# Mean of an increasing function over exponential distribution

I came across the following problem in my research

I have two random variables $X, Y$ which are exponentially distributed and $Y$ has a higher mean than $X$.

Then I have a function, say $f(z)$, which is known to be concave non negative and increasing in $z$. Can I claim that $$\mathbb{E}[f(Y)] > \mathbb{E}[f(X)]?$$

I tried with Jensen's inequality but it doesn't help to compare between two different random variables. If not general it's sufficient for me to know if the claim holds fo $f(z) = \log(1+z)$.

Thank you

• Do you know that $f(z) > 0$ or at least that $\int_\mathbb{R} f > 0$? Otherwise, likely there would be a counterexample... – gt6989b Jun 5 '13 at 22:41
• the actual function is of the form f(z) = log (1 + z). – triomphe Jun 5 '13 at 22:42
• I tried with linear i.e., f(z) = mz where m is a constant and the claim holds. I don't understand the counter example gt6989b, could you please explain a little more – triomphe Jun 5 '13 at 22:43
• take $-x^2$, which is concave and negative and a counterexample. – Seyhmus Güngören Jun 5 '13 at 22:49
• Sorry for the omission but now I updated the question as f(z) being non negative and the actual function is like f(z) = log(1+z). thanks Seyhmus – triomphe Jun 5 '13 at 22:54

A useful idea here is called coupling. Let us start from the fact that every exponential random variable $Z$ with mean $z$ is distributed like $zU$, where $U$ is a standard exponential random variable. Since expectations depend only on distributions, one is asked to prove that, for every $x\leqslant y$, $E[f(xU)]\leqslant E[f(yU)]$.

Since $xU\leqslant yU$ almost surely, this holds true for every nondecreasing function $f$ (such that the two expectations are finite).

• Concavity and nonnegativity are not needed. – Did Jun 6 '13 at 6:45
• Thank you Did. Could you please explain to me what you mean by " fact that every exponential random variable Z with mean z is distributed like zU, where U is a standard exponential random variable."? – triomphe Jun 6 '13 at 14:00
• I understood it Did, thanks a lot again. I have a result that depends on this claim. I would like to acknowledge you in my paper, if you wish. – triomphe Jun 6 '13 at 14:28
• Hello Did. Would coupling work even if $X, Y$ are defined in the same probability space? Thank you – triomphe Jun 11 '13 at 11:31
• No, coupling provides random variables X' and Y' (possibly defined on another probability space Omega') such that X' is distributed like X, Y' is distributed like Y, and (X',Y') has nice almost sure properties. But the original (X,Y) can be ugly. – Did Jun 11 '13 at 12:32

I also want to share my opinion although a very good answer is already available by @did. The $n$th moment of an exponential random variable is

$$E[X^n]=\frac{n!}{\lambda^n}$$

That is for every $n$, since $Y$ has a smaller $\lambda$ compared to $X$, we have a greater moment under $Y$ than under $X$. Since any linear scaling will not change the result, one can create an arbitrary function by the superposition of the scaled versions of $X^n$ for some set $n\in{\cal{N}}$. From the Taylor series expansion we can verify the claim.