0
$\begingroup$

[Spoiler: it's not true (see comments)]

I believe the following to be true, but am not sure how to prove it:

$$ \mathbb{E}_1[x] > \mathbb{E}_2[x] \implies \mathbb{E}_1 [g(x)] > \mathbb{E}_2 [g(x)] $$ when $g$ is an increasing, convex function. (We can also restrict ourselves to $x \geq 0$ for my problem.) By the subscripts $1$ and $2$, I refer to two different probability distributions over x.

In the continuous case, we'd have

$$ \int x f_1(x) dx > \int x f_2(x) dx \implies \int g(x) f_1(x) dx > \int g(x) f_2(x) dx $$

for two pdfs $f_1$ and $f_2$.

Question: Is my conjecture right, and if so, what's the proof?

What I've tried: We might like to write it this way:

$$ \int x (f_1(x) - f_2(x)) dx > 0 \implies \int g(x) (f_1(x) - f_2(x)) dx > 0 $$

I've tried using integration by parts to rewrite it in terms of $g^{\prime}(x)$ and the cdfs, but I wasn't able to carry that through. It seems promising, but I wasn't sure how to use it.

I also thought about assuming that $g(x)$ is analytic and using the Taylor series, but again, I'm not sure where to take that either.

Sidenote: It would also be cool to show whether $g$ has to be strictly increasing everywhere for this to be true.

Thanks for any advice!!

  • 0
    My advice: do not try to prove this *result*, it does not hold (and counterexamples abound).2012-04-23
  • 0
    OK, thanks, but I'm having trouble coming up with any counter-examples (pictorially, it seems counterintuitive). Could you give one or give me an idea of what to look for? Thank you!2012-04-23
  • 0
    Compare uniform distributions on intervals with nearly equal midpoints.2012-04-23
  • 0
    OK, I think I see by taking one distribution to be uniform on [0,1] and the other to be uniform on [0.5,0.5+epsilon]. The same argument with a minus epsilon would show a counterexample for concave functions. Thanks you, that answers my question!2012-04-23
  • 1
    Indeed. A common practice in such cases is that the OP (you!) writes an answer which, after a while so that others can check it, may be accepted. This allows to close the question.2012-04-23
  • 0
    OK, thank you -- I will!2012-04-23

1 Answers 1