density function, \(p(\theta) p(y \mid \theta)\), on a grid of points \(\theta=0, \frac{1}{m}, \frac{2}{m}, \ldots, 1\), for some large integer \(m\). Using the grid approximation, compute and plot the normalized posterior density function, \(p(\theta \mid y)\), as a function of \(\theta\). (b) Sample 1000 draws of \(\theta\) from the posterior density and plot a histogram of the draws. (c) Use the 1000 samples of \(\theta\) to obtain 1000 samples from the predictive distribution of a future observation, \(y_{6}\), and plot a histogram of the predictive draws.

Setting parameters for a beta prior distribution: suppose your prior distribution for \(\theta\), the proportion of Californians who support the death penalty, is beta with mean \(0.6\) and standard deviation \(0.3 .\) (a) Determine the parameters \(\alpha\) and \(\beta\) of your prior distribution. Sketch the prior density function. (b) A random sample of 1000 Californians is taken, and \(65 \%\) support the death penalty. What are your posterior mean and variance for \(\theta\) ? Draw the posterior density function.

Predictive distributions: let \(y\) be the number of 6 's in 1000 independent rolls of a particular real die, which may be unfair. Let \(\theta\) be the probability that the die lands on '6.' Suppose your prior distribution for \(\theta\) is as follows: $$ \begin{aligned} \operatorname{Pr}(\theta=1 / 12) &=0.25 \\ \operatorname{Pr}(\theta=1 / 6) &=0.5 \\ \operatorname{Pr}(\theta=1 / 4) &=0.25 \end{aligned} $$ (a) Using the normal approximation for the conditional distributions, \(p(y \mid \theta)\), sketch your approximate prior predictive distribution for \(y\). (b) Give approximate \(5 \%, 25 \%, 50 \%, 75 \%\), and \(95 \%\) points for the distribution of \(y .\) (Be careful here: \(y\) does not have a normal distribution, but you can still use the normal distribution as part of your analysis.)

Predictive distributions: consider two coins, \(C_{1}\) and \(C_{2}\), with the following characteristics: \(\operatorname{Pr}\left(\right.\) heads \(\left.\mid C_{1}\right)=0.6\) and \(\operatorname{Pr}\left(\right.\) heads \(\left.\mid C_{2}\right)=0.4\). Choose one of the coins at random and imagine spinning it repeatedly. Given that the first two spins from the chosen coin are tails, what is the expectation of the number of additional spins until a head shows up?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.