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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) \ ? $

Example:

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.

Ethan Bolker
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Kusavil
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  • Should I maybe take product of characteristic functions and try to compute inverse Fourier transformat? – Kusavil Feb 16 '15 at 12:15
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    I don't see any profit in this perspective for $XY$. Especially because $Z_1$ and $Z_2$ are not independent. – drhab Feb 16 '15 at 12:38
  • 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? – Kusavil Feb 16 '15 at 13:26
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    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. – drhab Feb 16 '15 at 14:21

1 Answers1

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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

Sextus Empiricus
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