# Generalized $\chi^2$ distribution?

Consider each $X_i \sim N(0,1)$. Then the random variable $Y=\sum_{i=1}^n X_i^2$ is a $\chi^2$ distribution with $n$ degree of freedom. Is there any probability distribution about a weighted sum of the square of standard normal random variables $Z=\sum_{i=1}^n w_i X_i^2$?

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It is quite easy to find the characteristic function of $Z$, since $$\mathbb{E} \left( \mathrm{e}^{i t X_i^2} \right) = \frac{1}{\sqrt{1-2 i t}}$$ Therefore $$\mathbb{E} \left( \mathrm{e}^{i t Z} \right) = \left( \prod_{i=1}^n 1-2 i t w_i \right)^{-1/2}$$ Inversion in closed form is possible only in special cases.
Thank you for the quick answer! It seems that the paper you mentioned is dealing with a general case where $X_i$'s are not necessarily independent. By the way, can you give me a reference on the inversion in closed form for special cases? – Endo Nov 23 '11 at 17:21
@user19736 When all $w_i$ are equal, it is a rescaled $\chi^2)n$. If $n$ is even, and $Z=\sum_{s=1}^{2m} w_s^2(X_{2m-1}^2+X_{2m}^2)$. Then $Z$ follows hypoexponential distribution, being the sum of exponentials. – Sasha Nov 23 '11 at 17:40
I don't fully understand your formulation of hypoexponential distribution. So, do we have only two random variables $X_{2m-1}$ and $X_{2m}$ both from $N(0,1)$? Then I'm not sure why you use the subscript $2m-1$ and $2m$ in $X$. – Endo Nov 23 '11 at 19:41
@user19736 I am sorry. I mistyped indexes.I intended $Z = \sum_{s=1}^m w_s( X_{2s-1}^2 + X_{2s}^2 )$. – Sasha Nov 23 '11 at 19:52
I still don't understand your formulation $Z=\sum_{s=1}^m w_s (X_{2s-1}^2 + X_{2s}^2)$. We can regard $Y_s = X_{2s-1}^2 + X_{2s}^2\sim Exp(1/2)$ since $X_{2s-1}^2 + X_{2s}^2$ has a $\chi^2$ distribution of 2 degree of freedom, which equals to an exponential distribution with $\lambda=1/2$. So your formulation can be written as $Z=\sum_{s=1}^m w_s Y_s$ where each $Y_s$ has the same parameter $\lambda=1/2$. Meanwhile, the hypoexpoential distribution is defined as $Z=\sum_{s=1}^m Y_s$ where each exponential random variable has its own param $\lambda_i$. Am I missing something in your argument? – Endo Nov 25 '11 at 3:27