Problem on Gamma Distribution

Question

Suppose $X_1, X_2, . . . , X_n$ is a random sample of size $n$ from the probability density

$$f(x) = \frac{\alpha^p}{\Gamma(p)}x^{(p−1)}\exp(−\alpha x)$$

with $x > 0$ where $p$ is a known positive constant and $\alpha > 0$ is an unknown parameter. Let $\hat{\alpha} = p/\bar{X}$ be a proposed estimator of $\alpha$. Then,

(A) $\mathbb{E}[\hat{\alpha}] = \alpha$
(B) $\mathbb{E}[\hat{\alpha}] = \alpha/(1− 1/np)$
(C) $\mathbb{E}[\hat{\alpha}] = \alpha/(1−np)$
(D) none of the above statements is true.

Which will be the correct option?

For solving this problem I was assuming summation of $X_i$ would follow gamma distribution with parameter alpha and $np$ . For calculating $\mathbb{E}[1/\bar{X}]$ I calculated $n\mathbb{E}[T^{-1}]$ where $T=\sum_i X_i$ and I got option B but I am not sure about it. Can you please help me out?

Answer 1

As best I can tell, your reasoning is:

Since $X_i\sim\Gamma(p,\,\alpha)$, $\sum_iX_i\sim\Gamma(np,\,\alpha)$ so $\bar{X}\sim\Gamma(np,\,n\alpha)$. So$$\Bbb E[p/\bar{X}]=\int_0^\infty\frac{p}{x}\frac{(n\alpha)^{np}}{\Gamma(np)}x^{np-1}\exp(-n\alpha x)dx=\frac{p(n\alpha)^{np}}{\Gamma(np)}\frac{\Gamma(np-1)}{(n\alpha)^{np-1}}=\frac{\alpha}{1-1/np},$$viz. (B).

This reasoning is correct. You can prove the distribution of $\sum_iX_i$ with characteristic functions; you can then prove that of $\bar{X}$ by rescaling the PDF.