Estimator (preferably unbiased) of $\ln (\text{E}[X])$ Given the distribution function of random variable $X$ I know how to estimate its mean. What would be an estimator (preferably unbiased) of $\ln(\text{E}[X])$ ?
 A: Following Glen's comment, if you have independent and identically distributed positive random variables $\{X_i\}_{i\geq 1}$ for which $\mu=\mathrm{E}[X_1]$ exists, then the sample mean
$$
  \bar{X}_n = \frac{1}{n}\sum_{i=1}^n X_i
$$
is clearly an unbiased estimator of $\mu$, because you can easily show that $\mathrm{E}[\bar{X}_n]=\mu$ for every $n\geq 1$. The natural logarithm is a concave function, and Jensen's inequality tells us that $\ln \bar{X}_n$ is a biased estimator of $\ln \mu$ because (supposing that $\bar{X}_n$ isn't degenerate for each $n\geq 1$)
$$
  \mathrm{E}[\ln \bar{X}_n] < \ln\mathrm{E}[\bar{X}_n]=\ln \mu \, .
$$
On the other hand, $\ln \bar{X}_n$ is a strongly consistent estimator of $\ln \mu$ since the Strong Law of Large Numbers says that
$$
 \ln \bar{X}_n \to \ln\mu \, ,
$$
almost surely when $n\to\infty$. With additional information about the distribution of the $X_i$'s you may try to get an upper bound for
$$
  \Pr(|\ln\bar{X}_n-\ln\mu|\geq\epsilon) \, ,
$$
for every $\epsilon>0$.
