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Let $X_1,...,X_n$ be i.i.d. with common density $$f(x)=\theta x^{\theta -1}I\{x \in [0,1]\}$$ where $\theta >0$.

e) Determine whether the MLE is unbiased for $\theta$. If not unbiased, could you redefine it to make it unbiased?

I found the MLE in a previous part of the problem to be $\frac{n}{-\sum_{i=1}^n \ln(x_i)}$ (I hope I'm correct on this; please inform me if I'm not). I know for an estimator to be unbiased, $E(\theta)=\theta$, but taking the expectation of the MLE is proving challenging. Any help? Or did I just find the MLE completely incorrectly?

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  • $\begingroup$ This is answered here and here. $\endgroup$ Commented Mar 15, 2020 at 8:13

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I think your MLE derivation is correct.

Calculating the expectation of MLE is a bit tricky, but once you figure out the distribution of $-\ln(X_i)$, there are ways to get the expectation without a whole lot of computation.

Let $Z = -\ln(X)$, and the CDF of $Z$ would be:

$F(z) = P(-\ln(X) \leq z) $

$= P(X \geq e^{-z})$

$= \int_{e^{-z}}^1 \theta x^{\theta-1} dx \qquad$ (for $z \geq 0$)

$= 1 - e^{-z\theta}$

Looks familiar? Yes, that's the CDF for exponential distribution $Exp(\theta)$. So $-\ln(X) \sim Exp(\theta)$.

Note that $Exp(\theta)$ is also a gamma distribution $Gamma(1, \theta)$, and the sum of a bunch of independent gamma distributions is also gamma (wiki: Gamma distribution). Therefore,

$-\sum_i \ln(X_i) \sim Gamma(n, \theta)$

$\Rightarrow \frac{1}{-\sum_i \ln(X_i)} \sim InverseGamma(n, \theta)$

$\Rightarrow E\frac{n}{-\sum_i \ln(X_i)} = n \frac{\theta}{n-1}$

It is indeed biased, but can be easily corrected by multiplying $\frac{n-1}{n}$

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