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$

  • 1
    $\begingroup$ @Łukasz Can you say anything more about the parameters $n$, $\alpha_i$, and $\beta_i$? It's possible to obtain exact expressions for $\sum_j{X_j}$ and thereby approximate the expectations of the ratios, but for certain combinations of the parameters one could exploit Normal or saddlepoint approximations with less work. I don't think there will be a universal approximation method, which is why additional restrictions would be welcome. $\endgroup$
    – whuber
    Commented Feb 8, 2011 at 6:17
  • $\begingroup$ $X_1$ and $\sum_j X_j$ are correlated so we have to approximate the integral itself. $\alpha_i$ is often a small number like 1 or 2 and sometimes as big as 10000. Similarly wih $\beta_i$ but it is usually 10 times bigger than $\alpha_i$. $\endgroup$ Commented Feb 9, 2011 at 13:50
  • $\begingroup$ The problem is with small $\alpha_i$. If all $\alpha_i$ are big then the good approxmiation of the whole integral is: $\frac{\alpha_1/\beta_1}{\sum_j \alpha_j/\beta_j}$ $\endgroup$ Commented Feb 9, 2011 at 13:51
  • $\begingroup$ @Łukasz If you need to evaluate the expression of the expectation, why do you require an algebraic formula? I am thinking in applying some numerical trick to get the expectation but I need some feedback :) $\endgroup$
    – deps_stats
    Commented Feb 10, 2011 at 16:58
  • $\begingroup$ I need to evaluate it many times in my program. It has to be very fast, i.e. no loops and preferably not too many divisions. $\endgroup$ Commented Feb 10, 2011 at 21:23

1 Answer 1


Just an initial remark, if you want computational speed you usually have to sacrifice accuracy. "More accuracy" = "More time" in general. Anyways here is a second order approximation, should improve on the "crude" approx you suggested in your comment above:

$$E\Bigg(\frac{X_{j}}{\sum_{i}X_{i}}\Bigg)\approx \frac{E[X_{j}]}{E[\sum_{i}X_{i}]} -\frac{cov[\sum_{i}X_{i},X_{j}]}{E[\sum_{i}X_{i}]^2} +\frac{E[X_{j}]}{E[\sum_{i}X_{i}]^3} Var[\sum_{i}X_{i}] $$ $$= \frac{\alpha_{j}}{\sum_{i} \frac{\beta_{j}}{\beta_{i}}\alpha_{i}}\times\Bigg[1 - \frac{1}{\Bigg(\sum_{i} \frac{\beta_{j}}{\beta_{i}}\alpha_{i}\Bigg)} + \frac{1}{\Bigg(\sum_{i} \frac{\alpha_{i}}{\beta_{i}}\Bigg)^2}\Bigg(\sum_{i} \frac{\alpha_{i}}{\beta_{i}^2}\Bigg)\Bigg] $$

EDIT An explanation for the above expansion was requested. The short answer is wikipedia. The long answer is given below.

write $f(x,y)=\frac{x}{y}$. Now we need all the "second order" derivatives of $f$. The first order derivatives will "cancel" because they will all involve multiples $X-E(X)$ and $Y-E(Y)$ which are both zero when taking expectations.

$$\frac{\partial^2 f}{\partial x^2}=0$$ $$\frac{\partial^2 f}{\partial x \partial y}=-\frac{1}{y^2}$$ $$\frac{\partial^2 f}{\partial y^2}=2\frac{x}{y^3}$$

And so the taylor series up to second order is given by:

$$\frac{x}{y} \approx \frac{\mu_x}{\mu_y}+\frac{1}{2}\Bigg(-\frac{1}{\mu_y^2}2(x-\mu_x)(y-\mu_y) + 2\frac{\mu_x}{\mu_y^3}(y-\mu_y)^2 \Bigg)$$

Taking expectations yields:

$$E\Big[\frac{x}{y}\Big] \approx \frac{\mu_x}{\mu_y}-\frac{1}{\mu_y^2}E\Big[(x-\mu_x)(y-\mu_y)\Big] + \frac{\mu_x}{\mu_y^3}E\Big[(y-\mu_y)^2\Big]$$

Which is the answer I gave. (although I initially forgot the minus sign in the second term)

  • $\begingroup$ This looks like exactly what I need. Can you explain how you got this expansion? I tried in a lot of ways and was unable to do that ... $\endgroup$ Commented Feb 26, 2011 at 10:16

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