An iid sample $X_1,\dots\,X_n$ have pdf $f(x;\theta)=\theta x^{\theta-1},\,\,\,\, 00$. The MLE is $$\hat{\theta}=\frac{-n}{\sum_{i=1}^{n}\log(X_i)}$$. I need to find the expectation of $\hat{\theta}$.

Question

An iid sample $X_1,\dots\,X_n$ have pdf $f(x;\theta)=\theta x^{\theta-1},\,\,\,\, 0<x\le 1, \,\,\theta >0$. The MLE is $$\hat{\theta}=\frac{-n}{\sum_{i=1}^{n}\log(X_i)}$$. I need to find the expectation of $\hat{\theta}$.

kirbykirby · Answer

It suggests showing $-\log X_i$~EXP$(1,\theta)$, then $-\sum_{i=1}^n \log X_i$~GAM$(n, 1/\theta)$, which I did and is ok.

The next step though says $$E\left[\left(-\sum_{i=1}^n \log X_i\right)^{-1}\right]=\left(\frac{1}{\theta}\right)^{-1}\frac{\Gamma(n-1)}{\Gamma(n)}=\frac{\theta}{n-1}$$

I understand that the expected value for GAM$(n, 1/\theta)$ is $n/\theta$, so I'm not sure how they got the line above?? Do we apply a 1-1 transformation Y=1/X, and then calculate the expectation by definition, or is there a faster way to do this?

kirbykirby · Answer

Well a transformation Y = 1/W, where W = $-\sum$log Xi, I should say.

anonymous · Answer

The question is very unclear.

kirbykirby · Answer

How is it unclear?

kirbykirby · Answer

@wio

kirbykirby · Answer

iid: independent and identically distributed,
pdf: probability density function
MLE: maximum likelihood estimator
is this better?