TITLE: What are the mean and variance of the log of a random variable?
QUESTION [5 upvotes]: Here's the problem.
We have a random variable X that follows a Poisson law. If we take the log of this variable, what are the first two moments (mean and variance) of the law it follows?
This looks like a simple question, but I can't find anything about it. Any idea?
EDIT: In order to prevent the X=0 case, we bias the Poisson law. Our random variable becomes log(X+epsilon) with X~Poisson(λ).

REPLY [2 votes]: This is also called delta method. If $g(X)$ is a function of random variable $X$, and $g(X)$ is differentiable, in practice we can use first-order Tylor expansion at $\bar{X}$, the mean of $X$: $$g(X) \approx g(\bar{X}) + \frac{d}{dX}g(\bar{X})(X-\bar{X})$$
Therefore, $$E[g(X)] \approx g(\bar{X}) + \frac{d}{dX}g(\bar{X})E(X-\bar{X}) = g(\bar{X}),$$ because the second term is 0.
Similarly, the variant is:
$$Var[g(X)] \approx Var(\frac{d}{dX}g(\bar{X})X) = [\frac{d}{dX}g(\bar{X})]^2Var(X)$$
In your case $X\sim Poisson(\lambda)$, $g(X) = logX$; plug in and you got:
$$E(logX) \approx log\lambda$$
$$Var(logX) \approx \frac{1}{\lambda}$$
You have to be careful with X=0 case, as other folks have already stated.