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Prove $E[X/Y]\ge1$ for X,Y iid positive random variables.
My attempt:
Let $Z=X/Y$ $Z\in(0,\infty)$. $$E[Z]=\int_0^\infty P(Z\gt z)dz=\int_0^\infty P(Y\lt{1\over z}X)dz$$ $$=\int_0^\infty\int_0^\infty\int_0^{{1\over z}X}f(x)f(y)dydxdz$$ $$=\int_0^\infty\int_0^\infty f(x)F({1\over z}X)dxdz$$ $$=\int_0^\infty (F(x)F({1\over z}X)|_0^{\infty}-\int_0^\infty F(x)f({1\over z}X){1\over z}dx)dz$$ $$=\int_0^\infty(1-\int_0^\infty F(x)f({1\over z}X){1\over z}dx)dz$$

This is how I get 1 in the equation. I hope it has something to do with the question. I know its a mess. Can anyone help me out?

clement
  • 587

2 Answers2

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Integrals are overkill. When $X,Y$ are i.i.d., we have $E[X/Y] = E[Y/X]$, and $X/Y + Y/X$ is always at least $2$ provided $X$ and $Y$ are positive. (For instance, by the AM-GM inequality, $\frac{X/Y+Y/X}{2} \ge \sqrt{X/Y \cdot Y/X} = 1$.)

Therefore $2 E[X/Y] = E[X/Y] + E[Y/X] = E[X/Y + Y/X] \ge 2$.

Misha Lavrov
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3

If $Z$ is a positive random variable, then $E(\frac{1}{Z})\ge \frac{1}{E(Z)}$ (Jensen's equality applied to $g(z)=\frac{1}{z}$ , convex in $\mathbb{R}_{>0}$).

Hence

$$E[X/Y] = E[X] \, E[1/Y] \ge \frac{E[X]}{E[Y]}=1$$

where the first equality comes from independence, the second one from identically distributed.

Actually independence is not required, it's enough to have the joint density symmetric in the variables $X,Y$ (as Misha Lavrov's answer shows). With this weaker assumption, we can still apply Jensen inequality for the same function, but now for the variable $Z=X/Y$. We get $E(X/Y) E(Y/X) \ge 1$, and since (by symmetry) $E(X/Y)=E(Y/X)$ :

$$E(X/Y)^2 \ge 1\implies E(X/Y) \ge 1$$

leonbloy
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