# Solved – Expected value of a natural logarithm

mathematical-statistics

I know $E(aX+b) = aE(X)+b$ with $a,b$ constants, so given $E(X)$, it's easy to solve. I also know that you can't apply that when its a nonlinear function, like in this case $E(1/X) \neq 1/E(X)$, and in order to solve that, I've got to do an approximation with Taylor's.
So my question is how do I solve $E(\ln(1+X))$?? do I also approximate with Taylor?

In the paper

Y. W. Teh, D. Newman and M. Welling (2006), A Collapsed Variational Bayesian Inference Algorithm for Latent Dirichlet Allocation, NIPS 2006, 1353–1360.

a second order Taylor expansion around $x_0=\mathbb{E}[x]$ is used to approximate $\mathbb{E}[\log(x)]$:

$$\mathbb{E}[\log(x)]\approx\log(\mathbb{E}[x])-\frac{\mathbb{V}[x]}{2\mathbb{E}[x]^2} \>.$$

This approximation seems to work pretty well for their application.

Modifying this slightly to fit the question at hand yields, by linearity of expectation,

$$\mathbb{E}[\log(1+x)]\approx\log(1+\mathbb{E}[x])-\frac{\mathbb{V}[x]}{2(1+\mathbb{E}[x])^2} \>.$$

However, it can happen that either the left-hand side or the right-hand side does not exist while the other does, and so some care should be taken when employing this approximation.