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Chapter 5: Problem 3

Show that the geometric density $$ f(y ; \pi)=\pi(1-\pi)^{y}, \quad y=0,1, \ldots, 0<\pi<1 $$ is an exponential family, and give its cumulant-generating function. Show that \(S=Y_{1}+\cdots+Y_{n}\) has negative binomial density $$ \left(\begin{array}{c} n+s-1 \\ n-1 \end{array}\right) \pi^{n}(1-\pi)^{s}, \quad s=0,1, \ldots $$ and that this is also an exponential family.

Short Answer

Expert verified
The geometric density is an exponential family with cumulant function \(-\log(1-e^{\eta})\). The negative binomial density is also an exponential family.

Step by step solution

01

Identify the Natural Parameter

To show that the geometric density is an exponential family, we start by comparing it to the general form of an exponential family distribution:\[f(y; \eta) = h(y) \exp(\eta T(y) - A(\eta))\]For the given geometric density, \(f(y; \pi) = \pi(1-\pi)^y\), rewrite it as:\[f(y; \pi) = \exp\left(\log \pi + y \log (1-\pi)\right)\]Thus, the natural parameter is \(\eta = \log(1-\pi)\).
02

Identify the Sufficient Statistic

Comparing the rewritten density with the general exponential form, the term that multiplies \(\eta\) in the exponent being \(y\), we identify the sufficient statistic as \(T(y) = y\).
03

Derive the Cumulant Function

The term \(A(\eta)\) in the exponential family form corresponds to the log-partition function or cumulant-generating function:\[A(\eta) = - \log \pi = -\log(1-e^{\eta})\] This follows because \(\pi = 1-e^{\eta}\).
04

Verify Negative Binomial Is Exponential Family

Next, consider the negative binomial distribution:\[P(S=s) = \binom{n+s-1}{n-1} \pi^n (1-\pi)^s\]This can be rewritten in terms of an exponential family:\[P(S=s) = \exp\left(\log \binom{n+s-1}{n-1} + n\log \pi + s \log(1-\pi)\right)\]Here, the natural parameter is \(\eta = \log(1-\pi)\), and \(T(s) = s\).
05

Derive the Negative Binomial Cumulant Function

The cumulant function or log-partition function for the negative binomial corresponds to the coefficient of the natural parameter's normalization:\[A(\eta) = -n \log(1-\pi) = -n \log(1-e^\eta)\]This ensures the distribution is in the canonical form of exponential family distributions.

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Key Concepts

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

Geometric Distribution
The geometric distribution is a fascinating concept often encountered in statistics. It models the number of trials needed for the first success in a sequence of independent and identically distributed Bernoulli trials. Here, each trial has two potential outcomes: success with probability \(\pi\) and failure with probability \(1-\pi\). Thus, the density function is given by:
\[f(y; \pi) = \pi (1-\pi)^y\]\for \(y = 0, 1, 2, \ldots\) and \(0 < \pi < 1\).
The geometric distribution is part of the exponential family. This is because it can be transformed to match the exponential family form:
  • Natural Parameter: \(\eta = \log(1-\pi)\)
  • Sufficient Statistic: \(T(y) = y\)
  • Cumulant Function: \(A(\eta) = -\log(1-e^\eta)\)
By understanding these transformations, we grasp why the geometric distribution can fit within the exponential family, highlighting its versatility and the theoretical underpinning.
Negative Binomial Distribution
The negative binomial distribution extends the idea of the geometric distribution by modeling the number of failures \(s\) before an \(n\)th success occurs. This can be represented by the density function:
\[P(S = s) = \binom{n+s-1}{n-1} \pi^n (1-\pi)^s\]for \(s = 0, 1, 2, \ldots\). Here, \(\pi\) is still the probability of success in each trial.
Just like the geometric distribution, the negative binomial distribution is also part of the exponential family:
  • Natural Parameter: \(\eta = \log(1-\pi)\)
  • Sufficient Statistic: \(T(s) = s\)
  • Cumulant Function: \(A(\eta) = -n \log(1-e^\eta)\)
Understanding these components helps us recognize the distribution's underlying structure, and its place within the exponential family framework. This understanding is crucial for various statistical applications, including regression analysis and Bayesian statistics.
Cumulant-Generating Function
The cumulant-generating function (CGF) is an essential tool in understanding exponential family distributions. A CGF provides valuable insight into the statistical properties of a distribution by describing the distribution's moments, such as mean and variance.
For an exponential family distribution, the CGF is equivalent to the log-partition function \(A(\eta)\). For the geometric and negative binomial distributions discussed earlier, the specific form of \(A(\eta)\) plays a crucial role:
  • For the geometric distribution: \(A(\eta) = -\log(1-e^\eta)\)
  • For the negative binomial distribution: \(A(\eta) = -n \log(1-e^\eta)\)
By deriving and analyzing these cumulative generating functions, statisticians can ascertain key properties of these distributions. These include tail behavior and moment generating properties, making CGFs a powerful tool in the analysis of statistical models.

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Most popular questions from this chapter

Suppose \(Y=\tau \varepsilon\), where \(\tau \in \mathbb{R}_{+}\)and \(\varepsilon\) is a random variable with known density \(f\). Show that this scale model is a group transformation model with free action \(g_{\tau}(y)=\tau y\). Show that \(s_{1}(Y)=\bar{Y}\) and \(s_{2}(Y)=\left(\sum Y_{j}^{2}\right)^{1 / 2}\) are equivariant and find the corresponding maximal invariants. Sketch the orbits when \(n=2\).

Let \(Y_{1}, \ldots, Y_{n}\) be independent exponential variables with hazard \(\lambda\) subject to Type I censoring at time \(c\). Show that the observed information for \(\lambda\) is \(D / \lambda^{2}\), where \(D\) is the number of the \(Y_{j}\) that are uncensored, and deduce that the expected information is \(i(\lambda \mid c)=n\\{1-\exp (-\lambda c)\\} / \lambda^{2}\) conditional on \(c\) Now suppose that the censoring time \(c\) is a realization of a random variable \(C\), whose density is gamma with index \(v\) and parameter \(\lambda \alpha\) : $$ f(c)=\frac{(\lambda \alpha)^{v} c^{v-1}}{\Gamma(v)} \exp (-c \lambda \alpha), \quad c>0, \alpha, v>0 $$ Show that the expected information for \(\lambda\) after averaging over \(C\) is $$ i(\lambda)=n\left\\{1-(1+1 / \alpha)^{-v}\right\\} / \lambda^{2} $$ Consider what happens when (i) \(\alpha \rightarrow 0\), (ii) \(\alpha \rightarrow \infty\), (iii) \(\alpha=1, v=1\), (iv) \(v \rightarrow \infty\) but \(\mu=v / \alpha\) is held fixed. In each case explain qualitatively the behaviour of \(i(\lambda)\).

Show that the inverse Gaussian density $$ f(y ; \mu, \lambda)=\left(\frac{\lambda}{2 \pi y^{3}}\right)^{1 / 2} \exp \left\\{-\lambda(y-\mu)^{2} /\left(2 \mu^{2} y\right)\right\\}, \quad y>0, \lambda, \mu>0 $$ is an exponential family of order \(2 .\) Give a general form for its cumulants.

Let \(X_{1}, \ldots, X_{n}\) be an exponential random sample with density \(\lambda \exp (-\lambda x), x>0, \lambda>0\) For simplicity suppose that \(n=m r\). Let \(Y_{1}\) be the total time at risk from time zero to the \(r\) th failure, \(Y_{2}\) be the total time at risk between the \(r\) th and the \(2 r\) th failure, \(Y_{3}\) the total time at risk between the \(2 r\) th and \(3 r\) th failures, and so forth. (a) Let \(X_{(1)} \leq X_{(2)} \leq \cdots \leq X_{(n)}\) be the ordered values of the \(X_{j}\). Show that the joint density of the order statistics is $$ f_{X_{(1)}, \ldots, X_{(n)}}\left(x_{1}, \ldots, x_{n}\right)=n ! f\left(x_{1}\right) f\left(x_{2}\right) \cdots f\left(x_{n}\right), \quad x_{1}

Find the exponential families with variance functions (i) \(V(\mu)=a \mu(1-\mu), \mathcal{M}=(0,1)\), (ii) \(V(\mu)=a \mu^{2}, \mathcal{M}=(0, \infty)\), and (iii) \(V(\mu)=a \mu^{2}, \mathcal{M}=(-\infty, 0)\).
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