4
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

Hint: Substitute $x_2=(1-x_1)u$ in the last integral.


You should obtain that $X_i \sim \operatorname{Beta}(α_i, \displaystyle\sum_{i\neq j=1}^{3}α_j)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.