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?
Best Answer
As Glen_b noted in the comments, if the variances are all the same you end up with a scaled noncentral chi-squared.
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 $A = I$.
There has been some work on computing things with this distribution:
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:
Bausch (2013) gives a more computationally efficient algorithm for the linear combination of central chi-squareds; his work might be extensible to noncentral chi-squareds, and you might find some interesting pointers in the related work section.