/*! 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 Suppose that the cumulant-genera... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 7

Suppose that the cumulant-generating function of \(X\) can be written in the form \(m\\{b(\theta+\) \(t)-b(\theta)\\}\). Let \(\mathrm{E}(X)=\mu=m b^{\prime}(\theta)\) and let \(\kappa_{2}(\mu)\) and \(\kappa_{3}(\mu)\) be the variance and third cumulant respectively of \(X\), expressed in terms of \(\mu ; \kappa_{2}(\mu)\) is the variance function \(V(\mu)\). (a) Show that $$ \kappa_{3}(\mu)=\kappa_{2}(\mu) \kappa_{2}^{\prime}(\mu) \quad \text { and } \quad \frac{\kappa_{3}}{\kappa_{2}^{2}}=\frac{d}{d \mu} \log \kappa_{2}(\mu) $$ Verify that the binomial cumulants have this form with \(b(\theta)=\log \left(1+e^{\theta}\right)\). (b) Show that if the derivatives of \(b(\theta)\) are all \(O(1)\), then \(Y=g(X)\) is approximately symmetrically distributed if \(g\) satisfies the second-order differential equation $$ 3 \kappa_{2}^{2}(\mu) g^{\prime \prime}(\mu)+g^{\prime}(\mu) \kappa_{3}(\mu)=0 $$ Show that if \(\kappa_{2}(\mu)\) and \(\kappa_{3}(\mu)\) are related as in (a), then $$ g(x)=\int^{x} \kappa_{2}^{-1 / 3}(\mu) d \mu $$ (c) Hence find symmetrizing transformations for Poisson and binomial variables. (McCullagh and Nelder, 1989 , Section 4.8)

Short Answer

Expert verified
Derive the cumulant forms and solve the differential equations for Poisson and binomial laws to find symmetrizing functions.

Step by step solution

01

Understanding the cumulant-generating function

The cumulant-generating function of a random variable \( X \) is given as \( m\{b(\theta+t)-b(\theta)\} \). This expression helps us find the cumulants, namely the variance (second cumulant) and third cumulant of \( X \). Here, \( \mathrm{E}(X)=\mu=m b^{\prime}(\theta) \). The challenge is to establish the relationship between the third cumulant, \( \kappa_{3}(\mu) \), the variance, \( \kappa_{2}(\mu) = V(\mu) \), and their derivative with respect to \( \mu \).
02

Deriving the second and third cumulant

From the expression given for the cumulant-generating function, differentiating with respect to \( t \) and evaluating at \( t = 0 \) will give us the cumulants. First derivative gives us the mean, second gives the variance, and third gives the third cumulant. Thus, the variance \( \kappa_{2}(\mu) \) can be calculated as \( mb^{\prime\prime}(\theta) \) and the third cumulant as \( mb^{\prime\prime\prime}(\theta) \).
03

Verify \(\kappa_{3}(\mu)=\kappa_{2}(\mu) \kappa_{2}^{\prime}(\mu)\)

Given from the cumulant expressions, \( \kappa_{3}(\mu) = mb^{\prime\prime\prime}(\theta) \) and \( \kappa_{2}(\mu) = mb^{\prime\prime}(\theta) \). Differentiating \( \kappa_{2}(\mu) \) with respect to \( \mu \), we derive \( \kappa_{2}^{\prime}(\mu) = b^{\prime\prime\prime}(\theta) / b^{\prime\prime}(\theta) \). Thus, \( \kappa_{3}(\mu) = \kappa_{2}(\mu) \kappa_{2}^{\prime}(\mu) \). This expression simplifies to the given relation.
04

Simplifying the logarithmic derivative

The fractional expression \( \frac{\kappa_{3}}{\kappa_{2}^{2}} \) can be rewritten using the expressions in terms of derivatives. If we differentiate \( \log \kappa_{2}(\mu) \) using the chain and quotient rules, this results in \( \frac{d}{d \mu} \log \kappa_{2}(\mu) = \frac{\kappa_{3}}{\kappa_{2}^{2}} \) confirming the given relation.
05

Verifying the binomial cumulant with given function

With \( b(\theta) = \log(1 + e^{\theta}) \), calculate derivatives: \( b^{\prime}(\theta) = \frac{e^{\theta}}{1+e^{\theta}} \), \( b^{\prime\prime}(\theta) = \frac{e^{\theta}}{(1+e^{\theta})^2} \), and \( b^{\prime\prime\prime}(\theta) = \frac{e^{\theta}(1-e^{\theta})}{(1+e^{\theta})^3} \). These derivatives confirm that the binomial cumulants satisfy \( \kappa_{3}(\mu) = \kappa_{2}(\mu) \kappa_{2}^{\prime}(\mu) \).
06

Solve the differential equation for symmetry

The differential equation is \( 3\kappa_{2}^{2}(\mu) g^{\prime\prime}(\mu) + g^{\prime}(\mu) \kappa_{3}(\mu) = 0 \). Substitute \( \kappa_{3}(\mu) = \kappa_{2}(\mu) \kappa_{2}^{\prime}(\mu) \) into the equation, finding that \( g(\mu) = \int^{\mu} \kappa_{2}^{-1/3}(u) du \) serves as a solution. This is derived by setting the power terms such that the integral becomes tractable.
07

Determine transformations for Poisson and Binomial

For Poisson variables where variance \( V(\mu) = \mu \), \( g(x) = \int^{x} \mu^{-1/3} d\mu \). For binomial variables with variance \( np(1-p) \), \( g(x) = \int^{x} (n \hat{p}(1-\hat{p}))^{-1/3} d\mu \). Evaluate these integrals based on specific parameters (mean or function settings) to achieve symmetry.

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Key Concepts

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

Variance Function
The variance function uniquely describes how data dispersion varies with the mean level of data. It is commonly denoted as \( V(\mu) \), where \( \mu \) represents the mean or expected value of a random variable. In statistical analysis, particularly in generalized linear models, understanding the variance function is crucial for assessing variability around the mean.
In the context of the cumulant-generating function, the variance \( \kappa_2(\mu) = V(\mu) \) is derived from the second derivative of the cumulant function with respect to the transformation variable \( t \).
  • The variance function helps determine the stability or spread of a dataset.
  • It's essential for identifying patterns or changes in the variance with different mean levels.
  • An accurate variance function can assist in developing fitting models and predictions.
In our example, the exploration of \( V(\mu) \) is part of deriving relationships between cumulants, particularly the second and third cumulants.
Differential Equations
Differential equations play a pivotal role in deriving relationships in variably changing data. They are equations that involve derivatives of a function and help describe how a particular quantity changes over time.
In the exercise, the focus is on solving a second-order differential equation that determines symmetrical distribution transformations. This equation, \( 3 \kappa_2^2(\mu) g''(\mu) + g'(\mu) \kappa_3(\mu) = 0 \), incorporates variance and third cumulant functions to devise a transformation function \( g(x) \).
  • To solve this, identification of relations between \( \kappa_3(\mu) \) and \( \kappa_2(\mu) \) simplifies the process.
  • The solution is obtained by integrating \( \kappa_2^{-1/3}(\mu) \), indicating how transformations are symmetrized.
  • The mathematical handling of such equations is crucial in creating models that transform and normalize asymmetric data.
This equation showcases the power of differential equations in statistical transformations.
Symmetrizing Transformations
Symmetrizing transformations aim to adjust a given dataset or distribution to become more symmetrical. This increases model viability when symmetry assumptions are necessary, for example, in linear regression.
In the context of the exercise, this is accomplished through a specific transformation function \( g(x) \), derived as \( \int^x \kappa_2^{-1/3}(\mu) d\mu \). Solving differential equations such as \( 3 \kappa_2^2(\mu) g''(\mu) + g'(\mu) \kappa_3(\mu) = 0 \) establishes how \( g(x) \) transforms the random variable \( X \) into a symmetrically distributed variable \( Y \).
  • This approach can help models achieve normality and improve statistical inference accuracy.
  • Symmetrizing is crucial in handling skewed data, ensuring more reliable statistical analysis.
  • The derived function \( g(x) \) thus grants flexibility in data transformation to half-off effects of skewness or kurtosis.
Understanding and applying such transformations can have significant practical implications for data handling and model formulation.
Poisson and Binomial Variables
Poisson and binomial variables are two fundamental types of discrete random variables often encountered in probability and statistics. They describe, respectively, events occurring independently over a fixed interval and the number of successes in a sequence of binary experiments.
For Poisson variables, the variance function is equal to the mean, \( V(\mu) = \mu \), which represents the natural variance in count data often applicable in real-life scenarios like customer arrival rates or event occurrences.
For binomial variables, the variance function is \( V(\mu) = np(1-p) \), where \( n \) is the number of trials and \( p \) is the probability of success in each trial.
  • These variance functions guide accurate modeling for different types of data.
  • Transformations for symmetry in these variables involve calculating integrals like \( \int^x \mu^{-1/3} d\mu \) (Poisson) or \( \int^x (n \hat{p}(1-\hat{p}))^{-1/3} d\mu \) (binomial).
  • Such transformations help refine statistical models by accounting for distribution-specific dispersion effects.
Understanding these functions is essential for practical applications across various data analysis settings, from quality assurance testing to biostatistics.

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Most popular questions from this chapter

At each of the doses \(x_{1}0\) is used, show that $$ \widehat{\beta}=\frac{1}{x_{0}} \Phi^{-1}(r / m), \quad \operatorname{var}(\widehat{\beta}) \doteq \frac{\Phi\left(\beta x_{0}\right)\left\\{1-\Phi\left(\beta x_{0}\right)\right\\}}{m x_{0}^{2}\left\\{\phi\left(\beta x_{0}\right)\right\\}^{2}} $$ where \(\phi\) and \(\Phi\) are the standard normal density and distribution functions. Plot the function \(\Phi(\eta)\\{1-\Phi(\eta)\\} / \phi(\eta)^{2}\) for \(\eta\) in the range \(-3 \leq \eta \leq 3\), and comment on the implications for the choice of \(x_{0}\) if there is some prior knowledge of the likely value of \(\beta\).

By writing \(\sum\left\\{y_{j}-\widehat{g}\left(x_{j}\right)\right\\}^{2}=(y-\widehat{g})^{\mathrm{T}}(y-\widehat{g})\) and recalling that \(y=g+\varepsilon\) and \(\widehat{g}=S y\), where \(S\) is a smoothing matrix, show that $$ \mathrm{E}\left[\sum_{j=1}^{n}\left\\{y_{j}-\widehat{g}\left(x_{j}\right)\right\\}^{2}\right]=\sigma^{2}\left(n-2 v_{1}+v_{2}\right)+g^{\mathrm{T}}(I-S)^{\mathrm{T}}(I-S) g $$ Hence explain the use of \(s^{2}(h)\) as an estimator of \(\sigma^{2}\). Under what circumstances is it unbiased?

Data \(y_{1}, \ldots, y_{n}\) are assumed to follow a binary logistic model in which \(y_{j}\) takes value 1 with probability \(\pi_{j}=\exp \left(x_{j}^{\mathrm{T}} \beta\right) /\left\\{1+\exp \left(x_{j}^{\mathrm{T}} \beta\right)\right\\}\) and value 0 otherwise, for \(j=1, \ldots, n\). (a) Show that the deviance for a model with fitted probabilities \(\widehat{\pi}_{j}\) can be written as $$ D=-2\left\\{y^{\mathrm{T}} X \widehat{\beta}+\sum_{j=1}^{n} \log \left(1-\hat{\pi}_{j}\right)\right\\} $$ and that the likelihood equation is \(X^{\mathrm{T}} y=X^{\mathrm{T}} \widehat{\pi}\). Hence show that the deviance is a function of the \(\widehat{\pi}_{j}\) alone. (b) If \(\pi_{1}=\cdots=\pi_{n}=\pi\), then show that \(\widehat{\pi}=\bar{y}\), and verify that $$ D=-2 n\\{\bar{y} \log \bar{y}+(1-\bar{y}) \log (1-\bar{y})\\} $$ Comment on the implications for using \(D\) to measure the discrepancy between the data and fitted model. (c) In (b), show that Pearson's statistic (10.21) is identically equal to \(n\). Comment.

A positive stable random variable \(U\) has \(\mathrm{E}\left(e^{-s U}\right)=\exp \left(-\delta s^{\alpha} / \alpha\right), 0<\alpha \leq 1\) (a) Show that if \(Y\) follows a proportional hazards model with cumulative hazard function \(u \exp \left(x^{\mathrm{T}} \beta\right) H_{0}(y)\), conditional on \(U=u\), then \(Y\) also follows a proportional hazards model unconditionally. Are \(\beta, \alpha\), and \(\delta\) estimable from data with single individuals only? (b) Consider a shared frailty model, as in the previous question, with positive stable \(U\). Show that the joint survivor function may be written as $$ \mathcal{F}\left(y_{1}, y_{2}\right)=\exp \left(-\left[\left\\{-\log \mathcal{F}_{1}\left(y_{1}\right)\right\\}^{1 / \alpha}+\left\\{-\log \mathcal{F}_{2}\left(y_{2}\right)\right\\}^{1 / \alpha}\right]^{\alpha}\right), \quad y_{1}, y_{2}>0 $$ in terms of the marginal survivor functions \(\mathcal{F}_{1}\) and \(\mathcal{F}_{2}\). Show that if the conditional cumulative hazard functions are Weibull, \(u H_{r}(y)=u \xi_{r} y^{\gamma}, \gamma>0, r=1,2\), then the marginal survivor functions are also Weibull. Show also that the time to the first event has a Weibull distribution.

Develop the details of local likelihood smoothing when a linear polynomial is fitted to Poisson data, using link function \(\log \mu=\beta_{0}+\beta_{1}\left(x-x_{0}\right)\).
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