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?

  • 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. $\endgroup$ 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


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.