# Expectation of a strictly increasing function

Assume that $X_1$ and $X_2$ are two i.i.d. random variables with pdf $f$. Also, assume that $a$ and $b$ are two fixed real numbers such that $a>b$. If $g$ is a strictly increasing function, do I have: $$E[g(a+X_1)] > E[g(b+X_2)],$$ where the expectations are with respect to the distribution of $X_1$ and $X_2$? To me, this seems trivial, because: $$E[g(a+X_1)] = \int g(a+x) f(x) dx > \int g(b+x) f(x) dx = E[g(b+X_2)],$$ where the inequality holds because for every $x$, $g(a+x)>g(b+x)$.

The reviewer of my paper says that's not always correct. Am I missing something? Are there situations where $E[g(a+X_1)] > E[g(b+X_2)]$ does not hold?

• This is true even when $X_1,X_2$ do not have pdf but are general random variables. But I think you can write $\geq$ instead of strict inequality. There may be bad sets. I am not sure about it though. Dec 28, 2016 at 18:20
• Indeed, since the integration operator is linear and $g(a+x) > g(b+x)$ for all $x$, I agree with your conclusion, as long as the integrals converge and the expected values exist. Perhaps the condition on $g$ could be stated more clearly for the reviewer? Dec 28, 2016 at 21:23