1
$\begingroup$

The Dirichlet distribution is defined as: $$ p(\vec{\mu}_M|\vec{\alpha}_M) = c(\vec{\alpha}_M) \Pi_{k=1}^M\mu_k^{\alpha_k-1} $$ where $\vec{\mu}_M, \vec{\alpha}_M$ is a vector of length $M$ and $\sum_{k=1}^M\mu_k=1$.

I want to show that: $$ c(\vec{\alpha}_M)=\frac{\Gamma(\alpha_1+\alpha_2+\cdots+\alpha_M)}{\Gamma(\alpha_1)\Gamma(\alpha_2)\cdots\Gamma(\alpha_M)} $$

Prove by induction: when $M=2$, the distribution is the same as Beta distribution, the relation holds.

Suppose it holds $M=N-1$, when $M=N$:

\begin{align} 1=&\int p(\vec{\mu}_N|\vec{\alpha}_N) \\ =& \vec{c}(\vec{\alpha}_N) \int \mathrm{d}\vec{\mu}_N \Pi_{k=1}^N\mu_k^{\alpha_k-1} \\ =& \vec{c}(\vec{\alpha}_N) \int_{0}^{1} \mathrm{d}\mu_N \mu_N^{\alpha_N-1}\int \mathrm{d}\vec{\mu}_{N-1}\Pi_{k=1}^{N-1}\mu_k^{\alpha_k-1} \end{align}

For $\sum_{k=1}^{N-1}\mu_k = 1-\mu_N$, if we change the variable $u_k=\mu_k/(1-\mu_N)$, then $\sum_{k=1}^{N-1}u_k = 1$.

Consider: \begin{align} &\int \mathrm{d}\vec{\mu}_{N-1}\Pi_{k=1}^{N-1}\mu_k^{\alpha_k-1} \\ =&(1-\mu_N)^{\sum_{k=1}^{k=N-1}\alpha_k} \int \mathrm{d}\vec{u}_{N-1}\Pi_{k=1}^{N-1}u_k^{\alpha_k-1} \\ =&(1-\mu_N)^{\sum_{k=1}^{k=N-1}\alpha_k} \frac{1}{c(\vec{\alpha}_{N-1})} \end{align} where I have used the assumption holds when $M=N-1$.

Therefore: \begin{align} 1=\frac{c(\vec{\alpha}_{N})}{c(\vec{\alpha}_{N-1})} \int_{0}^{1} \mathrm{d}\mu_N \mu_N^{\alpha_N-1} (1-\mu_N)^{\sum_{k=1}^{k=N-1}\alpha_k} \end{align}

It seems that I am very close to the desired result but missed a factor of $1/(1-\mu_N)$ in the final integrand, how to fix this?

$\endgroup$
2

1 Answer 1

0
$\begingroup$

So I guess I figured it out. In the three lines you have written after "Consider" you are actuall integrating over the following set:

$$S_{N-1}:=\left\{ \vec{\mu}_{N-1} \in \mathbb R^{N-1} \big\vert \mu_i > 0 \ \mathrm{for} \ i=1,...,N-1\, , \ \sum_{i=1}^{N-1} \mu_i = 1-\mu_{N-1} \right\} $$

That means you can either use the transformation theorem to also properly transform this (meaning the boundaries would look somewhat different) or insert that information to the integral via a $\delta$-function (which I would do because I was raised to be a physicist, sorry guys.). So the integral should acutally look like:

$$I:=\int \mathrm{d}\vec{\mu}_{N-1} \, \delta\left( \sum_{i=1}^{N-1} \mu_i - (1 - \mu_{N-1}) \right) \prod_{i=1}^{N-1}\mu_i^{\alpha_i - 1} \, .$$

Now using $\delta(\alpha x) = \delta(x)/\alpha$ (in an integral over $x$, see https://en.wikipedia.org/wiki/Dirac_delta_function under "Properties") and the rescaling you proposed $u_k = \mu_k / (1-\mu_N)$ you eventually get

$$I = \int \mathrm{d}\vec{\mu}_{N-1} \, \delta\left((1 - \mu_{N-1}) \left( \sum_{i=1}^{N-1} \frac{\mu_i}{1 - \mu_{N-1}} - 1 \right) \right) \prod_{i=1}^{N-1}\mu_i^{\alpha_i - 1} \\ = \int \mathrm{d}\vec{u}_{N-1} \, \delta\left( \sum_{i=1}^{N-1} u_i - 1 \right) \frac{1}{1 - \mu_{N-1}} (1-\mu_{N-1})^{\sum_{i=1}^{N-1}\alpha_i-(N-1)+(N-1)} \prod_{i=1}^{N-1} u_i^{\alpha_i - 1}\\ = (1-\mu_{N-1})^{\sum_{i=1}^{N-1}\alpha_i - 1} \int_{S_{N-1}} \mathrm d\vec{u}_{N-1} \prod_{i=1}^{N-1} u_i^{\alpha_i - 1} $$

The rest should be rewriting it to the gamma functions. Maybe this still bothers you or anybody else here. I know I joined the party a little late.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .