# Show multivariate Beta integrates to 1

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 $x_1, x_2, ... x_k$?

• Your second an third integrals, at least, are incorrect. The Dirichlet distribution with $n$ parameters is a distribution over an $(n-1)$-dimensional space (a simplex, to be precise). So you want to express one variable as 1 less the sum of the others. Commented Aug 4, 2013 at 8:37
• math.stackexchange.com/questions/207073/… Commented Dec 30, 2019 at 12:46

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 $$p^{n-2}$$.