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:
-
What are the limits of integration here?
-
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.
Best Answer
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}