# Expectation of function of stochast

I've got a general question regarding a certain sticking point I often encounter. When tackling questions where for example an UMVUE (uniformly minimum-variance unbiased estimator) has to found I get stuck at the part where an expectation has to be determined. Maybe I'm overlooking certain theorems.

Example 1. Consider a random sample of size $n$ with $X_i\tilde{}Poi(\mu)$.

I am able to find the UMVUE of $\mu$ and the MLE (Maximum Likelihood Estimator) of $\theta=e^{-\mu}$. But then for example follows the question: is this MLE unbiased for $\theta$?

If I want to determine the expectation, should I just write out the sum as follows from the definition of the expected value of a discrete distributed variable? This seems to lead to quite an ugly expression...

(For this particular problem I found that Jensen's Inequality could come in handy to show biasedness (or actually non-unbiasedness))

But if I could summarize the problem I've got: how to rewrite expected values of functions of stochast in general? Which theorems or properties should I use... for example I do use $E(aX)=aE(X)$ for $a$ a constant and $var(X)=E(X^2)-E(X)^2$ to rewrite certain expectations to a combination of simpler/known ones.

Example 2. Take a random sample $n$ of a distribution with pdf $f(x;\theta)=\theta x^{\theta -1}$ if $0<x<1$ and else $0$ and with $\theta > 0$.

Find the UMVUE of $\theta$

Here I get stuck again at the expectation. I've got a sufficient and complete statistic in $\sum ln(X_i)$.

The expected value, i'd figure, is equal to $nE(ln(X_i))=-n/ \theta$ $\,\,\,\,$(where $[-ln(X)]=\frac{1}{\theta}$ was given as a hint).

But what if I want to determine the expected value of $\frac{1}{\sum_{i}ln(X_i)}$ ?

In general: how to determine the expectations of functions of stochasts? For example $E[ln(X)]$or $E[e^{\bar{X}}]$...?

If $X$ has density $f_X$, then for every function $u$, $E[u(X)]=\displaystyle\int u(x)f_X(x)\mathrm dx$. For example, in your last case, $$E_\theta\left[\frac1{\sum\limits_i\ln X_i}\right]=\iint\cdots\int\frac1{\sum\limits_i\ln x_i}\prod_if(x_i;\theta)\prod_i\mathrm dx_i.$$ If $X$ is discrete and $p_x=P[X=x]$ for every $x$, this should read $E[u(X)]=\displaystyle\sum\limits_xu(x)p_x$.