# Marginal of Dirichlet distribution is Beta (integral)

Just for the sake of simplicity, take $K=3$ then a random vector $(X_1,X_2,X_3)$ has a Dirichlet distribution, i.e. $(X_1,X_2,X_3)\sim Dirichlet(\alpha_1,\alpha_2,\alpha_3)$ if the density takes the form: $$p(x_1, x_2) \propto x_1^{\alpha_1-1} x_2^{\alpha_2-1} (1-x_1-x_2)^{\alpha_3-1}.$$ The support is the $2$-dimensional simplex: $0<x_1,x_2,x_3<1$, $x_1+x_2+x_3=1$.

The dependence is just on two variables because of the restriction $x_1+x_2+x_3=1$. So one can ask about the distribution of $X_1$ and $X_2$.

My question is: How to prove (easily) that the marginals are Beta distributed. For instance, $X_1\sim Beta ( \tilde{\alpha}_1, \tilde{\alpha}_2)$ and what are the parameters?

It should be: $$p(x_1) = \int_0^1 p(x_1,x_2)dx_2 \propto x_1^{\alpha_1-1}\int_0^1 x_2^{\alpha_2-1} (1-x_1-x_2)^{\alpha_3-1} dx_2.$$

It is the latter integral I struggle with: I tried expanding the power using Newton's binomial but I get weird things. Does anyone see the mistake/way to proceed or have an easy proof? Thanks a lot!

Hint: Substitute $x_2=(1-x_1)u$ in the last integral.
You should obtain that $X_i \sim \operatorname{Beta}(α_i, \displaystyle\sum_{i\neq j=1}^{3}α_j)$.