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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

Did
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triomphe
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  • 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 05 '13 at 22:41
  • the actual function is of the form f(z) = log (1 + z). – triomphe Jun 05 '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 05 '13 at 22:43
  • take $-x^2$, which is concave and negative and a counterexample. – Seyhmus Güngören Jun 05 '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 05 '13 at 22:54
  • however $-x^2$ is of course not increasing. so you need a general answer or only an answer for the $log(1+z)$ case.- – Seyhmus Güngören Jun 05 '13 at 23:00
  • I only need for the log(1+z) case, but I have a feeling it will hole generally for the defined f(z), thanks – triomphe Jun 05 '13 at 23:07
  • Are $X,Y$ independent? If so you can easily calculate the distribution of $(1+X)/(1+Y)$ and its density, then compute $E[\log((1+X)/(1+Y))]$ explicitly. See http://math.stackexchange.com/questions/33778/cdf-of-a-ratio-of-exponential-variables to get started. – nullUser Jun 06 '13 at 01:04

2 Answers2

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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).

Did
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  • Concavity and nonnegativity are not needed. – Did Jun 06 '13 at 06:45
  • 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 06 '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 06 '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
  • ?? To get a coupling one must define X and Y on a common probability space. – Did Jun 11 '13 at 11:34
  • Sorry the question should have been would coupling work even if $X, Y$ were in same probability space initially (before coupling)? – triomphe Jun 11 '13 at 11:53
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    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
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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.

Seyhmus Güngören
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