# 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?

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Yes you can apply the delta method in this case. –  Michael Chernick Sep 29 '12 at 23:45
You should also look into the Jensen Inequality. –  kjetil b halvorsen Sep 30 '12 at 19:58
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## 3 Answers

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.

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Interestingly, This can be used to get an approximation to the digamma function. –  probabilityislogic Oct 3 '12 at 22:53
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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.

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If the second integral exists. It needs not to. Take Cauchy distribution and $g(x)=x^2$. –  mpiktas Sep 30 '12 at 13:34
I would add a second layer of pedantry by saying that you actually need $E[|g(X)|]<\infty$ for the expectation to be well defined. –  probabilityislogic Oct 3 '12 at 22:45
@mpiktas - This expectation actually does exist but it is infinite. A better example is $g(x)=x$ for the Cauchy distribution. This expectation depends on how the lower and upper limits of integration tend to infinity. –  probabilityislogic Oct 3 '12 at 22:49
@prob: No, you don't need that condition in your first comment, and even in a situation that may be very relevant to this question! (+1 to your second comment, though, which was something I had been meaning to comment on as well.) –  cardinal Oct 3 '12 at 22:51
@prob: It is sufficient, but if you compare your first comment to your second one, you'll see why it's not necessary! :-) –  cardinal Oct 3 '12 at 23:02
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There are two usual approaches:

1. If you know the distribution of $X$, you may be able to find the distribution of $\ln(1+X)$ and hence 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 $x$).

2. As you suggest, if you know the first few moments you can compute a Taylor approximation.

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