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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!

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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