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This question is in a way duplicating this question, but I'm not happy with the final answer and feel like I need a deeper dive.

Assume that $X_j \sim \Gamma(\alpha_j,\beta_j)$, where $\alpha, \beta$ denotes shape and rate.

I'm interested in deriving the form of $GG(\gamma, \alpha, \beta)$ of the following sum, where $\gamma$ is the form/power parameter:

$$S = \sum_{j = 1}^{d} X_j^2$$

From the previous answer I understand, that $Y_j=X_j^2$ has a Generalized Gamma distribution with $\gamma = 1/2$. It follows, that $\sum_{j=1}^{d} Y_j^{1/2}$ has gamma distribution, therefore $S$ is also Generalized Gamma variable.

However, how would the final parameters of the distribution look like?

Assuming $\beta_j = \beta, \forall j$, it is known that $Z = \sum_{j=1}^{d} Y_j^{1/2} \sim \Gamma(\sum \alpha_j, \beta)$. Knowing $Z$, how should we proceed in deriving the form of $S$? The previous answer claims that it is straightforward from there as if it's $GG(\gamma, \sum \alpha_j, \beta)$, which doesn't feel true since $S^{1/2} \neq Z$.

runr
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2 Answers2

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Knowing $Z$, how should we proceed in deriving the form of $S$? The previous answer claims that it is straightforward from there as if it's $GG(\gamma, \sum \alpha_j, \beta)$, which doesn't feel true since $S^{1/2} \neq Z$.

You can not apply the result from the other question.

The situation in the question that you reference is a lot different from your case

  1. In the other question, the sum is of gamma distributed variables and not of generalised gamma distributed variables. $$z^d = \sum_i x_i^d = \underbrace{ \sum_i y_i}_{\substack{\text{sum of}\\\text{gamma distributed}\\\text{ variables}}}$$ If $X$ is generalized gamma distributed with parameter $d$ as power, then $Y=X_i^d$ is gamma distributed.

  2. In the other question, the sum is of independent and identical distributed variables.

    The parameters of $X$ and the associated $Y$ are the same. For independent and identical distributed gamma distributions you get that their sum is again a gamma distributed variable. So in that other question $z^d$ is gamma distributed and that makes $z$ generalized gamma distributed.

In your case

  • You seem to describe a case with non-identical distributed gamma distributions (or at least you define parameters with subscripts $\alpha_i$ and $\beta_i$ which suggests that the distributions can be different).

    In this case you get that the sum follows a distribution that can be described as a mixture of exponential distributions. (see Generic sum of Gamma random variables)

  • You describe a sum of generalised gamma distributions.

    If $X$ is gamma distributed, then $X_j^2$ is generalised gamma distributed. This is an opposite situation from the referenced question.

    You have a sum of powers, but note that the power operation is both a transformation that can transform a generalized gamma distributions to a gamma distributed variable as the other way around and transform a gamma distributed variable to a generalized gamma distributed variable.

    If $X_j$ is gamma distributed then $X^{1/p}$ is generalized gamma distributed with parameter $\gamma = p$. If $X$ is generalized gamma distributed with with parameter $\gamma = p$. then $X^{p}$ is gamma distributed.

    So your analogy with the other question does not work

    It follows, that $\sum_{j=1}^{d} Y_j^{1/2}$ has gamma distribution, therefore $S$ is also Generalized Gamma variable.

    You indeed have that $\sum_{j=1}^{d} Y_j^{1/2}$ has a gamma distribution (but as the previous point notes only for iid variables), but your case is the sum $\sum_{j=1}^{d} Y_j$

  • The above answers the question about the connection with the other question. But it is a bit of an XY-problem and the underlying problem is not solved. To get that underlying question solved it might help to make a question that describes the problem more precisely. Depending on the situation one can solve it differently, for example one might use approximations. – Sextus Empiricus Jun 09 '23 at 08:11
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For the sum of independent Gamma random variables with different shape and rate/scale parameters, the answer has been given in

Moschopoulos, P. G. (1985). The distribution of the sum of independent gamma random variables. Annals of the Institute of Statistical Mathematics, 37(3), 541-544.

The density can be expressed as an infinite sum involving some complicated expressions of the underlying parameters.

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Perhaps this can be a base to work the sum of squared variables.

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    It's hard to see how this result (which is demonstrated at https://stats.stackexchange.com/questions/72479/generic-sum-of-gamma-random-variables/72486#72486 here on CV) can be applied to the sum of squared Gamma variables. Do you have a suggestion about how to proceed? – whuber Nov 26 '20 at 16:51
  • @whuber I leave that to the OP who, if interested enough they can lookup the reference, see there that the expression comes from working with the MGF of the sum, and likewise explore whether the same technique can be applied to the MGF of the sum of squares. – Alecos Papadopoulos Nov 26 '20 at 17:02