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Chapter 11: Problem 15

Show that the \((1-2 \alpha)\) HPD credible interval for a continuous unimodal posterior density \(\pi(\theta \mid y)\) is the shortest credible interval with level \((1-2 \alpha)\).

Short Answer

Expert verified
The HPD interval is the shortest credible interval for a unimodal posterior because it contains the most probable values, strictly above a threshold density, exploiting the unimodality.

Step by step solution

01

Define the Problem

We need to demonstrate that the Highest Posterior Density (HPD) credible interval is the shortest interval for a given level \(1-2\alpha\) when considering a continuous unimodal posterior density \(\pi(\theta \mid y)\). The HPD interval contains the most probable values of \(\theta\) and is defined by the set of values greater than a specific cutoff density.
02

Define the HPD Interval

The HPD interval is defined by \(C = \{ \theta : \pi(\theta \mid y) \geq c \}\) such that \( \int_C \pi(\theta \mid y) \, d\theta = 1 - 2\alpha \). The value \(c\) is chosen so that the posterior distribution integral over \(C\) is exactly \(1-2\alpha\). This ensures that all values inside the interval have a posterior density greater than any value outside of it.
03

Use the Unimodality Property

Since the posterior \(\pi(\theta \mid y)\) is unimodal, it has a single peak. The values of \(\theta\) with the highest posterior density are most centrally located around this peak. The values that fall within the HPD interval are all greater than or equal to \(c\), exploiting the unimodality to minimize the range of the interval.
04

Compare with Other Credible Intervals

Any other interval of the same level \(1-2\alpha\) must have some portions with posterior density less than \(c\) or have an equal density but cannot exclude higher density values included in the HPD interval. This makes any alternative intervals that are not HPD intervals have an equal or longer length.

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Key Concepts

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

Unimodal Posterior Distribution
Understanding the unimodal posterior distribution is key to grasping the concept of credible intervals. A distribution is unimodal when it has a single peak or a highest point, allowing it to neatly summarize which values are most likely. In the context of Bayesian statistics, a posterior distribution like \( \pi(\theta \mid y) \) provides a snapshot of how likely different parameter values \( \theta \) are, given some observed data \( y \). This single peak feature is beneficial because it means that we can assume that as we move away from this peak, the probabilities or likelihood for a given parameter \( \theta \) decrease. The highest probability density close to the peak assures us confidence about our parameter estimates. By knowing that the distribution is unimodal, we also infer that credible intervals around the mode (the location of the peak) will effectively capture the most probable values of \( \theta \). This property is essential for the formation of the HPD credible interval, which we will explore further.
Shortest Credible Interval
The concept of the shortest credible interval plays a crucial role in making parameter estimations more precise and efficient. A credible interval is a range within which an unobserved parameter value falls with a certain probability, as determined by the posterior distribution.The shortest credible interval, especially in the context of the HPD credible interval for a unimodal posterior distribution, is the interval with the smallest possible width that still captures the desired level of probability, \( 1-2\alpha \). Achieving the shortest interval is significant because it ensures that we are reporting the most concise range of parameter values that are most likely given the observed data. This concept is desirable in statistical inference since it provides a tight, reliable estimation which can be crucial when making predictions or decisions based on the data.
Posterior Density
The posterior density function \( \pi(\theta \mid y) \) plays a pivotal role in Bayesian analysis, providing a complete picture of what is known about the parameter \( \theta \) after observing the data \( y \). It combines information from the prior distribution and the likelihood of the observed data. In the context of HPD intervals, the posterior density determines which values of \( \theta \) are included within the credible interval. Specifically, in the HPD interval, we include those values of \( \theta \) where the posterior density is above a certain cutoff level \( c \). This ensures that the interval captures the most credible values, forming the central part of the distribution. The use of posterior density in constructing the HPD interval also means that any point outside of this interval has a lower probability density, making it less likely that such points correspond to the true value of \( \theta \). This effectively isolates and highlights the core range of values that are most supported by the data.
Credible Interval Level
The credible interval level \( 1-2\alpha \) is a parameter in Bayesian statistics that denotes the probability mass that the interval is designed to contain. It's essentially the confidence level of the interval, tailored to Bayesian inference under the posterior distribution. Choosing this level means deciding what percentage of the probability mass of the distribution we are interested in capturing within our interval. The value \( 1-2\alpha \) signifies the portion of the distribution that we consider to be most credible, providing a way to express certainty and manage uncertainty.An interval with a high credible interval level will generally be broader because it aims to cover more of the distribution to ensure that the true parameter lies within it. However, the HPD interval is special in its ability to provide the shortest possible interval for any given credible interval level, thanks to its nature of encompassing the most densely packed, highest-probability values of the distribution.

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Most popular questions from this chapter

Consider predicting the outcome of a future random variable \(Z\) on the basis of a random sample \(Y_{1}, \ldots, Y_{n}\) from density \(\lambda^{-1} e^{-u / \lambda}, u>0, \lambda>0\). Show that \(\pi(\lambda) \propto \lambda^{-1}\) gives posterior predictive density $$ f(z \mid y)=\frac{\int f(z, y \mid \lambda) \pi(\lambda) d \lambda}{\int f(y \mid \lambda) \pi(\lambda) d \lambda}=n s^{n} /(s+z)^{n+1}, \quad z>0 $$ where \(s=y_{1}+\cdots+y_{n}\) Show that when Laplace's method is applied to each integral in the predictive density the result is proportional to the exact answer, and assess how close the approximation is to a density when \(n=5\).

A forensic laboratory assesses if the DNA profile from a specimen found at a crime scene matches the DNA profile of a suspect. The technology is not perfect, as there is a (small) probability \(\rho\) that a match oocurs by chance even if the suspect was not present at the scene, and a (larger) probability \(\gamma\) that a match is reported even if the profiles are different; this can arise due to laboratory error such as cross-contamination or accidental switching of profiles. (a) Let \(R, S\), and \(M\) denotes the events that a match is reported, that the specimen does indeed come from the suspect, and that there is a match between the profiles, and suppose that $$ \operatorname{Pr}(R \mid M \cap S)=\operatorname{Pr}(R \mid M \cap \bar{S})=\operatorname{Pr}(R \mid M)=1, \operatorname{Pr}(\bar{M} \mid S)=0, \operatorname{Pr}(R \mid S)=1 $$ Show that the posterior odds of the profiles matching, given that a match has been reported, depend on $$ \frac{\operatorname{Pr}(R \mid S)}{\operatorname{Pr}(R \mid \bar{S})}=\frac{\operatorname{Pr}(R \mid M \cap S) \operatorname{Pr}(M \mid S)+\operatorname{Pr}(R \mid \bar{M} \cap S) \operatorname{Pr}(\bar{M} \mid S)}{\operatorname{Pr}(R \mid M \cap \bar{S}) \operatorname{Pr}(M \mid \bar{S})+\operatorname{Pr}(R \mid \bar{M} \cap \bar{S}) \operatorname{Pr}(\bar{M} \mid \bar{S})} $$ and establish that this equals \(\\{\rho+\gamma(1-\rho)\\}^{-1}\) (b) Tabulate \(\operatorname{Pr}(R \mid S) / \operatorname{Pr}(R \mid \bar{S})\) when \(\rho=0,10^{-9}, 10^{-6}, 10^{-3}\) and \(\gamma=0,10^{-4}\), \(10^{-3}, 10^{-2}\) (c) At what level of posterior odds would you be willing to convict the suspect, if the only evidence against them was the DNA analysis, and you should only convict if convinced of their guilt 'beyond reasonable doubt'? Would your chosen odds level depend on the likely sentence, if they are found guilty? How does your answer depend on the prior odds of the profiles matching, \(\operatorname{Pr}(S) / \operatorname{Pr}(\bar{S}) ?\)

Show that if \(y_{1}, \ldots, y_{n}\) is a random sample from an exponential family with conjugate prior \(\pi(\theta \mid \lambda, m)\), any finite mixture of conjugate priors, $$ \sum_{j=1}^{k} p_{j} \pi\left(\theta, \lambda_{j}, m_{j}\right), \quad \sum_{j} p_{j}=1, p_{j} \geq 0 $$ is also conjugate. Check the details when \(y_{1}, \ldots, y_{n}\) is a random sample from the Bernoulli distribution with probability \(\theta\).

Two balls are drawn successively without replacement from an urn containing three white and two red balls. Are the outcomes of the first and second draws independent? Are they exchangeable?

Suppose that \(\left(U_{1}, U_{2}\right)\) lies in a product space, of form \(U_{1} \times U_{2}\) (a) Show that $$ \pi\left(u_{1}\right)=\frac{\pi\left(u_{1} \mid u_{2}\right)}{\pi\left(u_{2} \mid u_{1}\right)} \pi\left(u_{2}\right), \quad \text { for any } u_{1} \in \mathcal{U}_{1}, u_{2} \in \mathcal{U}_{2} $$ and deduce that for each \(u_{2} \in \mathcal{U}_{2}\) and an arbitrary \(u_{1}^{\prime} \in \mathcal{U}_{1}\), $$ \pi\left(u_{2}\right)=\left\\{\int \frac{\pi\left(u_{1} \mid u_{2}\right)}{\pi\left(u_{2} \mid u_{1}\right)} d u_{1}\right\\}^{-1}=\frac{\pi\left(u_{2} \mid u_{1}^{\prime}\right)}{\pi\left(u_{1}^{\prime} \mid u_{2}\right)}\left\\{\int \frac{\pi\left(u_{2} \mid u_{1}^{\prime}\right)}{\pi\left(u_{1}^{\prime} \mid u_{2}\right)} d u_{2}\right\\}^{-1} $$ (b) If \(U_{2}^{1}, \ldots, U_{2}^{S}\) is a random sample from \(\pi\left(u_{2} \mid u_{1}^{\prime}\right)\), show that $$ \widehat{\pi}\left(u_{2}\right)=\frac{\pi\left(u_{2} \mid u_{1}^{\prime}\right)}{\pi\left(u_{1}^{\prime} \mid u_{2}\right)}\left\\{S^{-1} \sum_{s=1}^{s} \pi\left(u_{1}^{\prime} \mid U_{2}^{s}\right)^{-1}\right\\}^{-1} \stackrel{P}{\longrightarrow} \pi\left(u_{2}\right) \text { as } S \rightarrow \infty $$ (c) Verify that the code below applies this approach to the bivariate normal model in Example \(11.21\) Does this work well? Why not? Try with \(u_{1}^{\prime}=-2,-1,0\). What lesson does this example suggest for the use of this approach in general?
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