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Chapter 12: Problem 1

Suppose that \(Y_{1}, \ldots, Y_{n}\) are independent Poisson variables with means \(\psi \pi_{j}\), where \(0<\) \(\pi_{j}<1\) and \(\sum \pi_{j}=1\). Find a marginal likelihood for \(\psi\) based on \(Y_{1}, \ldots, Y_{n}\), and show that no information about \(\psi\) is lost by using the marginal likelihood rather than the full likelihood.

Short Answer

Expert verified
The marginal likelihood for \( \psi \) is \( L(\psi | \sum Y_j) = \psi^{\sum Y_j} e^{-\psi} \). No information about \( \psi \) is lost.

Step by step solution

01

Define the Likelihood Function

The likelihood function for independent Poisson variables is given by: \[ L(Y_1, ext{...}, Y_n | heta) = \prod_{j=1}^{n} \frac{(\psi \pi_j)^{Y_j} e^{-\psi \pi_j}}{Y_j!} \]Given that these are Poisson variables with means \( \psi \pi_j \), the overall likelihood can be expressed in terms of these parameters.
02

Simplifying the Exponential Component

The product of the exponential terms in the likelihood becomes:\[ e^{-\psi \sum_{j=1}^{n} \pi_j} = e^{-\psi} \]because \( \sum_{j=1}^{n} \pi_j = 1 \). This simplification shows that this term depends only on \( \psi \).
03

Consider the Joint Distribution of Y

The joint distribution function of \( Y_1, \ldots, Y_n \) is:\[ \prod_{j=1}^{n} \frac{(\psi \pi_j)^{Y_j}}{Y_j!} \]Reducing the expression yields:\[ \psi^{\sum Y_j} \prod_{j=1}^{n} \frac{\pi_j^{Y_j}}{Y_j!} \]This simplifies the likelihood function into a combination of \( \psi \) and the probabilities \( \pi_j \).
04

Derive the Marginal Likelihood for \( \psi \)

The marginal likelihood of \( \psi \) in terms of summed observations is:\[ L(\psi | \sum Y_j) = \psi^{\sum Y_j} e^{-\psi} \prod_{j=1}^{n} \int \pi_j^{Y_j} \, d\pi_j \]Since \( \psi \) has been separated, this demonstrates that marginal likelihood can be computed independently of specific \( \pi_j \) values.
05

Information Retention Comparison

By expressing the likelihood explicitly in terms of \( \psi \) and knowing that the family of distributions remains Poisson, we see that all information about \( \psi \) is captured. The transformation to the marginal likelihood, which aggregates evidence over \( \pi_j \) but retains dependency on \( \psi \), ensures that no information is truly lost.

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Key Concepts

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

Poisson Variables
To understand Poisson variables, let's simplify. A Poisson variable represents counts of events happening in a fixed interval, where these events occur with a known constant mean rate. These events happen independently of the time since the last event. For example, counting the number of cars passing through a street in one hour can be modeled as a Poisson process if they happen randomly and independently.

Each independent Poisson variable in our scenario has a mean that is adjusted by the parameter \( \psi \) and the corresponding weight \( \pi_j \). The mean of each Poisson variable in this setup is \( \psi \pi_j \). Therefore:
  • \(Y_j\) are the observed counts, like cars per street in each interval.
  • \(\pi_j\) represents the proportion of overall weight, kind of like how much of \(\psi\) goes to each \(Y_j\).
Understanding these variables is foundational for setting up the likelihood function. As they are independent, their joint behavior conveniently simplifies mathematical workings.
Independent Variables
When we say variables are independent, it means the occurrence or result of one variable does not influence another. In the context of Poisson variables, this independence implies that the behavior of one Poisson variable does not affect another. Each variable is like an isolated experiment.

For example, if you are counting the number of people entering different store sections, the count in one section doesn't affect others.
  • This means each \( Y_j \) is independent from any of the others.
  • Mathematically, independence allows us to use products over probabilities or likelihood functions.
This is powerful because it reduces complexity. The likelihood of observing a specific series of counts from these Poisson variables becomes a simple product of individual probabilities.
Likelihood Function
A likelihood function is a way to assess how 'likely' a set of observed data is given specific parameters. In the context of Poisson variables, our objective is to express the probability of observing the data, based on specific assumed means.

In this setup:
  • The likelihood function is a product of Poisson probabilities. Each probability is calculated based on a hypothetical true mean \( \psi \pi_j \).
  • You multiply each probability of the observed \(Y_j\), leading to the overall likelihood function.
The likelihood function for independent Poisson variables simplifies as shown:\[ L(Y_1, \ldots, Y_n | \theta) = \prod_{j=1}^{n} \frac{(\psi \pi_j)^{Y_j} e^{-\psi \pi_j}}{Y_j!} \] When deriving the marginal likelihood, we aim to focus solely on \( \psi \) by integrating out the \( \pi_j \). This helps find the best estimate for \( \psi \) and ensures that all relevant information regarding \( \psi \) is retained, even though the full likelihood considers specific \(\pi_j\).

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Most popular questions from this chapter

$$ \text { Let } Y_{1}, \ldots, Y_{n} \stackrel{\text { iid }}{\sim} N\left(\mu, c^{2} \mu^{2}\right) \text {, with } c \text { known. Show that } \bar{Y} / S \text { is ancillary for } \mu \text {. } $$

Let \(Y_{1}, \ldots, Y_{n} \stackrel{\mathrm{iid}}{\sim} N\left(\mu, \sigma^{2}\right)\) with \(\sigma^{2}\) known. Show that \(\left(Y_{1}-\bar{Y}, \ldots, Y_{n}-\bar{Y}\right)\) is distribution constant, and deduce that \(\bar{Y}\) and \(\sum\left(Y_{j}-\bar{Y}\right)^{2}\) are independent.

Let \(Y\) and \(X\) be independent exponential variables with means \(1 /(\lambda+\psi)\) and \(1 / \lambda\). Find the distribution of \(Y\) given \(X+Y\) and show that when \(\psi=0\) it has mean \(s / 2\) and variance \(s^{2} / 12 .\) Construct an exact conditional test of the hypothesis \(\mathrm{E}(Y)=\mathrm{E}(X)\).

A Poisson variable \(Y\) has mean \(\mu\), which is itself a gamma random variable with mean \(\theta\) and shape parameter \(v\). Find the marginal density of \(Y\), and show that \(\operatorname{var}(Y)=\theta+\theta^{2} / v\), and that \(v\) and \(\theta\) are orthogonal. Hence show that \(v\) is orthogonal to \(\beta\) for any model in which \(\theta=\theta\left(x^{\mathrm{T}} \beta\right), x\) being a covariate vector. Is the same true for the model in which \(v=\theta / \kappa\), so that \(\operatorname{var}(Y)=(1+\kappa) \mu ?\) Discuss the implications for inference on \(\beta\) when the variance function is unknown.

Let \(X\) and \(Y\) be independent exponential variables with means \(\gamma^{-1}\) and \((\gamma \psi)^{-1}\). Show that the parameter \(\lambda(\gamma, \psi)\) orthogonal to \(\psi\) is the solution to the equation \(\partial \gamma / \partial \psi=-\gamma /(2 \psi)\), and verify that taking \(\lambda=\gamma / \psi^{-1 / 2}\) yields an orthogonal parametrization. Investigate how this solution changes when \(X\) and \(Y\) are subject to Type I censoring at \(c\).
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