Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It is easy to produce a random variable with Dirichlet distribution using Gamma variables with the same scale parameter. If:

$ X_i \sim \text{Gamma}(\alpha_i, \beta) $


$ \left(\frac{X_1}{\sum_j X_j},\; \ldots\; , \frac{X_n}{\sum_j X_j}\right) \sim \text{Dirichlet}(\alpha_1,\;\ldots\;,\alpha_n) $

Problem What happens if the scale parameters are not equal?

$ X_i \sim \text{Gamma}(\alpha_i, \beta_i) $

Then what is the distribution this variable?

$ \left(\frac{X_1}{\sum_j X_j},\; \ldots\; , \frac{X_n}{\sum_j X_j}\right) \sim \; ? $

For me it would be sufficient to know the expected value of this distribution.
I need a approximate closed algebraic formula that can be evaluated very very quickly by a computer.
Let's say approximation with accurancy of 0.01 is sufficient.
You can assume that:

$ \alpha_i, \beta_i \in \mathbb{N} $

Note In short, the task is to find an approximation of this integral:

$ f(\vec{\alpha}, \vec{\beta}) = \int_{\mathbb{R}^n_+} \;\frac{x_1}{\sum_j x_j} \cdot \prod_j \frac{\beta_j^{\alpha_j}}{\Gamma(\alpha_j)} x_j^{\alpha_j - 1} e^{-\beta_j x_j} \;\; dx_1\ldots dx_n$

share|cite|improve this question
You can find a relatively simple expression for the pdf of $X_1+\cdots+X_n$ on p. 3 in – Shai Covo Feb 7 '11 at 21:44
Also, that pdf in the case of integer shape parameters is obtained in… – Shai Covo Feb 7 '11 at 22:15
It is well known that sum of Gammas is a Gamma distributed as well, but it doesn't hel here – Łukasz Lew Feb 8 '11 at 11:11
$X_1+\cdots+X_n$ is Gamma distributed if all the scale parameters are equal. – Shai Covo Feb 8 '11 at 19:35
Indeed. Anyway the distribution of sum doesn't help as it is correlated with $X_1$ – Łukasz Lew Feb 9 '11 at 2:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.