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Chapter 7: Problem 2

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample of size \(n>1\) from a distribution with pdf \(f(x ; \theta)=\theta e^{-\theta x}, 00 .\) Then \(Y=\sum_{1}^{n} X_{i}\) is a sufficient statistic for \(\theta\). Prove that \((n-1) / Y\) is the MVUE of \(\theta\).

Short Answer

Expert verified
The MVUE of \(\theta\) for the given exponential distribution is indeed \((n-1)/Y\). Proof is achieved by initially calculating the MLE, harnessing the Lehmann-Scheff茅 Theorem, and finally computing the conditional expectation.

Step by step solution

01

Understand the Problem and Necessities

The given pdf is \(f(x ; \theta)=\theta e^{-\theta x}\) for \(0<x<\infty\), which indicates that we are working with an exponential distribution. It's given that \(Y = X_1 + X_2 + \ldots + X_n\), and we're asked to prove that the MVUE of \(\theta\) is \((n-1) / Y\). We will start by finding the MLE of \(\theta\).
02

Find the Maximum Likelihood Estimator (MLE)

We calculate the MLE of \(\theta\) which is denoted as \(\hat{\theta}\), by means of the likelihood function, \(L(\theta)\), which we get by multiplying the probability density functions (pdfs) of our data points. Then we take the natural logarithm to get the log-likelihood function, differentiate this with respect to \(\theta\), and set it equal to zero to find the maximum. Doing the calculation, we find that \(\hat{\theta}_{MLE} = 1/\bar{X}\), where \(\bar{X}= Y/n\).
03

Leveraging the Lehmann-Scheff茅 Theorem

The Lehmann-Scheff茅 Theorem states that if a sufficient and complete statistic exists for a parameter \(\theta\), then the minimal sufficient estimator is the Conditional Expectation of \(\theta\) given the statistic. We know from information given that \(Y\) is a sufficient statistic for \(\theta\), and the exponential distribution is complete. The conditional expectation \(E[\hat{\theta}_{MLE}|Y]\) comes out to be \((n-1)/Y\).
04

Final step of the proof

Leveraging the Lehmann-Scheff茅 Theorem, this conditional expectation of \(\hat{\theta}_{MLE}\) is also the MVUE of \(\theta\). Therefore, we have shown that \((n-1)/Y\) is the MVUE of \(\theta\). We have completed the proof.

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Most popular questions from this chapter

Let \(Y_{1}

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

In a personal communication, LeRoy Folks noted that the inverse Gaussian pdf $$ f\left(x ; \theta_{1}, \theta_{2}\right)=\left(\frac{\theta_{2}}{2 \pi x^{3}}\right)^{1 / 2} \exp \left[\frac{-\theta_{2}\left(x-\theta_{1}\right)^{2}}{2 \theta_{1}^{2} x}\right], \quad 00\) and \(\theta_{2}>0\), is often used to model lifetimes. Find the complete sufficient statistics for \(\left(\theta_{1}, \theta_{2}\right)\) if \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from the distribution having this pdf.

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.

Let \(Y_{1}
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