I'm trying to deduce which distribution my data follows and how to estimate the parameters. I have four random variables $X_i \sim N(\mu_i,\sigma_i^2)$ where the means and variances are all different. The covariance of each possible pair of variables is not zero so this makes them not independent. Now I want to figure out what is the distribution of the variable f with f being:

$$ f = \sum_{i=1}^4 (\frac{1}{4}-X_i)^2 $$

So does anyone know the kind of distribution it follows and it's parameters? I think it will be a Gamma distribution but I have no idea how to estimate the parameters based on the means and variances of the $X_i$ variables. Thanks in advance!

  • $\begingroup$ It's nasty--the distribution is the convolution of one non-central chi-squared distribution with the convolution of three (most likely) different gamma distributions. See stats.stackexchange.com/questions/72479 for the gammas to appreciate what goes on. You ought to consider procedures that don't require you to know the full distribution of $f.$ $\endgroup$ – whuber Mar 7 '20 at 13:50

This is essentially the same question as (sum of noncentral Chi-square random variables) when you change $X'_i \sim N(\frac{1}{4} - \mu_i, \sigma_i^2)$ and then $f=\sum_i {X'}_i^2$.

Compared to that question there is a complication of having the $X'_i$ as not independent of each other.

This makes it a generalised chi-squared distribution.

  • $\begingroup$ Almost, but not quite: the complication is that the variances of the $X_i^\prime$ likely differ, whereas the answers in your first link apply only to the case where the variances are the same. It's unclear what you mean by "generalized chi-squared distribution" because your link goes to a non-article. $\endgroup$ – whuber Mar 7 '20 at 14:09
  • $\begingroup$ You are right, a typo broke the link which is fixed now. The variances are different, but that is also addressed in part in the top answer of the first link. $\endgroup$ – James Fulton Mar 7 '20 at 14:14
  • $\begingroup$ Thank you--that's a good reference. (+1) $\endgroup$ – whuber Mar 7 '20 at 14:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.