I am currently trying to determine what sorts of general conditions will allow the inequality, $E\left(\frac{X}{Y}\right) \geq 1$ to hold for positive r.v.'s $X$ and $Y$. One condition is to let $X$ and $Y$ be exchangeable, that is, the joint distribution of $(X,Y)$ is the same as the joint distribution of $(Y,X)$. However, I read in a book that an even more general condition than exchangeability holds. The hint given was to determine a concave function and to use Jensen's Inequality. One idea I had was to define the function $g(x,y) = \frac{x}{y}$. If this function is convex, then we have that: $$E\left(g(X,Y)\right) = E\left(\frac{X}{Y}\right) \geq g(E(X,Y)) = g(E(X), E(Y)) \geq \frac{E(X)}{E(Y)}$$ which then implies that the condition is the the means are equal. However, $g(x,y) = \frac{x}{y}$ is not convex and I am not able to think of another example that works. Would anyone have any ideas on whether I am approaching this problem correctly?
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Davide Giraudo
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user154510
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$\frac{X}{Y}$ is what is called a Ratio Distribution (see here: http://mathworld.wolfram.com/RatioDistribution.html and here: http://math.stackexchange.com/questions/77873/how-calculate-the-probability-density-function-of-z-x-1-x-2/77875#77875). The pdfs are of these distributions are known, so perhaps that may help – Brenton Dec 14 '15 at 19:10
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The function $\phi\colon t\mapsto t+1/t$ is convex on $(0,+\infty)$, since its second derivative is $2/t^3$. Consequence, applying Jensen's inequality to $X/Y$, we get $$\phi\left(\mathbb E\left[\frac XY\right]\right)\leqslant \mathbb E\left[\frac XY+\frac YX\right].$$ Since $\phi(t)\geqslant 2$ for each $t\gt 0$, we have $$2\leqslant \mathbb E\left[\frac XY+\frac YX\right],$$ hence we have $\mathbb E\left[\frac XY\right]\geqslant 1$ as long as $$\mathbb E\left[\frac XY\right]\leqslant \mathbb E\left[\frac YX\right],$$ which is weaker than exchangeability.
Davide Giraudo
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