[Math] Show multivariate Beta integrates to 1

integrationprobability distributions

I am trying to either find or construct a concrete example of a multivariate Beta distribution (Dirichlet) that integrates to $1$.

From the definition of the Beta distribution, we have
$$
\int \frac{\Gamma(\alpha_1 + \alpha_2)}{\Gamma(\alpha_1)\Gamma(\alpha_2)}x^{\alpha_1-1}(1-x)^{\alpha_2-1} \rm \, dx = 1
$$

The Dirichlet is multivariate generalization of the Beta, so it seems to me that the same can be expressed alternatively as follows (where $\sum_{i=1}^k x_i=1$), or am I mistaken? If so, what are the correct integration bounds?
$$
\iint \frac{\Gamma(\alpha_1 + \alpha_2)}{\Gamma(\alpha_1)\Gamma(\alpha_2)}x_1^{\alpha_1-1}x_2^{\alpha_2-1} \rm \, dx_2 \, \rm dx_1 = 1?
$$

Now suppose an example with $k = 3; \mathbf{x} = (x_1, x_2, x_3)$.

$$
\iiint \frac{\Gamma(\alpha_1 + \alpha_2 + \alpha_3)}{\Gamma(\alpha_1)\Gamma(\alpha_2)\Gamma(\alpha_3)}x_1^{\alpha_1-1}x_2^{\alpha_2-1}x_3^{\alpha_3-1} \rm \, dx_3 \rm \, dx_2 \, \rm dx_1
$$

How would you go about evaluating this integral? For example, as follows?

$$
=\frac{\Gamma(\alpha_1 + \alpha_2 + \alpha_3)}{\Gamma(\alpha_1)\Gamma(\alpha_2)\Gamma(\alpha_3)} \iint x_1^{\alpha_1-1}x_2^{\alpha_2-1} \frac{1^{\alpha_3}}{\alpha_3} \rm \, \, dx_2 \, \rm dx_1
$$
$$
=\frac{\Gamma(\alpha_1 + \alpha_2 + \alpha_3)}{\Gamma(\alpha_1)\Gamma(\alpha_2)\Gamma(\alpha_3)} \int_{0}^{1} x_1^{\alpha_1-1} \frac{1}{\alpha_2} \frac{1}{\alpha_3} \rm \, \rm dx_1
$$

Finally, is there a less tedious way to do so, besides evaluating the iterated integral over the $k$ variables $x_1, x_2, … x_k$?

Best Answer

You can follow the proof in https://en.wikipedia.org/wiki/Beta_function#Relationship_between_gamma_function_and_beta_function. Here is an example to demonstrate $\Gamma(x) \Gamma(y) \Gamma(z) = \Gamma(x+y+z) B(x,y,z)$:

\begin{align} \Gamma(x) \Gamma(y) \Gamma(z) = & \int_{u=0}^{\infty}\int_{v=0}^{\infty}\int_{w=0}^{\infty} e^{-(u+v+w)} u^{x-1} v^{y-1} w^{z-1} dudvdw \\\\ = & \int_{p=0}^{\infty} \iint_{q+r\leq 1} e^{-p} p^{x+y+z-3} q^{x-1} r^{y-1} (1-q-r)^{z-1} |J(p,q,r)|dpdqdr \\\\ \end{align} In the transformation, we let $u=pq, v=pr, w=p(1-q-r)$ and the Jacobian determinant evaluate to $p^2$. It is easy to simplfy the result and get $\Gamma(x+y+z) B(x,y,z)$ from the definition for multivariate Beta function.

For $n$ variables, the key step is to compute the Jacobian determinant, which equals $p^{n-2}$.

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