Conditional covariance.

Question

Suppose we have two random variables, $X$ and $Y$, defined over nonnegative reals. Obviously, the following formula holds:

$$\mathrm{Cov}(X,Y)=\mathbb{E}\left[XY\right]-\mathbb{E}\left[X\right]\mathbb{E}\left[Y\right].$$

However intuitional it may be, I am not so sure whether another formula that I stumbled upon holds:

$$\mathrm{Cov}(X,Y|X>0\,\wedge\,Y>0)=\mathbb{E}\left[XY|X>0\,\wedge\,Y>0\right]-\mathbb{E}\left[X|X>0\,\wedge\,Y>0\right]\cdot\mathbb{E}\left[Y|X>0\,\wedge\,Y>0\right].$$

I am not looking for a proof, but I would be thankful for a short note why or why not the above formula holds. Thank you in advance.

No, essentially I am asking if an analogous formula for covariance holds if we put conditions on it. — Johnny Westerling, Dec 20 '13 at 05:04
Are you asking if the well-known formula, which you wrote at the top, holds under the condition that $X\gt 0, Y\gt 0$? — mathlove, Dec 20 '13 at 05:17

iballa · Answer 1 · 2013-12-20T19:53:40.553

6

Assuming we define $\mathrm{Cov}[X,Y | A] = E[(X - E[X|A])(Y - E[Y|A])|A]$, we will have (for any nonzero event $A$ including "$X > 0 \wedge Y > 0$")

$$ \mathrm{Cov}[X,Y | A] = E[XY|A] - E[X|A] E[Y|A] - E[X|A] E[Y | A] + E[X|A]E[Y|A]$$ $$ = E[XY|A] - E[X|A]E[Y|A]$$

by linearity of conditional expectation. So your formula holds!

edited Dec 20 '13 at 19:53

answered Dec 20 '13 at 05:20

iballa

603

1

sign typo at the last? – mathlove Dec 20 '13 at 05:26

score 1 · Accepted Answer · answered Dec 20 '13 at 05:31

1

The formula which you wrote at the top holds for any two random variable $X,Y$. This means your formula holds under the condition of $X\gt0, Y\gt0$.

answered Dec 20 '13 at 05:31

mathlove

139,939

Thanks, I guess I should be getting more sleep... – Johnny Westerling Dec 20 '13 at 06:17

Conditional covariance.

2 Answers2