More or less as the title says. I was writing up a proof and I realized I had intuitively assumed this to be the case without proof.
Let $X$, $Y$, $T$, and $U$ be random variables taking values in the nonnegative reals. Assume they all have well-defined expectations and $\Pr(T = 0) = \Pr(U = 0) = 0$. Assume finally that $E(X) > E(Y)$ and $E(T) \le E(U)$. Clearly we have
$$ \frac{E(X)}{E(T)} > \frac{E(Y)}{E(U)}. $$
My question is: do we also have
$$ E\left(\frac{X}{T}\right) > E\left(\frac{Y}{U}\right)? $$
As mentioned it seems intuitively true to me but I don't even know where to begin to try to prove it.
If it turns out to be false I'd be interested to know if it's perhaps true when $X$ and $T$ are assumed to be independent and $Y$ and $U$ are assumed to be independent as well.
Thanks in advance for any insight!
No. Choose $Y = T = 1$, $X = 2$ and let $U$ assume the values $\varepsilon$ and $2-\varepsilon$ with equal probability. Now choose $\varepsilon$ sufficiently small.
Moments don't behave well at all in conjunction with quotients.