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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?

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