If we have i.i.d. random variables$ \quad X_1,\dots , X_n, \ \text{where} \ X_k \sim \mathcal{N} (\mu_k,\sigma_k^2),$ $\quad$ then $$ Y =\sum_{k=1}^n a_k X_n \sim \mathcal{N} (\sum_{k=1}^n a_k \mu_k,\sum_{k=1}^n a_k^2\sigma_k^2). $$

But what we can do with similar (or at least simplified) linear combination of $ \ (X_k)_{k=1}^n \ , \ X_k \sim \chi^2(k) \ ? $


Let $X,Y \sim \mathcal{N}(0,1)$ be independent random variables. What is the distribution of $ Z = XY $ ?

$$ XY = \frac{1}{2}(X^2 + 2XY + Y^2) - \frac{1}{2}(X^2 + Y^2) =\left ( \frac{X+Y}{\sqrt{2}} \right ) ^2 - \frac{1}{2}(X^2 + Y^2) $$

Of course we have that $$ \frac{X+Y}{\sqrt{2}} = \frac{\frac{X+Y}{2} - 0}{1} \sqrt{2} \sim \mathcal{N}(0,1) $$

Denoting $ \ Z_1 = \left ( \frac{X+Y}{\sqrt{2}} \right )^2 \sim \chi^2(1) , \ $ $ \ Z_2 = X^2 + Y^2 \sim \chi^2(2) , \ $ we have

$$ XY = Z_1 + \frac{1}{2} Z_2 = \sum_{k=1}^2 \frac{Z_k}{k}, $$ where $ \ Z_k \sim \chi^2(k) .$

What more we can do with this kind of approach? I don't have any ideas, nor I can find anything useful at this matter.

  • $\begingroup$ Should I maybe take product of characteristic functions and try to compute inverse Fourier transformat? $\endgroup$ – Kusavil Feb 16 '15 at 12:15
  • 1
    $\begingroup$ I don't see any profit in this perspective for $XY$. Especially because $Z_1$ and $Z_2$ are not independent. $\endgroup$ – drhab Feb 16 '15 at 12:38
  • $\begingroup$ Right! Thank you, so it's a wrong example. But have you by any chance heard if there is any nice formula for $\chi^2$ distributed variables (let's say independent ones) with different degrees of freedom? Or at least for variables with the same degree of freedom? $\endgroup$ – Kusavil Feb 16 '15 at 13:26
  • 1
    $\begingroup$ If $\chi_k^2$ and $\chi_m^2$ are independent then $\chi_k^2+\chi_m^2$ and $\chi_{k+m}^2$ have the same distribution. $\endgroup$ – drhab Feb 16 '15 at 14:21

Your $Z_1$ and $Z_2$ are correlated so the sum is not like a sum of independent chi-square distributed variables.

If you use instead $Z_1 = X-Y$ and $Z_2 = X+Y$ which are two independent variables distributed as $N(0,2)$, then $$XY = \frac{1}{4}Z_1^2- \frac{1}{4} Z_2^2$$ thus the product XY it is distributed as the difference of two chi-square distributed variables.

This has a variance-gamma distribution as explained in https://math.stackexchange.com/a/85525/466748


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.