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Two independent samples \(Y_{1}, \ldots, Y_{n} \stackrel{\text { iid }}{\sim} N\left(\mu, \sigma^{2}\right)\) and \(X_{1}, \ldots, X_{m} \stackrel{\text { iid }}{\sim} N\left(\mu, c \sigma^{2}\right)\) are available, where \(c>0\) is known. Find posterior densities for \(\mu\) and \(\sigma\) based on prior \(\pi(\mu, \sigma) \propto 1 / \sigma\).

Short Answer

Expert verified
The posterior for \( \mu \) is normal, and for \( \sigma^2 \), it is inverse gamma.

Step by step solution

01

Define the Likelihoods

The likelihood for the first sample \( Y_1, \ldots, Y_n \) is a normal distribution \( N(\mu, \sigma^2) \) for each \( Y_i \). Therefore, the joint likelihood for all \( Y_i \) is: \[ L_Y(\mu, \sigma^2) = \prod_{i=1}^{n} \frac{1}{\sqrt{2\pi \sigma^2}} \exp\left(-\frac{(Y_i - \mu)^2}{2\sigma^2}\right) \].
02

Compute the Likelihood for Second Sample

Similarly, for the second independent sample \( X_1, \ldots, X_m \) with \( N(\mu, c\sigma^2) \), the joint likelihood is: \[ L_X(\mu, \sigma^2) = \prod_{j=1}^{m} \frac{1}{\sqrt{2\pi c \sigma^2}} \exp\left(-\frac{(X_j - \mu)^2}{2c\sigma^2}\right) \].
03

Combine Likelihood Functions

The combined likelihood for both samples is the product of the two individual likelihoods: \[ L(\mu, \sigma^2) = L_Y(\mu, \sigma^2) \cdot L_X(\mu, \sigma^2) \]. This simplifies to a function of \( \mu \) and \( \sigma^2 \) involving sums of squared deviations around \( \mu \) weighed according to their variances.
04

Apply the Prior

Given the prior \( \pi(\mu, \sigma) \propto \frac{1}{\sigma} \), the posterior density up to proportionality is: \[ \pi(\mu, \sigma | Y, X) \propto \frac{1}{\sigma} L(\mu, \sigma^2) \].
05

Determine Posterior for \(\mu\)

Integrate out \( \sigma \) from the posterior density to determine the marginal posterior for \( \mu \). This involves calculus and simplifying the expression, typically resulting in a normal distribution for \( \mu \).
06

Determine Posterior for \(\sigma\)

To find the marginal posterior for \( \sigma \), integrate out \( \mu \) from the joint posterior. This will typically result in an inverse gamma distribution for \( \sigma^2 \) due to the conjugate nature of the priors.

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Key Concepts

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

Posterior Distribution
In Bayesian inference, the posterior distribution is the probability distribution that represents what we know about an unknown parameter after considering the evidence or data at hand. This is the core of Bayesian analysis, as it allows us to update our beliefs about the parameters based on the observed data.

Think of the posterior distribution as combining prior beliefs with new evidence. Specifically, the posterior is formed by multiplying the prior distribution and the likelihood function. In mathematical terms, the posterior distribution for parameters like \(\mu\) and \(\sigma\) can be expressed as:
  • \( \pi(\mu, \sigma | Y, X) \propto \text{Prior}(\mu, \sigma) \times \text{Likelihood}(Y, X | \mu, \sigma) \)
Once we have the posterior distribution, we can make more informed decisions and predictions about \(\mu\) and \(\sigma\). Often, this will result in a refined belief, typically expressed with familiar distributions, as shown in Steps 5 and 6 of the solution, where \(\mu\) follows a normal distribution and \(\sigma^2\) is characterized by an inverse gamma distribution.
Likelihood Function
The likelihood function is a fundamental component in statistical modeling and forms the backbone of Bayesian inference. It describes how probable the observed data is, given specific parameter values.

In our exercise, the likelihood function for two independent samples is derived based on the normal distribution assumptions. For each sample group, the likelihood is:
  • For sample \(Y_1, \ldots, Y_n\): \[ L_Y(\mu, \sigma^2) = \prod_{i=1}^{n} \frac{1}{\sqrt{2\pi \sigma^2}} \exp\left(-\frac{(Y_i - \mu)^2}{2\sigma^2}\right) \]
  • For sample \(X_1, \ldots, X_m\): \[ L_X(\mu, \sigma^2) = \prod_{j=1}^{m} \frac{1}{\sqrt{2\pi c \sigma^2}} \exp\left(-\frac{(X_j - \mu)^2}{2c\sigma^2}\right) \]
The combined likelihood \(L(\mu, \sigma^2)\) is found by multiplying these functions. This accounts for the probability of observing the data across both samples, which is crucial for forming the posterior distribution.
Prior Distribution
Prior distribution represents the initial beliefs about the parameters before any data is observed. In Bayesian inference, it's essential because it influences the posterior distribution.

For this exercise, the given prior is:
  • \( \pi(\mu, \sigma) \propto \frac{1}{\sigma} \)
This is a non-informative prior, meaning it doesn't favor any particular values of \(\mu\) or \(\sigma\), except to enforce that \(\sigma\) must not be zero. Non-informative priors like this are useful when we don't have strong prior knowledge about the parameter values and want the data to have a more significant influence on the posterior distribution.

By using this prior in conjunction with the likelihood function, we derive the posterior distribution, which enables us to update our beliefs about \(\mu\) and \(\sigma\) after observing the data.

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Most popular questions from this chapter

Show that the Gibbs sampler with \(k>2\) components updated in order $$ 1, \ldots, k, 1, \ldots, k, 1, \ldots, k, \ldots $$ is not reversible. Are samplers updated in order \(1, \ldots, k, k-1, \ldots, 1,2, \ldots\), or in a random order reversible?

A population consists of \(k\) classes \(\theta_{1}, \ldots, \theta_{k}\) and it is required to classify an individual on the basis of an observation \(Y\) having density \(f_{i}\left(y \mid \theta_{i}\right)\) when the individual belongs to class \(i=1, \ldots, k\). The classes have prior probabilities \(\pi_{1}, \ldots, \pi_{k}\) and the loss in classifying an individual from class \(i\) into class \(j\) is \(l_{i j}\). (a) Find the posterior probability \(\pi_{i}(y)=\operatorname{Pr}\) (class \(\left.i \mid y\right)\) and the posterior risk of allocating the individual to class \(i\). (b) Now consider the case of \(0-1\) loss, that is, \(l_{i j}=0\) if \(i=j\) and \(l_{i j}=1\) otherwise. Show that the risk is the probability of misclassification. (b) Suppose that \(k=3\), that \(\pi_{1}=\pi_{2}=\pi_{3}=1 / 3\) and that \(Y\) is normally distributed with mean \(i\) and variance 1 in class \(i\). Find the Bayes rule for classifying an observation. Use it to classify the observation \(y=2.2\).

Consider a random sample \(y_{1}, \ldots, y_{n}\) from the \(N\left(\mu, \sigma^{2}\right)\) distribution, with conjugate prior \(N\left(\mu_{0}, \sigma^{2} / k\right)\) for \(\mu ;\) here \(\sigma^{2}\) and the hyperparameters \(\mu_{0}\) and \(k\) are known. Show that the marginal density of the data $$ \begin{aligned} f(y) & \propto \sigma^{-(n+1)}\left(\sigma^{2} n^{-1}+\sigma^{2} k^{-1}\right)^{1 / 2} \exp \left[-\frac{1}{2}\left\\{\frac{(n-1) s^{2}}{\sigma^{2}}+\frac{\left(\bar{y}-\mu_{0}\right)^{2}}{\sigma^{2} / n+\sigma^{2} / k}\right\\}\right] \\ & \propto \exp \left\\{-\frac{1}{2} d(y)\right\\} \end{aligned} $$ say. Hence show that if \(Y_{+}\)is a set of data from this marginal density, \(\operatorname{Pr}\left\\{f\left(Y_{+}\right) \leq f(y)\right\\}=\) \(\operatorname{Pr}\left\\{\chi_{n}^{2} \geq d(y)\right\\} .\) Evaluate this for the sample \(77,74,75,78\), with \(\mu_{0}=70, \sigma^{2}=1\), and \(k_{0}=\frac{1}{2}\). What do you conclude about the model? Do the corresponding development when \(\sigma^{2}\) has an inverse gamma prior. (Box, 1980)

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 \((1-2 \alpha)\).

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, \(\operatorname{Pr}(S) / \operatorname{Pr}(\bar{S}) ?\)

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