# Dirichlet sample by normalising Gamma RVs

I know that if you sample $K$ random variables $(X_1, X_2, \dots, X_K)$ from Gamma distributions using shape parameters $(\alpha_1, \alpha_2, \dots \alpha_K)$ and a scale parameter $\theta = 1$ such that $X_i \sim \Gamma(\alpha_i,\theta) = \Gamma(\alpha_i,1)$ then $\left(\frac{X_1}{\sum_{i = 1}^K X_i}, \frac{X_2}{\sum_{i = 1}^K X_i}, \dots, \frac{X_K}{\sum_{i = 1}^K X_i}\right) \sim \textrm{Dir}(\alpha_1, \alpha_2, \dots, \alpha_K)$ where $\textrm{Dir}(\alpha_1, \alpha_2, \dots, \alpha_K)$ is a Dirichlet distribution with a concentration parameter $(\alpha_1, \alpha_2, \dots, \alpha_K)$.

My question is does this result hold for any scale parameter $\theta > 0$? The proofs I've read (e.g. http://mayagupta.org/publications/FrigyikKapilaGuptaIntroToDirichlet.pdf), which use the change-of-variables formula, seem to set $\theta = 1$.

• As long as its the same scale. The reasoning is simple, since dividing a variable with scale $\theta$ by $\theta$ gives scale $1$. In the ratios, the $\theta$s in the numerator and denominator will cancel, making it the same for any other particular value $\theta$ might be. – Glen_b Jun 26 '14 at 2:51

The result for scale 1 straightforwardly implies the generalized result for arbitrary scale $\theta>0$ as follows. Let
$$X_k \sim \Gamma(\alpha_i,\theta),~i\in \{1,\ldots,K\},$$ and the $X_i$s be mutually independent. Now, let us define variables $(Z_1,\ldots,Z_k)$ by scaling the $X$s: $$Z_i = \frac{X_i}{\theta},~i\in \{1,\ldots,K\}.$$ The $Z_i$s are mutually independent and $Z_i\sim \Gamma(\alpha_i,1)$. Then, we express $$\left(\frac{X_1}{\sum_{i = 1}^K X_i}, \frac{X_2}{\sum_{i = 1}^K X_i}, \dots, \frac{X_K}{\sum_{i = 1}^K X_i}\right)$$ in terms of the $Z_i$s: $$=\left(\frac{Z_1\theta}{\sum_{i = 1}^K( Z_i\theta)}, \frac{Z_2\theta}{\sum_{i = 1}^K (Z_i\theta)}, \dots, \frac{Z_K\theta}{\sum_{i = 1}^K (Z_i\theta)}\right).$$ Divide both numerator and denominator by $\theta$: $$=\left(\frac{Z_1}{\sum_{i = 1}^K Z_i}, \frac{Z_2}{\sum_{i = 1}^K Z_i}, \dots, \frac{Z_K}{\sum_{i = 1}^K Z_i}\right).$$ By construction the $Z_i$s have $\Gamma(\alpha_i,1)$ distributions, and thus the last expression has the $\mathrm{Dir}(\alpha_1,\alpha_2,\ldots,\alpha_K)$ distribution. That is, it has been shown that $$\left(\frac{X_1}{\sum_{i = 1}^K X_i}, \frac{X_2}{\sum_{i = 1}^K X_i}, \dots, \frac{X_K}{\sum_{i = 1}^K X_i}\right) \sim \mathrm{Dir}(\alpha_1,\alpha_2,\ldots,\alpha_K).$$
• Thanks for your answer. I know the result is the same but did you mean to write $\theta Z_i$ rather than $Z_i / \theta$ when replacing $X_i$ (and then multiply numerator and denominator by $\theta^{-1}$)? – Richy Jun 27 '14 at 10:33