Is $\hat{\theta}$ unbiased?

Question

The random variable $X$ has pdf $f_{\theta}(x)=\theta x^{\theta-1}, \ > 0<x<1,$ for an unknown parameter $\theta>0.$ Assume that $x_1,...,x_n$ measure data from a sample on $X$.

a) Compute the ML-estimator $\hat{\theta}$ of $\theta.$

b) Compute $E[\ln X]$ as a function of $\theta$ if $X$ is distributed according to the above.

c) Is $\hat{\theta}$ unbiased?

Long story short, a) and b) I have computed to be

a): $$\hat{\theta}=-\frac{n}{\sum_{k=1}^n\ln{X_k}},$$

b): $$E[\ln X]=-\frac{1}{\theta}.$$

This is correct as far as the book. Now to the real problem. Here is what I did on c):

$\hat{\theta}$ is unbiased iff $E[\hat{\theta}]=\theta.$ Let's check if it's true. We have that

$$E[\hat{\theta}]=E\left[\frac{-n}{\sum_{k=1}^n\ln{X_k}}\right]=\frac{-n}{\sum_{k=1}^nE[\ln{X_k}]}=\frac{-n}{-\frac{n}{\theta}}=\theta.\tag1$$

Yes, it is unbiased.

Correct answer: No, it is not unbiased. Take for example $\theta = 1$ and $n=1$, then $\hat{\theta}=1/\ln{X}$ and

$$E[\hat{\theta}]=-E\left[\frac{1}{\ln{X}}\right]=\infty.$$

Question:

What is wrong with $(1)?$

Answer 1

$E(\frac{1}{X})$ is generally not equal to $\frac{1}{E(X)}$.

Answer 2

$f(x)=1/x$ is strictly convex for $x>0$, so by Jensen's inequality, $E[f(X)]\ge f(E[X])$ for a positive r.v. $X$, moreover, it the inequality is strict unless $X$ is a constant. In our case, $$ E\left[\frac{1}{\frac1n \sum_{i=1}^n \big(-\ln X_i\big)}\right]> \frac{1}{E\left[\frac1n \sum_{i=1}^n \big(-\ln X_i\big)\right]} = \theta, $$so the estimator has a positive bias for any $n\ge 1$.