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I have a problem in comparing expected value of two functions.

Let $X,Y$ be two i.i.d. random variables, exponentially distributed with mean $\lambda$. I want to claim that $E [\log(1+ \frac{1}{X})]\geqslant E [\log(1+\frac{1}{X+Y})]$.

I have a feeling this is correct. But I have no idea how to prove it.

Thank you.

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You need a bit more information about the distribution. If $X$ and $Y$ are uniformly distributed between -4 and -5, then it seems unlikely that the inequality holds. – deinst Jun 10 '13 at 22:36
Hello deinst, I corrected it. Both $X, Y$ are nonnegative. – triomphe Jun 10 '13 at 23:17
up vote 1 down vote accepted

Let $g(x,y)$ denote the joint probability density function for $X$ and $Y$ - so that for "nice" subsets $A\subseteq[0,\infty)\times[0,\infty)$, $$ P((X,Y)\in A)=\iint\limits_A g(x,y)\,dA. $$ Then $$ \mathbb{E}\left[\log\left(1+\frac{1}{X+Y}\right)\right]=\iint\limits_{[0,\infty)^2}\log\left(1+\frac{1}{x+y}\right)g(x,y)\,dA. $$ Now, for any $x,y\geq 0$, $$ \log\left(1+\frac{1}{x+y}\right)\leq\log\left(1+\frac{1}{x}\right), $$ because the logarithm is an increasing function on positive numbers. So, $$ \iint\limits_{[0,\infty)^2}\log\left(1+\frac{1}{x+y}\right)g(x,y)\,dA\leq\iint\limits_{[0,\infty)^2}\log\left(1+\frac{1}{x}\right)g(x,y)\,dA. $$ This function is non-negative; so, by Tonelli's theorem, $$ \iint\limits_{[0,\infty)^2}\log\left(1+\frac{1}{x}\right)g(x,y)\,dA=\int_0^{\infty}\int_0^{\infty}\log\left(1+\frac{1}{x}\right)g(x,y)\,dy\,dx. $$ Since the log term doesn't depend on $y$, it can be pulled out of the inside integral to get $$ \int_0^{\infty}\log\left(1+\frac{1}{x}\right)\left[\int_0^{\infty}g(x,y)\,dy\right]\,dx=\int_0^{\infty}\log\left(1+\frac{1}{x}\right)f_X(x)\,dx, $$ where $f_X(x)$ is the (marginal) probability density function for $X$. But, then this last expression is exactly $\mathbb{E}[\log(1+\frac{1}{X})]$! This proves the inequality.

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Thanks a lot for the answer. – triomphe Jun 11 '13 at 2:04

I don't know what your background is; to prove the result in the stated generality, you really need to be coming at this from measure theory. (If you want a more elementary result, we need more assumptions; are the variables uniform on some set? Are they discrete? Etc.)

In a measure-theoretic setting, this result is actually very quick to prove. You need to combine the following facts:

1) Because $X, Y\geq 0$, then $\frac{1}{X}\geq\frac{1}{X+Y}$. (To be more specific: if $X$ and $Y$ are random variables on the measurable space $(\Omega,\mathcal{F},P)$, and $X(\omega),Y(\omega)\geq0$ for all $\omega\in\Omega$, then $\frac{1}{X(\omega)}\geq\frac{1}{X(\omega)+Y(\omega)}$ for all $\omega\in\Omega$.)

2) The logarithm is an increasing function on positive inputs. So, bringing (1) back in to this, we know that $$ \log\left(1+\frac{1}{X}\right)\geq\log\left(1+\frac{1}{X+Y}\right), $$ in the same sense as above.

3) This implies the desired inequality on the expectations. Remember, random variables are just functions on $\Omega$ that are measurable with respect to $\mathcal{F}$, and the expectation is simply the integral with respect to probability measure. All we need here is the fact that $f(\omega)\leq g(\omega)$ for all $\omega\in\Omega$ implies $$ \int_{\Omega}f\,dP\leq\int_{\Omega}g\,dP. $$

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Hello nrperterson. Thank you. Could you please explain what you mean by "are the variables uniform on some set"? May be I can provide more information. I know very little about measure theory. But I am not sure if you can assume that both $X, Y$ are defined in the same measurable space. In my case these two random variables are not said to be in same measurable space. I only know they are independent. – triomphe Jun 10 '13 at 23:50
I was just asking if you were given any more assumptions related to the variables X and Y. By asking if they were uniform, I was asking "Do they have the uniform distribution?", whether a continuous or discrete one. As for the other: the statement "$X$ and $Y$ are independent random variables" requires that they be defined on the same measure space. – Nick Peterson Jun 10 '13 at 23:57
Yes I have that information. They are exponentially distributed. Thanks – triomphe Jun 11 '13 at 0:00
Well, that changes things, and allows a much more direct proof. In the future, please include ALL of the relevant information! One moment. – Nick Peterson Jun 11 '13 at 0:01
@MLT I added an answer that will work for any two IID random variables with a continuous distribution (that is, a distribution that has a probability density function). – Nick Peterson Jun 11 '13 at 1:24

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