# sum of noncentral Chi-square random variables

I need to find the distribution of the random variable $$Y=\sum_{i=1}^{n}(X_i)^2$$ where $X_i\sim{\cal{N}}(\mu_i,\sigma^2_i)$ and all $X_i$s are independent. I know that it is possible to first find the product of all moment generating functions for $X_i$s, and then transform back to obtain $Y$'s distribution. However, I wonder whether there is a general form for $Y$ like the Gaussian case: we know the sum of independent Gaussian is still a Gaussian, and thus we only need to know the summed mean and summed variance.

How about all $\sigma^2_i=\sigma^2$? Will this condition make a general solution?

• Looking at the first paragraph under here, clearly the final condition yields a scaled noncentral chi-square (divide through by $\sigma^2$ (the scale factor you take out the front) and make $\sigma_i=1$ in $\sum_{i=1}^k (X_i/\sigma_i)^2$). The more general form you started with looks like a linear combination or scaled-weighted-average, with coefficients $\sigma^2_i$ rather than a plain sum of scaled squares ... and I believe that won't generally have the required distribution. Aug 16 '13 at 3:22
• Depending on what you need it for, in specific cases you may be able to do numerical convolution, or simulation. Aug 16 '13 at 22:25
• This is generalized by the 'weighted sum of log chi-squares to power' distribution. My R package sadists provides approximate 'dpqr' functions for $Y$; c.f. github.com/shabbychef/sadists Mar 21 '15 at 4:37
• @shabbychef thank you for your work!
– runr
Oct 19 '20 at 3:33

If not, there is a concept of a generalized chi-squared distribution, i.e. $x^T A x$ for $x \sim N(\mu, \Sigma)$ and $A$ fixed. In this case, you have the special case of diagonal $\Sigma$ ($\Sigma_{ii} = \sigma_i^2$), and $A = I$.
You can also write it as a linear combination of independent noncentral chi-squared variables $Y = \sum_{i=1}^n \sigma_i^2 \left( \frac{X_i^2}{\sigma_i^2} \right)$, in which case: