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
 A: 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).
A: 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.
