# Expected Value for a combination of IID from $f(x,\theta) = \frac{ \theta }{x^{ \theta + 1}}$

Let $X$ be a random variable with density function $$f(x,\theta) = \frac{ \theta }{x^{ \theta + 1}}$$ with $x \geq1$ and $\theta > 0$; $\theta$ is an unknown parameter not a variable. Suppose to draw i.i.d. $X_1, X_2, \cdots, X_n$. Calculate the Maximum Likelihood Estimator for $\theta$ and its bias if any (or, restated for mathematicians: show that $E(\frac{n}{log(X_1X_2\cdots X_n)}) \not = \theta$ and evaluate it).

Attempt: Since $f(t ) = \frac 1 t$ is convex and $log(X_1X_2\cdots X_n) = log(X_1) + \cdots log(X_n)$ using Jensen's Inequality we get $$E(\frac{n}{log(X_1X_2\cdots X_n)}) > \frac{1}{E(log(X))} = \theta$$ the former equation calculated with integration. I don't know how to calculate or estimate it more precisely.

Thanks!

Note that $$\log \left(\prod_{i=1}^n X_i \right) = \sum_{i=1}^n \log X_i,$$ so we are motivated to find the distribution of $Y_i = \log X_i$. We have $$f_{Y_i}(y) = f_{X_i}(e^y) e^y = \frac{\theta}{e^{y(\theta+1)}} e^y = \theta e^{-\theta y},$$ hence $Y_i \sim \operatorname{Exponential}(\theta)$, and $$\sum_{i=1}^n \log X_i \sim \operatorname{Gamma}(n, \theta).$$ Then the reciprocal of this is inverse gamma distributed, and we can explicitly compute the expectation, which I leave to you as an exercise.