Let's say that $(X_1, \dots, X_4) \sim Dirichlet(\alpha_1, \dots, \alpha_4)$ with $\sum_{i=1}^4 X_i=1$. I want to find the distribution of $(X_1, X_2)$.

I know the marginal distributions of the $X_i$ (answered in other questions on this site) but it doesn't seem that this has been answered.

Denote the full joint pdf by $f_0$ so that $f_0(x_1, x_2, x_3, x_4) = \frac{1}{B(\vec \alpha)} \prod_{i=1}^4 x_i^{\alpha_i-1}$. I'll denote the joint pdf of $(X_1, X_2)$ by $g$.

First of all, since $X_4 = 1 - X_1 - X_2 - X_3$ it seems that I should only consider $X_1$, $X_2$, and $X_3$ and therefore I'll only need to do a single integration.

This means that I'll be working with $$ f(x_1, x_2, x_3) := \frac{x_1^{\alpha_1 - 1} x_2^{\alpha_2 - 1} x_3^{\alpha_3 - 1}(1-x_1-x_2-x_3)^{\alpha_4-1}}{B(\vec \alpha)}. $$

From this it follows that $$ g(x_1, x_2) = \int_{x_3} f(x_1, x_2, x_3) dx_3 $$

$$ = \frac{x_1^{\alpha_1-1} x_2^{\alpha_2-1}}{B(\vec \alpha)} \int_{x_3} x_3^{\alpha_3-1} (1-x_1-x_2-x_3)^{\alpha_4-1} dx_3. $$

My questions:

  1. What are the limits of integration here?

  2. How do I actually do this integral? I assume that I need to shoehorn it into a beta integral but I don't see how.

For the limits of integration, certainly $x_3 \geq 0$, but I don't know what the upper bound is. Would it just be $0 \leq x_3 \leq 1 - x_1 - x_2$?

As for the actual integral, if I'm correct about $0 \leq x_3 \leq 1 - x_1 - x_2$ then I just need to be able to do $$ I = \int_{0}^{1-k} t^{\alpha-1}(1-k-t)^{\beta-1} dt $$ where $k = x_1 + x_2$, $\alpha = \alpha_3$, and $\beta = \alpha_4$. This looks really close to an incomplete beta but not quite.

  • $\begingroup$ If I am not mistaking there is a factor $2$ missing in the integral. You lift $(x_1, x_2, x_3)$ to $\mathbb R^4$ with a function $u(x_1,x_2,x_3) = (x_1,x_2,x_3,1-x_1-x_2-x_3)$. So when you integrate the Dirichlet distribution on $\mathbb R^3$, which maps to the 3 simplex a 3 dimensional manifold embedded into $\mathbb R^4$, you have to account for the distortion done by $u$. This is a constant factor $\sqrt{Du^TDu}=2$ in the integral over $f(u(x_1,x_2,x_3))$ where $Du$ is the Jacobian matrix of $u$. Please do not mind if I am wrong, but I intend to do something very similar and the answer prov $\endgroup$
    – katosh
    Sep 5, 2018 at 16:29

1 Answer 1


By manipulating the integral you can obtain the beta function. For ease of notation, let $1-k = x$, then $I$ can be written as \begin{align} I &= \int_{0}^{x} t^{\alpha-1}(x-t)^{\beta-1}dt \\ &= x^{\beta-1} \int_{0}^{x} t^{\alpha-1} \left(1-\frac{t}{x}\right)^{\beta-1}dt. \end{align} Let $t/x = u$ then $dt = x du$. If $t=x$ then $u=1$, hence the above integral reduces to $$\int_{0}^{1} x^{\alpha-1} u^{\alpha-1} (1-u)^{\beta-1} x du.$$ Therefore, \begin{align} I &= x^{\alpha+\beta-1} \int_{0}^{1} u^{\alpha-1}(1-u)^{\beta-1}du \\ &= x^{\alpha+\beta-1} \beta(\alpha,\beta). \end{align}

  • $\begingroup$ Thanks a lot! Regarding my question (1), did I get the limits of integration correct? $\endgroup$
    – alfalfa
    Apr 6, 2016 at 22:00
  • $\begingroup$ Regarding the definition of the Dirichlet distribution, I think the bounds for $x_3$ are indeed correct. $\endgroup$
    – Cavents
    Apr 6, 2016 at 22:06

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