# Sum of Chi-squared Variables

What is the distribution of the variable $Y$ which is given by $$Y = \sum^N_{i=1} Z_i X_i$$ where :

$Z_1,\dotsc,Z_N$ are independent and identically distributed and the density function $f_Z(z)$ is known (a shifted chi-square distribution),

$X_i \sim\chi^2_L(\lambda)$, where $\lambda$ is the noncentrality paramter and $X_1,\dotsc,X_N$ are i.i.d. as well.

The question is:

(a) what is the distribution of $Y$ when $\lambda=0$.

(b) what is the distribution of $Y$ when $\lambda\ne 0$.

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You are asking "What is the distribution of the variable $Y$..." but then saying "..and the density function $f_Y(y)$... is known (a shifted chi-square distribution)" so you know the density function of $Y$ already. What is the question that you really meant to ask? –  Dilip Sarwate Mar 8 '12 at 15:57
Did you mean the $Z$'s have a shifted $\chi^2$ distribution? Also, in the case of a $\chi^2$ distribution (or any continuous distribution actually) the density function is not equivalent to ${\rm Prob}(Y = y)$. In fact, for a continuous random variable ${\rm Prob}(Y = y) = 0$ for all $y$. –  Macro Mar 8 '12 at 16:48
just fixed, the density $f(z)$ is known. –  Remy Mar 8 '12 at 16:54
Corsario, that is the correct answer (although it won't be pretty to implement). I suggest posting it as an answer. Note you've also assumed that each $X_i$ is independent of each each $Z_j$ for $i \neq j$ (for the convolution part), but I'd imagine that was intended as part of Remy's description. –  Macro Mar 8 '12 at 17:13
Perhaps Remy, but in this case the weights are random variables, so it's not that simple. I think the answer lies in Corsario's comment. –  Macro Mar 9 '12 at 0:02