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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...

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Make the substitution $$ Y_i = -\log X_i. $$ It is easy to show that $Y_i$ has density $$ f(y ; \theta) = (\theta + 1) e^{-(\theta + 1) y} \ I(y > 0). $$ 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 $(\theta + 1) / (n - 1)$.

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  • $\begingroup$ "it's easy to show..." Could you elaborate here a little bit? What was your thought process here? Is the original pdf of the x's some sort of standard distribution (with a known identity for Y = -logX) I ought to know? Or did you intuit trying the Y=-logX transformation and then worked out the distribution for Y from scratch? $\endgroup$ – bill_e Sep 16 '18 at 19:09
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    $\begingroup$ @bill_e My logic was that, because this looks like a homework problem, it must be the case that $\sum_i -\log (X_i)$ has a simple distribution. Hence $-\log (X_i)$ probably has some kind of gamma distribution, because if it didn't then the question would be difficult (intro Math Stat courses learn that sums of iid gammas are gammas, and that is the only type of identity of this sort that seems like it might be useful). Anyway, the original pdf is a $\operatorname{Beta}(\theta+1, 1)$, but that fact didn't motivate the approach I took. $\endgroup$ – guy Sep 16 '18 at 19:11
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    $\begingroup$ @bill_e Also, I worked out the density by doing a change of variables (i.e., substitue $e^{-y}$ for $x$ and multiply by $|\frac d {dy} e^{-y}|$). $\endgroup$ – guy Sep 16 '18 at 19:15

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