6
$\begingroup$

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?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ 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. $\endgroup$
    – Landon Carter
    Dec 28, 2016 at 18:20
  • 4
    $\begingroup$ 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? $\endgroup$
    – jwimberley
    Dec 28, 2016 at 21:23

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.