Expected value of a natural logarithm 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?
 A: Also, if you don't need an exact expression for $\text{E}[\log(X + 1)]$, oftentimes the bound given by Jensen's inequality is good enough:
$$
\log [\text{E}(X) + 1] \geq\text{E}[\log(X + 1)] 
$$
A: There are two usual approaches: 


*

*If you know the distribution of $X$, you may be able to find the distribution of $\ln(1+X)$ and from there find its expectation; alternatively you may be able to use the law of the unconscious statistician directly (that is, integrate $\ln(1+x) f_{X}(x)$ over the domain of $x$). 

*As you suggest, if you know the first few moments you can compute a Taylor approximation.
A: Suppose that $X$ has probability density $f_X$. Before you start approximating, remember that, for any measurable function $g$, you can prove that 
$$
  E[g(X)]=\int g(X)\,dP = \int_{-\infty}^\infty g(x)\,f_X(x)\,dx \, ,
$$
in the sense that if the first integral exists, so does the second, and they have the same value.
A: 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.
