# Bounds on integral for computing expectation

I have a discrete random variable $X$ with $P(X \geq x) = c^x$ and I would like to bound $E(\log{X})$. I can write this as follows I think $$E(\log{X}) = \sum_{x=1}^{\infty} c^x \log{x}.$$

We know that $0\leq c \leq 1$. I would like to bound $E(\log{X})$ above and below.

One would approach would be to replace the sum by an integral but I didn't get anywhere. Can anyone see how to get good bounds?

Question has been edited to make it clearer.

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What is $f(n)$ and how is $f(n)^x$ entering in the computation of $E(X)$? –  Did Nov 13 '12 at 19:54
Why should one use $f(n)^x$ to denote $P(X=x)$ eludes me, I must admit. Anyway, the RHS is $E(\log X)$, not $E(X)$. Are you looking for upper/lower bounds of $E(\log X)$ in terms of $E(X)$, or of other quantities? –  Did Nov 13 '12 at 20:01
But if $f(n)=1/n$, the sum of $f(n)^x$ on every positive integer $x$ is not $1$ (and the sum of $P(X=x)$ should be $1$). –  Did Nov 13 '12 at 20:09
@did, You are right that was a bad example to choose. –  Arloff Nov 13 '12 at 20:19
OK. So... what is the question, in the end? –  Did Nov 13 '12 at 20:21
Let us assume that $X\geqslant1$ is geometric with parameter $a$ in $(0,1)$, that is, that, for every $n\geqslant1$, $\mathbb P(X\geqslant n)=(1-a)^{n-1}$. Then $\mathbb E(X)=1/a$, hence Jensen inequality yields $$\mathbb E(\log X)\leqslant\log \mathbb E(X)=-\log a.$$ On the other hand, the function logarithm is nondecreasing hence, for every $n\geqslant1$, $$\mathbb E(\log X)\geqslant\log(n)\cdot \mathbb P(X\geqslant n)=\log(n)\cdot (1-a)^{n-1}.$$ In particular, for $n$ the integer part of $1/a$, one gets approximately $$\mathbb E(\log X)\stackrel{(\mathrm{approx.})}{\geqslant}-\log(a)\cdot(1-a)^{(1-a)/a}.$$ Note that when $a\to0$, $(1-a)^{(1-a)/a}\to\mathrm e^{-1}$ hence the lower bound is asymptotically of the order of the upper bound.