Given two continuous random variables $X$ and $Y$, suppose we know probability distributions $p_X, p_Y$ of them and $cov(X,Y)$. (Note we don't impose the independence between them.) Then can we calculate $p_{Z}(z)$ where $Z=XY$? If not, what more do we need? Can we calculate $p_Z$ if we have $P_{X,Y}(x,y)$ the joint distribution of $X$ and $Y$?

  • $\begingroup$ Are the random variables discrete or continuous ? $\endgroup$ – callculus Feb 27 '17 at 15:54
  • $\begingroup$ oh, they are continuous. $\endgroup$ – julypraise Feb 27 '17 at 15:57
  • 1
    $\begingroup$ Covariance: not enough. Joint distribution: suffices, the standard approach works (for example, using a change of variables to compute the joint distribution of (Z,Y), then computing the first marginal). $\endgroup$ – Did Feb 27 '17 at 16:02
  • $\begingroup$ @Did Thanks. Though, I'am quite new to the stuff and don't have a right reference. Where might I look for a explicit example? $\endgroup$ – julypraise Feb 27 '17 at 17:38
  • $\begingroup$ In your textbook, perhaps? $\endgroup$ – Did Feb 27 '17 at 17:44

You can calculate $E[XY]$ from just $\mbox{cov }(X,Y)$ and the individual $E[X]$ and $E[Y]$:

$$ E[XY] = \mbox{cov }(X,Y)+E[X]E[Y]$$

Since you can easily get $E[X]$ from $p_X$ (similarly for $Y$), the information you propose is enough to determine $E[XY]$.

But it is insufficient, in general, to determine $p_Z(z)$. Rather surprisingly, if you restrict the form of $p_{XY}(x,y)$ to a second degree expression on the unit square, and zero outside, then in fact the marginal distributions and the covariance together determine a unique joint probability function of that form. But if you relax that restriction, you can find cases that agree in marginal distributions and in covariances, but are not identical joint distributions.

  • $\begingroup$ Not so surprisingly: enumerate the free parameters and compare to the number of constraints. $\endgroup$ – Did Feb 27 '17 at 17:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.