2

Suppose $X$ and $Y$ are arbitrary random variables and $E[X]>E[Y]$ where $X>0$ and $Y>0$.

Then would the following inequality hold? $\operatorname E\left[\int_0^1 t^{X} \, dt \right]<E\left[\int_0^1 t^{Y} \, dt \right]$.

I tried to solve it using Jensen's inequality but Jensen's inequality only provides lower bound for LHS or RHS.

user3509199
  • 175
  • Something seems off here. Suppose that $X=-1$ with probability one. Then $E[\int_0^1 t^X,dt]=\int_0^{1} t^{-1}dt$ which is divergent. Is there a restriction on the domain of $X$? (The previous issue goes away if $X>-1$.) – Semiclassical Sep 25 '18 at 14:43
  • @Semiclassical Thank you very much for pointing out. Both X and Y are strictly greater than zero. I have updated the question. – user3509199 Sep 25 '18 at 14:44
  • 1
    Ok. Under that assumption, $\int_0^1 t^x,dt=\frac{1}{1+x}$. So the question can be equivalently posed as: If $E[X]>E[Y]$, must $E[1/(1+X)]<E[1/(1+Y)]$? – Semiclassical Sep 25 '18 at 14:46
  • @Semiclassical Yes, I think so. Then I think the inequality may not hold. – user3509199 Sep 25 '18 at 14:48
  • ...And you are right to think that -- but maybe you should have checked the answer below slightly more thoroughly before accepting it? – Did Sep 25 '18 at 15:50
  • @Did Thanks you so much for your suggestion. It seems that the answer is deleted. – user3509199 Sep 25 '18 at 16:26

0 Answers0