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I am trying to understand how my professor (see derivation below) has derived the multivariate marginal distribution of a subvector of $\theta_j$´s from a Dirichlet distribution. I understand everything really, until the third line. I understand that he wants the integral to represent a unnormalised beta integral and hence he have to divide $\theta^{\alpha_{k-1}-1}_{k-1}$ and $(1-\theta_1 ... - \theta_{k-1})^{\alpha_{k}-1}$ by their sum, i.e $\theta^{\alpha_{k-1}-1}_{k-1} + (1-\theta_1 ... - \theta_{k-1})^{\alpha_{k}-1}$

If I understand correctly, the sum is equal to $(1-\theta_1 ... - \theta_{k-1})^{\alpha_{k-2}+\alpha_{k-1}-2}$, but how do we get that this is in fact the sum?

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  • $\begingroup$ Thanks for the edit @Xi'an $\endgroup$ Commented Nov 10, 2020 at 13:08

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Since$$p(\theta_1,\ldots,\theta_{k-1})\propto (1-\theta_1-\cdots-\theta_{k-1})^{\alpha_{k}-1}\prod_{i=1}^{k-1} \theta_i^{\alpha_i-1}$$over the $\mathbb R^k$-simplex, $$\mathfrak S = \left\{(\theta_1,\ldots,\theta_{k-1})\in\mathbb R^{k-1}_+\,;\,\sum_{i=1}^{k-1} \theta_i\le 1\right\}$$ integrating out $\theta_{k-1}$ produces the marginal density of $(\theta_1,\ldots,\theta_{k-2})$: \begin{align} p(\theta_1,\ldots,\theta_{k-2})&\propto \int_0^{1-\theta_1-\cdots-\theta_{k-2}} (1-\theta_1-\cdots-\theta_{k-1})^{\alpha_k-1}\prod_{i=1}^{k-1} \theta_i^{\alpha_i-1}\,\text d\theta_{k-1}\\ &= \prod_{i=1}^{k-2} \theta_i^{\alpha_i-1}\, \int_0^{1-\theta_1-\cdots-\theta_{k-2}} (1-\theta_1-\cdots-\theta_{k-1})^{\alpha_{k}-1}\,\theta_{k-1}^{\alpha_{k-1}-1}\,\text d\theta_{k-1}\\ \end{align} Given the upper bound on $\theta_{k-1}$ found in the integral, one can rather naturally consider the change of variable $$\theta_{k-1}=(1-\theta_1-\cdots-\theta_{k-2})\eta$$ leading to \begin{align} p(\theta_1,\ldots,\theta_{k-2}) &= \prod_{i=1}^{k-2} \theta_i^{\alpha_i-1}\, \int_0^1 (1-\theta_1-\cdots-\theta_{k-2})^{\alpha_{k-1}-1}(1-\eta)^{\alpha_{k-1}-1}\\ &\qquad\times(1-\theta_1-\cdots-\theta_{k-2})^{\alpha_{k-1}-1}\eta^{\alpha_{k-1}-1}\, (1-\theta_1-\cdots-\theta_{k-2})\text d\eta\\ &=\prod_{i=1}^{k-2} \theta_i^{\alpha_i-1}\,(1-\theta_1-\cdots-\theta_{k-2})^{\alpha_{k-1}+\alpha_k-1}\, \underbrace{\int_0^1 (1-\eta)^{\alpha_{k}-1}\eta^{\alpha_{k-1}-1}\,\text d\eta}_\text{constant in $\theta$}\\ &\propto\prod_{i=1}^{k-2} \theta_i^{\alpha_i-1}\,(1-\theta_1-\cdots-\theta_{k-2})^{\alpha_{k-1}+\alpha_k-1}\, \end{align}

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    $\begingroup$ +1, I wonder if there's a typo at the first line after "leading to..."; shouldn't it be $(1 - \eta)^{\alpha_{k}-1}$ rather than $(1 - \eta)^{\alpha_{k-1}-1}$? $\endgroup$ Commented Apr 26, 2022 at 6:29

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