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I know that if $X_{1},X_{2},...X_{n}$ are independent $\mathrm{Gamma}(\alpha_{i},\theta)$ - distributed variables (notice they all have the same scale parameter $\theta$) and

$Y_{i}=\frac{X_{i}}{\sum_{j=1}^{n}X_{j}}$

then :

$Y=(Y_{1},Y_{2},...Y_{n})\;$~$\;\mathrm{Dirichlet}(\alpha_1,\alpha_2,...,\alpha_n)$

I'm interested in what happens if $X_{1},X_{2},...X_{n}$ are allowed to have different scale parameters. That is:

$X_i$~$\mathrm{Gamma}(\alpha_i,\theta_i)$

The problem is: find a closed-form solution for the expectation of $Y_i\ \; \forall i$.

Great if you can also tell me how this distribution is called and provide a reference (Textbook, paper).

Bounty's on. Thank you!

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    $\begingroup$ Related: stats.stackexchange.com/questions/6963/… $\endgroup$ Aug 3, 2021 at 21:43
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    $\begingroup$ There probably is no closed form for the mean and no common name for the distribution. $\endgroup$
    – Matt F.
    Aug 4, 2021 at 12:54
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    $\begingroup$ There actually is a common name: the scaled Dirichlet. See researchgate.net/publication/… (which addresses your question about the lack of closed form solution for the expectations) and Chapter 10, "Notes on the Scaled Dirichlet Distribution" in Compositional Data Analysis: Theory and Applications, ed. Vera Pawlowsky-Glahn and Antonella Buccianti. $\endgroup$ Jul 19, 2022 at 17:06
  • $\begingroup$ There are at least 3 questions about this type of distribution. It would be great to have a tag to identify them easily: scaled-dirichlet-distribution. Could somebody with the requisite points kindly create such a tag? TIA. $\endgroup$ Jul 20, 2022 at 18:19

1 Answer 1

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Fair warning- we will not get far. This is not even close to a full answer, but I think it is a bit more than a comment.

As mentioned by Matt F. in the comments, there is highly unlikely to be a closed-form solution. To see why, consider that the denominator's density itself "barely" has a closed-form, with a density given by the infinite series set out in equation 2.9 in Moschopoulos (1985) https://doi.org/10.1007/BF02481056. Specifically, the sum of n gamma distributions is given by: $$ \begin{aligned} g(y)=C\sum_{k=0}^\infty\frac{\delta_k y^{\rho+k-1}e^{-y/{\beta_1}}}{\Gamma(\rho+k)\beta_1^{\rho+k}}\\ \end{aligned} $$

Here, $\{\delta_k\}$ is an infinite series of constants, $\rho$ and $C$ are also constants, and $\beta_1$ is the smallest value of $\theta_i$

Because the support is positive, you can follow the procedure here to get the density of the reciprocal distribution of the denominator https://en.wikipedia.org/wiki/Inverse_distribution. But the dependence between the numerator and the denominator (and it's reciprocal) seems to preclude a "nice" answer, even with the moments.

So, obviously getting the moments numerically is not so bad. But a closed-form seems unlikely given the above.

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  • $\begingroup$ Thanks Matterhorn. What if we asssume large n? Then the numerator and the numerator should become approximately independent. Do we have an approximate closed-form expression for the mean, at least in the limit based on this? Best answer nonetheless, so you get the bounty :) $\endgroup$
    – bbecon
    Aug 8, 2021 at 11:36
  • $\begingroup$ Thanks! I thought maybe at first, but I'm no longer so sure. Let $Z=\sum X_i$. Since both the variance and expectation of $1/Z$ are declining with $n$, it's not at all clear to me that there is a nice limit. If I were going to try to show this, I might try to prove that for large $n$, $cov[X_i,1/Z]\to 0$ much faster than $E[X_i]E[1/Z]\to 0$, since then then $E[X_i/Z] \approx E[X_i]E[1/Z]$. But it is not at all clear to me that this is true. $\endgroup$
    – Matterhorn
    Aug 8, 2021 at 20:06

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