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There are three convexly decreasing functions $\mathbb{R^+}\rightarrow \mathbb{R^+}$. $f,g $ and $h$. I'm wondering whether it's true that

$E[f(x)^2]E[g(x)h(x)]<E[f(x)g(x)](1+E[f(x)h(x)])$

for an arbitrary probability distribution.

iridium
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  • @simeon, it seems to be true in your example. I'm not sure whether it's ture. Maybe I'm more curious in a possible condition on $f,g$ and $h$ for this inequality to hold. like they are bounded by certain value.. – iridium Mar 10 '15 at 22:15
  • No. See counter-example below. The idea is to try to perturb a situation where $E[f(X)^2]=\infty$. – Michael Mar 10 '15 at 22:57

1 Answers1

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No, here is a counter-example. Let $X$ be uniform over $[0,1]$. For $x \geq 0$ define:

\begin{align} &f(x) = \frac{1}{x+0.01} \\ &g(x) = h(x) = e^{-x} \end{align}

Then: \begin{align} &E[f(X)^2] = \int_0^1 \left(\frac{1}{x+.01}\right)^2dx \approx 99.0099 \\ &E[f(X)g(X)] = E[f(X)h(X)] = \int_0^1 \frac{e^{-x}}{x+.01} dx \approx 3.8606 \\ &E[g(X)h(X)] = \int_0^1 e^{-2x}dx \approx 0.43233 \end{align}

So: $$ E[f(X)^2]E[g(X)h(X)] \approx 42.8 > 18.76 \approx E[f(X)g(X)](1+E[f(X)h(X)])$$

Michael
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  • Thank you! I just post another question like this with the condition that f(x)h(x)<1. Your example is quite useful. – iridium Mar 10 '15 at 23:24
  • I see you have edited the question to add a condition $f(x)h(x)<1$, which was not there when I answered the question. That is kind of annoying since it makes my answer look incomplete. – Michael Mar 11 '15 at 18:53
  • sorry about that I posted a new question with the condition added. – iridium Mar 11 '15 at 18:57
  • Thanks, I will look at the other one if I have time. – Michael Mar 11 '15 at 18:58
  • Okay I answered that question here: http://math.stackexchange.com/questions/1184532/inequality-involving-taking-expectations – Michael Mar 11 '15 at 19:59