Let $(X1….Xn)$ be i.i.d.random variables with p.d.f.
$$f(x) =\frac{1}{\theta} (1-x)^{\frac{1-\theta}{\theta}} $$ $0<x<1$ $\theta>0$
- find MLE specifying asymptotic variance
I found $$\theta_{MLE} = -\frac{\sum_{i=1}^n log(1-X_i)}{n}$$
Since the second derivative of the likelihood is negative in $\theta_{MLE}$ What I found is a max
Then to specify the asymptotic variance I should find the fisher information and the asymptotic variance is equal to the inverse of the fisher
I got for the second derivative $$\frac{d^2logL(\theta)}{d^2(\theta)} = \frac{n}{\theta^2}+2\frac{\sum_{i=1}^n log(1-X_i)}{\theta^3}$$
To find the fisher information I should compute the expected value of the second derivative $$I_n = -E(\frac{d^2logL(\theta)}{d^2(\theta)})$$
I get stuck finding the expected value of $\sum_{i=1}^n log(1-X_i)$ I could find $E(X)$ by doing the $\int_{0}^1 x*f(x)$ but how Can I find the expected value of $\sum_{i=1}^n log(1-X_i)$
help would be greatly appreciated
Best Answer
Let $Y = -\log( 1 - X) $, then \begin{align} F_Y(y) &= P(Y\le y)\\ &= P(-\log(1-X)\le y)\\ &= P( e^{-y} \le 1 - X )\\ & = P ( X < 1- e^{-y})\\ & = F_X(1-e^{-y}), \end{align} hence, using the fact that $f_Y(y) = \frac{\partial }{ \partial y} F_Y(y)$, we have $$ f_Y(y) = f_X( 1- e^{-y}) e^y = \theta^{-1}e^{-y(1/\theta - 1)}e^y = \theta^{-1} e^{-y/\theta}, $$ where $ 0 < X < 1$ then $1 < 1 - X < 0$, hence $Y = - \ln ( 1- X) > 0$. So, as you can see, $Y \sim Exp(1/\theta)$, hence $E[Y] = \theta$.