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Chapter 7: Problem 2

Prove that the sum of the observations of a random sample of size \(n\) from a Poisson distribution having parameter \(\theta, 0<\theta<\infty\), is a sufficient statistic for \(\theta\).

Short Answer

Expert verified
Using the Factorization Theorem, it's demonstrated that the sum of the observations of a random sample of size \(n\) from a Poisson distribution is a sufficient statistic for the parameter \(\theta\), because the likelihood function can be factorized into a function of \(\theta\) and \(T\), and a function of \(T\) alone, where \(T\) is the sum of the observations.

Step by step solution

01

Understand the problem

We are investigating a random sample of size \(n\) from a Poisson distribution with parameter \(\theta\). We need to prove that the sum of the observations of the random sample is sufficient for \(\theta\). That can be done using the Factorization Theorem.
02

Express the likelihood function

Given a random sample \(X_{1}, X_{2}, ..., X_{n}\) following a Poisson distribution, we know the Probability Mass Function (PMF) for a Poisson distributed random variable \(X_i\) is given by \(P(X_{i} = x_{i}) = (e^{-\theta}\theta^{x_{i}}) / x_{i}!\). So, the joint PMF of all \(n\) variables, the likelihood function, is \(L(\theta) = \prod_{i=1}^{n}(e^{-\theta}\theta^{x_{i}}) / x_{i}! = e^{-n\theta}\theta^{T} / T!\), where \(T = \sum_{i=1}^{n}X_{i}\). We simplified products to a sum through the properties of exponentials.
03

Apply the Factorization Theorem

The Factorization Theorem states that a statistic is sufficient for a parameter if the likelihood can be factored such that one factor includes only this sample and the other depends only on the parameter. For our likelihood function, we can factor it as follows: \(L(\theta) = e^{-n\theta}\theta^{T} / T! = (e^{-n\theta}\theta^{T}) \cdot (1 / T!) \). Here, the first factor \(e^{-n\theta}\theta^{T}\) depends on the parameter \(\theta\) and the statistic \(T\), and the second factor \(1 / T!\) depends only on the statistic \(T\).
04

Finalize the result

Since the likelihood function for this Poisson distribution can be factorized into a function of \(\theta\) and \(T\), and a function of \(T\) alone, the Factorization Theorem states that \(T\) is a sufficient statistic for the parameter \(\theta\). Therefore, we have proven that the sum of observations from a random sample of size \(n\) from a Poisson distribution is a sufficient statistic for \(\theta\).

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a distribution with pdf \(f(x ; \theta)=\theta^{2} x e^{-\theta x}, 00\) (a) Argue that \(Y=\sum_{1}^{n} X_{i}\) is a complete sufficient statistic for \(\theta\). (b) Compute \(E(1 / Y)\) and find the function of \(Y\) which is the unique MVUE of \(\theta\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample from a distribution that is \(N(\theta, 1),-\infty<\theta<\infty\). Find the MVUE of \(\theta^{2}\). Hint: \(\quad\) First determine \(E\left(\bar{X}^{2}\right)\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample with the common pdf \(f(x)=\) \(\theta^{-1} e^{-x / \theta}\), for \(x>0\), zero elsewhere; that is, \(f(x)\) is a \(\Gamma(1, \theta)\) pdf. (a) Show that the statistic \(\bar{X}=n^{-1} \sum_{i=1}^{n} X_{i}\) is a complete and sufficient statistic for \(\theta\). (b) Determine the MVUE of \(\theta\). (c) Determine the mle of \(\theta\). (d) Often, though, this pdf is written as \(f(x)=\tau e^{-\tau x}\), for \(x>0\), zero elsewhere. Thus \(\tau=1 / \theta\). Use Theorem \(6.1 .2\) to determine the mle of \(\tau\). (e) Show that the statistic \(\bar{X}=n^{-1} \sum_{i=1}^{n} X_{i}\) is a complete and sufficient statistic for \(\tau\). Show that \((n-1) /(n \bar{X})\) is the MVUE of \(\tau=1 / \theta\). Hence, as usual, the reciprocal of the mle of \(\theta\) is the mle of \(1 / \theta\), but, in this situation, the reciprocal of the MVUE of \(\theta\) is not the MVUE of \(1 / \theta\). (f) Compute the variances of each of the unbiased estimators in parts (b) and (e).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) represent a random sample from the discrete distribution having the pmf $$ f(x ; \theta)=\left\\{\begin{array}{ll} \theta^{x}(1-\theta)^{1-x} & x=0,1,0<\theta<1 \\ 0 & \text { elsewhere } \end{array}\right. $$ Show that \(Y_{1}=\sum_{1}^{n} X_{i}\) is a complete sufficient statistic for \(\theta .\) Find the unique function of \(Y_{1}\) that is the MVUE of \(\theta\). Hint: \(\quad\) Display \(E\left[u\left(Y_{1}\right)\right]=0\), show that the constant term \(u(0)\) is equal to zero, divide both members of the equation by \(\theta \neq 0\), and repeat the argument.

If \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from a distribution that has a pdf which is a regular case of the exponential class, show that the pdf of \(Y_{1}=\sum_{1}^{n} K\left(X_{i}\right)\) is of the form \(f_{Y_{1}}\left(y_{1} ; \theta\right)=R\left(y_{1}\right) \exp \left[p(\theta) y_{1}+n q(\theta)\right]\). Hint: \(\quad\) Let \(Y_{2}=X_{2}, \ldots, Y_{n}=X_{n}\) be \(n-1\) auxiliary random variables. Find the joint pdf of \(Y_{1}, Y_{2}, \ldots, Y_{n}\) and then the marginal pdf of \(Y_{1}\).
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