Given a sample $x_1, x_2, \cdots x_n$ from the pdf:
$$ f(x ; \theta) = (\theta + 1) x^\theta $$ where $0 < x < 1$ and $\theta > -1$ is unknown. What is the bias of the MLE of $\theta$?
I've found the MLE to be
$$ \hat\theta = \frac{-n}{\sum_{i=1}^{n} \log(x_i)} - 1 $$
but I'm stuck on finding the bias of this estimator. The sum in the denominator makes it hard to take the expected value. I think there is something simple here that I am missing...
Make the substitution $$ Y_i = -\log X_i. $$ It is easy to show that $Y_i$ has density $$ f(y ; \theta) = (\theta + 1) e^{-(\theta + 1) y} \mathcal{I}(y > 0). $$ so $Y_i \sim \operatorname{Exponential}(\theta + 1)$. It follows that $\sum_i -\log (X_i) \sim \operatorname{Gamma}(n, \theta + 1)$. Recall that if $Z \sim \operatorname{Gamma}(\alpha, \beta)$ then $E Z^{-1} = \frac{\beta}{\alpha - 1}$. Therefore $$ E_\theta(\widehat \theta) = ... = \frac{n\theta + 1}{n - 1} $$ and the bias is then easily shown to be $(\theta + 1) / (n - 1)$.
