Friday, July 18, 2008

Formula for Var(XY)

The variance of two random variables, X+Y is in the beginning of every statistics book. The distribution of X/Y is a standard Cauchy variable. So when I tried to find the variance of X*Y, I figured no problem. But it is actually very difficult to find on the web, and tedious to derive. As a public service, here is the result when X and Y are both normally distributed:

Let V(x) and V(y) be the variance of X and Y respectively
Let C(x,y) be the covariance of X and Y

then the variance of the product XY, is


note: see derivation from anonymous commenter (I can't get math in html for my blog, so God bless him)


  1. Anonymous3:43 PM

    The result does not depend on X&Y being normally distd. You just need finite mean and variance.

  2. Well, there's a third moment, E(delta(x)^2*delta(y)), that's zero for normal, but not for all other distributions.

  3. Anonymous7:33 AM

    Is this calculation =

    E([{X-E(X)}{Y-E(Y)}]**2) ?

  4. Anonymous7:34 AM

    What's this info useful for?

  5. Why, to make gobs of money via the idea that XXXXXCENSOREDBYSPONSOR XXXXXXXXX so if you know V(xy) you can maximize the Sharpe ratio

  6. Anonymous11:53 PM

    Another cool result is that if X and Y are independant unit normal distributions, and Z=XY, then the probability density of Z is (1/Pi)K_0(|z|), where K_0 is a modified Bessel function of the second kind.

  7. Thank you!!!!

    I've been looking everywhere for this!

    I couldn't find it anywhere in any of my texts or on the internet.

  8. Hi,
    I found this article very intersting, but, by reading the proof i had a problem with the Cov(X^2,Y^2)
    Would you mind explaining me how did you get the result? (i understood that there is a delta method somewhere, but i've never heared of this method before)

    Thx in advance