# Upperbound for conditional expectation of $X/Y$

Suppose that $$X$$ and $$Y$$ are positive random variables and are negatively correlated, i.e. $$\text{cov}(X,Y)<0$$. Is the following inequality correct for a positive constant $$a$$?

$$\mathbb{E}\Big[\frac{X}{Y}\Big|Y>a\Big] \leq \mathbb{E}\Big[\frac{X}{Y}\Big].$$

My reasoning is that the left-hand side takes the expectation over larger values of $$Y$$. Furthermore, $$X$$ and $$Y$$ are negatively correlated so the expectation is taken over smaller values of $$X$$ as well. Since both $$X$$ and $$Y$$ are positive, the fraction $$\frac{X}{Y}$$ is smaller in this region and so is its expectation.

The conjecture is false. Here is a simple counter-example.

The joint distribution $$(X,Y)$$ has only three points. For some tiny $$\epsilon > 0$$:

• $$P((X,Y) = (1,2)) = \frac12 - \epsilon$$

• $$P((X,Y) = (2,1)) = \frac12 - \epsilon$$

• $$P((X,Y) = (1000, 10)) = 2\epsilon$$

Assuming $$\epsilon$$ is tiny, we have:

• $$E[X] = E[Y] \approx \frac32$$

• $$E[XY] \approx 2$$

• $$Cov(X,Y) \approx 2 - \frac32 \frac32 = -\frac14 < 0$$

• $$E[X/Y] \approx \frac12 (2 + \frac12) = \frac54$$

• But $$E[X/Y \mid Y > 5] = 1000/10 = 100$$

Obviously, this example works because the covariance is dominated by what happens when $$Y < 5$$, and when we condition on $$Y > 5$$ all that becomes irrelevant.

In my experience, there is very little one can conclude based on covariance (if the r.v.s are not otherwise restricted). Many things that sound "intuitively true" turn out to have counter-examples. Some similar examples here and here. This is about independent r.v.s but also share some similarity to your conjecture.

Now of course, if you know additional info about the joint distribution, e.g. it is linear or jointly-Gaussian, then covariance can mean a lot. But for general r.v.s, not so much (in my experience).

The wikipedia article on correlation is highly recommended - it is filled with counter-intuitive examples, i.e. where correlation doesn't mean what you might think it means.