$\mathbb{E}\left[\frac{X}{Y}\right]$ relationship to $\frac{\mathbb{E}[X]}{\mathbb{E}[Y]}$ for random variables X and Y that are not independent

Question

Suppose $Y > 0$ with probability 1, $\mathbb{E}[X]$ and $\mathbb{E}\left[\frac{1}{Y}\right]$ are finite, and $\mathbb{E}[X] > 0$.

I know that if X and Y are independent, then we have $\mathbb{E}\left[\frac{X}{Y}\right] = \mathbb{E}[X]\mathbb{E}\left[\frac{1}{Y}\right]$, and we can use Jensen's Inequality to say that $\mathbb{E}\left[\frac{X}{Y}\right] \geq \frac{\mathbb{E}[X]}{\mathbb{E}[Y]}$.

But what if X and Y are dependent? Can we still derive the same inequality?

Answer 1

For instance, let's specialize to the case $X=Y^2$. Is it true that $E[Y]\ge E[Y^2]/E[Y]$, i.e. that $\operatorname{Var}(Y)\le 0$ ? Well, only quite situationally.

Answer 2

For a positive result, note that if $X$ and $1/Y$ are square integrable and positively correlated, then \begin{align*} [X/Y] &= \mathrm{Cov}[X,1/Y] + [X][1/Y] \\&\geq \mathrm{Cov}[X,1/Y] + [X]/[Y] \\ &\geq [X]/[Y]. \end{align*}

For example this will be the case when $X$ is a decreasing function of $Y$ by the FKG inequality.