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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 $\theta = 1$.

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    $\begingroup$ 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. $\endgroup$
    – Glen_b
    Commented Jun 26, 2014 at 2:51

1 Answer 1

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The result for scale 1 straightforwardly implies the generalized result for arbitrary scale $\theta>0$ as follows. Let
\begin{equation} X_k \sim \Gamma(\alpha_i,\theta),~i\in \{1,\ldots,K\}, \end{equation} and the $X_i$s be mutually independent. Now, let us define variables $(Z_1,\ldots,Z_k)$ by scaling the $X$s: \begin{equation} Z_i = \frac{X_i}{\theta},~i\in \{1,\ldots,K\}. \end{equation} The $Z_i$s are mutually independent and $Z_i\sim \Gamma(\alpha_i,1)$. Then, we express \begin{equation} \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) \end{equation} in terms of the $Z_i$s: \begin{equation} =\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). \end{equation} Divide both numerator and denominator by $\theta$: \begin{equation} =\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). \end{equation} 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 \begin{equation} \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). \end{equation}

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  • $\begingroup$ 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 $\theta^{-1}$)? $\endgroup$
    – Richy
    Commented Jun 27, 2014 at 10:33
  • $\begingroup$ Indeed. Sorry about that, thanks for spotting the mistake. I have now edited the answer to fix that. $\endgroup$ Commented Jun 27, 2014 at 10:43

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