If $X_i\sim\Gamma(\alpha_i,\beta_i)$ for $1\leq i\leq n$, let $Y = \sum_{i=1}^n c_iX_i$ where $c_i$ are positive real numbers. Assume all the parameters $\alpha_i$'s and $\beta_i$'s are all known, what is $Y$'s distribution ?
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See Theorem 1 given in Moschopoulos (1985) for the distribution of a sum of independent gamma variables. You can extend this result using the scaling property for linear combinations.
ars
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According to reference, the distribution function is: $g(y) = C\sum_{k=1}^\infty\frac{\delta_k y^{\rho+k-1}e^{-\frac{y}{\beta_1}}}{\Gamma(\rho+k)\beta_i^{\rho+k}} $ – Gong-Yi Liao Aug 23 '10 at 20:53
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where $C = \sum_{i=1}^n\left(\frac{\beta_1}{\beta_i}\right)^{\alpha_i}$ – Gong-Yi Liao Aug 23 '10 at 20:59
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And $\gamma_k = \sum_{i=1}^n\alpha_i\frac{(1 - \frac{\beta_1}{\beta_i})^k}{k}$ – Gong-Yi Liao Aug 23 '10 at 21:19
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with $\rho = \sum_{i=1}^n\alpha_i$ – Gong-Yi Liao Aug 23 '10 at 21:19
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And finally $\delta_k = \frac{1}{k+1}\sum_{i=1}^{k+1}i\gamma_i\delta_{k+1-i}; \delta_0=1, k=0,1,2,3,\ldots$ – Gong-Yi Liao Aug 23 '10 at 21:19
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1@Gong-Yi: nice, thanks for updating with the answer! – ars Sep 14 '10 at 01:32
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Actually, for N=2, this result can be simplified, where the summation after C is finite. – May 20 '11 at 08:17
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@Tina Welcome to the site; but please don't use answers for comments. Please read the FAQ. – May 20 '11 at 09:30
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@Gong-YiLiao I think the formula for $\delta_k$ is actually for $\delta_{k+1}$, according to Eq 2.8 in the paper. – bcf Jun 30 '17 at 18:27