Let $X$ be a random variable with support $S$, let $f$ and $g$ be arbitrary positive functions of $X$. If $$\frac{\mathbb E(f(X))}{\mathbb E(g(X))} = r$$ then is it true that there exists $x\in S$ such that $$\frac{f(x)}{g(x)}\le r$$ I think this is true, and can't come up with a counterexample, but I'm not sure how to show it.
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It is true. It is a special case of the more general result:
If $E[h(X)]=0$ then there exists $x$ such that $h(x)\le 0$.
Your claim corresponds to $h(x):=f(x)-rg(x)$. You can prove the general result via contrapositive: If $h(x)>0$ for every $x$ then $h(X)$ is a positive random variable, hence its expectation is not zero. (This conclusion follows from the fact that a nonnegative random variable with zero expectation must be zero almost surely.)
grand_chat
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How is it a case of that? Doesn't your argument apply to $\mathbb E\left[\frac{f(X)}{g(X)}\right]$? The question is about $\frac{\mathbb E[f(X)]}{\mathbb E[g(X)]}$. – joriki Feb 08 '23 at 09:04
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I see now – you're applying this to $h(X)=f(X)-rg(X)$? Perhaps you should make that explicit. – joriki Feb 08 '23 at 09:30
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this makes sense, thanks for your help – AWhite Feb 08 '23 at 14:09
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1@joriki I added a note to make the connection explicit. – grand_chat Feb 08 '23 at 16:29