/*! 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 2 The Conway-Maxwell-Poisson distr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 2

The Conway-Maxwell-Poisson distribution has probability function: \\[ P(Y=y)=\frac{\lambda^{y}}{(y !)^{\mathrm{v}}} \frac{1}{Z(\lambda, v)} \quad y=0,1,2, \dots \\] where \\[ Z(\lambda, v)=\sum_{i=0}^{\infty} \frac{\lambda^{i}}{(i !)^{v}} \\] Place this in exponential family form, identifying all the relevant components necessary for use in a GLM.

Short Answer

Expert verified
The Conway-Maxwell-Poisson distribution can be written in exponential family form with natural parameter \( \eta = \log(\lambda) \), sufficient statistic \( T(y) = y \), and cumulant function \( A(\eta) = \log\left(\sum_{i=0}^{\infty} \frac{e^{\eta i}}{(i!)^v}\right) \).

Step by step solution

01

Write the probability function with required substitutions

Starting with the given probability function: \[P(Y=y) = \frac{\lambda^y}{(y !)^{\mathrm{v}}} \frac{1}{Z(\lambda, v)}\quad y=0,1,2, \dots\] One can observe that by isolating terms involving \(Y\), rewriting is easier.
02

Identify the natural parameter and sufficient statistic

Rewrite the probability function in the form: \[P(Y=y) = h(y)\exp(\eta T(y) - A(\eta))\] where \( T(y) = y \) is the sufficient statistic. Identify \( \eta = \log(\lambda) \) as the natural parameter, and \( h(y) = \frac{1}{(y!)^v} \) as the base measure.
03

Determine the cumulant function

The cumulant function is identified from the equation component: \[ e^{A(\eta)} = Z(\lambda, v) = \sum_{i=0}^{\infty} \frac{\lambda^i}{(i!)^v}\] Thus, the cumulant function is, \[ A(\eta) = \log\left(\sum_{i=0}^{\infty} \frac{e^{\eta i}}{(i!)^v}\right) \]
04

Write the exponential family form

Express the probability function in the exponential family form: \[ P(Y=y) = \frac{1}{(y!)^v} \exp(y \log(\lambda) - \log\left(\sum_{i=0}^{\infty} \frac{\lambda^i}{(i!)^v}\right)) \] Here, \( \eta = \log(\lambda) \), the sufficient statistic \( T(y) = y \), and cumulant function \( A(\eta) = \log\left(\sum_{i=0}^{\infty} \frac{e^{\eta i}}{(i!)^v}\right) \).

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 91影视!

Key Concepts

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

Generalized Linear Model
A Generalized Linear Model (GLM) is a flexible generalization of ordinary linear regression. It allows for the dependent variable to have a distribution other than normal. In simple terms, it鈥檚 a way to predict outcomes based on predictors when the outcome is not necessarily continuous, like counting results we see in the Conway-Maxwell-Poisson distribution.

GLMs consist of three components:
  • A linear predictor \(\boldsymbol{\beta}^T\boldsymbol{x}\), where \(\boldsymbol{\beta}\) is a vector of coefficients and \(\boldsymbol{x}\) is a vector of predictors.
  • A link function g that relates the mean of the dependent variable to the linear predictor, \(\theta = g^{-1}(\boldsymbol{\beta}^T\boldsymbol{x})\).
  • A probability distribution from the exponential family.
In this context, the Conway-Maxwell-Poisson distribution can be placed in an exponential family form for use in a GLM scenario by identifying its natural parameter, cumulant function, and sufficient statistic.
Exponential Family
The exponential family is a class of probability distributions that have a particular form. They are particularly useful because they allow for a unified approach to many statistical models.

A probability distribution belongs to the exponential family if it can be written as:
\[ P(Y=y) = h(y) \, exp(\theta \, T(y) - A(\theta)) \]

Here, h(y) is the base measure, 胃 is the natural parameter, T(y) is the sufficient statistic, and 础(胃) is the cumulant function.

Placing the Conway-Maxwell-Poisson distribution into this form involves identifying each of these components. For instance, our base measure for the Conway-Maxwell-Poisson is \( \frac{1}{(y!)^v} \).
Cumulant Function
The cumulant function (or log-partition function) 础(胃) is a critical component of the exponential family. It normalizes the probability distribution and ensures it sums to one.

For the Conway-Maxwell-Poisson distribution, the cumulant function is identified from the summation part of the expression:
\[ e^{A(\theta)} = Z(\theta, v) = \sum_{i=0}^{\infty} \frac{e^{\theta i}}{(i!)^v} \]

Thus, the cumulant function is:
\[ 础(胃) = \log\left(\sum_{i=0}^{\infty} \frac{e^{\theta \, i}}{(i!)^v}\right) \]

This function plays a role in defining the mean and variance of the distribution, which are derived by differentiating 础(胃).
Natural Parameter
The natural parameter (or canonical parameter) 胃 is part of how the exponential family is expressed.

For the Conway-Maxwell-Poisson distribution, the natural parameter is:
\( 胃 = \log(\lambda) \)

This parameter connects the distribution鈥檚 mean to the linear predictor in GLM models. It essentially re-parameterizes the distribution in a way that makes it easier to work with.
Sufficient Statistic
A sufficient statistic T(y) is a value that provides as much information about the parameter of the distribution as the entire data set.

For the Conway-Maxwell-Poisson distribution, the sufficient statistic is:
\( T(y) = y \)

This means the value y alone contains all the necessary information to make inferences about the natural parameter 胃 of the distribution. This helps simplify the calculations when modeling with the exponential family in GLMs.

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

Data is generated from the exponential distribution with density \(f(y)=\) \(\lambda \exp (-\lambda y)\) where \(\lambda, y>0\) (a) Identify the specific form of \(\theta, \phi, a(), b()\) and \(c()\) for the exponential distribution. (b) What is the canonical link and variance function for a GLM with a response following the exponential distribution? (c) Identify a practical difficulty that may arise when using the canonical link in this instance. (d) When comparing nested models in this case, should an \(F\) or \(\chi^{2}\) test be used? Explain. (e) Express the deviance in this case in terms of the responses \(y_{i}\) and the fitted values \(\hat{\mu}_{i}\).

Consider the following distributions and determine whether they can be put in exponential family form. If it is not possible, consider whether there is a special case which follows the form. \(\bullet\) The Uniform distribution: \\[ f(y | \alpha, \beta)=\frac{1}{\beta-\alpha} I(y \in[\alpha, \beta]) \\] \(\bullet\) The Weibull distribution: \((y>0)\) \\[ f(y | \alpha, \beta)=\alpha \beta^{-\alpha} y^{\alpha-1} e^{(y / \beta)^{\alpha}} \\] \(\bullet\) The Gumbel distribution: \((y>0)\) \\[ f(y | \alpha, \beta)=\frac{1}{\beta} \exp ((y-\alpha) / \beta-\exp ((y-\alpha) / \beta)) \\]
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation
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.