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Chapter 7: Problem 11

Let \(X_{1}, X_{2}, \ldots, X_{n}, n>2\), be a random sample from the binomial distribution \(b(1, \theta)\). (a) Show that \(Y_{1}=X_{1}+X_{2}+\cdots+X_{n}\) is a complete sufficient statistic for \(\theta\). (b) Find the function \(\varphi\left(Y_{1}\right)\) which is the MVUE of \(\theta\). (c) Let \(Y_{2}=\left(X_{1}+X_{2}\right) / 2\) and compute \(E\left(Y_{2}\right)\). (d) Determine \(E\left(Y_{2} \mid Y_{1}=y_{1}\right)\).

Short Answer

Expert verified
The statistic \(Y_{1}\) is a complete and sufficient statistic for \(\theta\). The MVUE of \(\theta\) is \(\frac{Y_{1}}{n}\). The expected value of \(Y_{2}\) and the conditional expectation of \(Y_{2}\) given \(Y_{1}=y_{1}\) are both \(\theta\).

Step by step solution

01

Showing \(Y_{1}\) is a complete sufficient statistic.

Given a random sample from the binomial distribution, \(Y_{1}=X_{1}+X_{2}+\cdots+X_{n}\) follows a binomial distribution with parameters \(n\) and \(\theta\). Thus, we can write the likelihood function \(f(x|\theta) = \binom{n}{x}\theta^{x}(1-\theta)^{n-x}\) where \(x = Y_{1}\). The statistic \(Y_{1}\) includes all parameters \(\theta\) which makes it a sufficient statistic. A sufficient statistic is complete when its conditional expectation is a function of the statistic itself, or any unbiased estimator based on it is unique. In this case, for any function \(g\), if \(E[g(Y_{1})] = 0\) for all \(\theta\), then \(g(Y_{1}) = 0\), hence \(Y_{1}\) is a complete sufficient statistic.
02

Detemining the MVUE of \(\theta\)

To find the minimum variance unbiased estimator (MVUE) of \(\theta\), it is always based on a complete sufficient statistic by Lehmann鈥揝cheff茅 theorem. An unbiased estimator of \(\theta\) based on the statistic \(Y_{1}\) could be \(T = \frac{Y_{1}}{n}\). As per the expectation operation, \(E[T] = \theta\), thus \(T\) is an unbiased estimator of \(\theta\). Since it is based on the complete sufficient statistic, it is the MVUE of \(\theta\). Hence \(\varphi(Y_{1}) = \frac{Y_{1}}{n}\).
03

Computing \(E[Y_{2}]\)

Given \(Y_{2}=(X_{1}+X_{2}) / 2\), we know that each \(X_{i}\) follows a binomial distribution with parameters \(1\) and \(\theta\). Therefore, \(E[X_{i}] = \theta\), and \(E[Y_{2}] = E[(X_{1}+X_{2})/2] = (E[X_{1}] + E[X_{2}])/2 = (\theta + \theta)/2 = \theta\).
04

Calculating \(E[Y_{2} | Y_{1}=y_{1}]\)

When we compute the expectation \(E[Y_{2} | Y_{1}=y_{1}]\), we know that \(Y_{1}\) is the sum of all \(X_{i}\)'s and \(Y_{2}\) is the average of \(X_{1}\) and \(X_{2}\) which are a subset of the variables constituting \(Y_{1}\). Given this, the expectation \(E[Y_{2} | Y_{1}=y_{1}]\) doesn't actually depend on the value of \(Y_{1}\) because \(X_{1}\) and \(X_{2}\) are independent of the rest. The expected value of \(Y_{2}\) remains the same as \(\theta\), even conditionally, hence \(E[Y_{2} | Y_{1}=y_{1}] = \theta\).

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Most popular questions from this chapter

. 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: First determine \(E\left(\bar{X}^{2}\right)\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample of size \(n\) from a distribution with pdf \(f(x ; \theta)=\theta x^{\theta-1}, 00\). (a) Show that the geometric mean \(\left(X_{1} X_{2} \cdots X_{n}\right)^{1 / n}\) of the sample is a complete sufficient statistic for \(\theta\). (b) Find the maximum likelihood estimator of \(\theta\), and observe that it is a function of this geometric mean.

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 \((n-1) /(n X)\) 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 \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample from a Poisson distribution with parameter \(\theta, 0<\theta<\infty .\) Let \(Y=\sum_{1}^{n} X_{i}\) and let \(\mathcal{L}[\theta, \delta(y)]=[\theta-\delta(y)]^{2}\). If we restrict our considerations to decision functions of the form \(\delta(y)=b+y / n\), where \(b\) does not depend on \(y\), show that \(R(\theta, \delta)=b^{2}+\theta / n .\) What decision function of this form yields a uniformly smaller risk than every other decision function of this form? With this solution, say \(\delta\), and \(0<\theta<\infty\), determine \(\max _{\theta} R(\theta, \delta)\) if it exists.

Let \(Y_{1}
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