# Expectation of MLE with a logarithm [duplicate]

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?

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

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 $$\frac{n-1}{n}$$