Question

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 \(\theta\).

Short answer

The statistic \(Y = \sum_{i=1}^{n} X_i\) is a complete sufficient statistic for \(\theta\). The expected value of \(1 / Y\) is \(1 / (n\theta)\). The function of \(Y\) that is the minimum variance unbiased estimator (MVUE) of \(\theta\) is \(n / Y\).

Step-by-step solution

Proving the Sufficiency of Y

We start by proving that \(Y\) is a sufficient statistic for \(\theta\). The joint probability density function of a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from a distribution with the given pdf can be written as: \(f(x_{1}, x_{2}, \ldots, x_{n}; \theta) = (\theta^n)(e^{-\theta\sum X_{i}})(\Pi X_i)\). The joint pdf can be factorized into two functions, one of which depends on \(\theta\) through \(\sum X_{i}\) only. Thus, according to the Factorization Theorem, \(\sum X_{i}\), or \(Y\), is a sufficient statistic for \(\theta\).

Proving the Completeness of Y

Next, we prove the completeness of \(Y\). For complete statistics, if the expectation of a function g(Y) equals zero for all \(\theta\), then \(P(g(Y) = 0) = 1\). As such, Y is a complete statistic for \(\theta\).

Calculation of E(1 / Y)

Now, for the second task, \(1 / Y\) does not follow a well-known distribution. The expectation of \(1 / Y\) equals \(\int_0^\infty (1/y)f_Y{(y)} dy\), which equals \(\int_0^\infty n\theta (n-1)!(\theta y)^n exp(-n\theta y)/(y(n\theta)^n)dy\). This simplifies to \(\int_0^\infty n(n-1)!exp(-n\theta y) dy\), which equals \(\frac{n(n-1)!}{(n\theta)^n}\int_0^\infty t^{n-1}exp(-t) dt\). By recognizing the integrand as the pdf of a Gamma distribution, the result of the integral equals \(n!\). Thus \(E(1 / Y) = \frac{n(n-1)!}{n!(\theta n)} = \frac{1}{n\theta}\), which is unbiased for \(1 / n\theta\).

Find the MVUE of \(\theta\)

To find the MVUE of \(\theta\), we need to consider functions of \(Y\) that are unbiased for \(\theta\). Note that \(var(n/Y)\) is a function of \(\theta\), and indeed, \(var(n/Y)\) = \(n^{2}\theta^{-2}var(1/Y)\) = \(n^{2}\theta^{-2}E((1/Y)^2) - [E(1/Y)]^2\) = \(\theta^{-2}(2n / n\theta^2) - (n / n\theta)^2\), which equals \(\theta^{-2}/n\). Therefore, the MVUE of \(\theta\) is \(n/Y\).