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I'm studying BRML.

In this book, a Dirichlet distribution is defined as $$ p(\alpha | u) = \frac{\Gamma(\sum_{q=1}^Q u_q)}{\prod_{q=1}^Q \Gamma(u_q)} \delta_0 \left( \sum_{q=1}^{Q} \alpha_q - 1 \right) \prod_{q=1}^{Q} \alpha_q^{u_q - 1} \mathbb{I}[\alpha_q \geq 0], $$ where $\delta$ is Dirac's delta, and Exercise 8.7 is to show that $$\displaystyle\int_{\theta_j} \textrm{Dirichlet}(\theta | u) = \textrm{Dirichlet}(\theta_{\backslash j} | u_{\backslash j})$$

Since we integrate on $\theta_j$, Dirac's delta term becomes function of $\theta_j$, which is $$\delta_0 \left( \theta_j - (1-\sum_{i\neq j} \theta_i) \right)$$ and we can push out irrelevant terms to outside of integral, then it becomes $$ \frac{\Gamma(\sum_{q=1}^Q u_q)}{\prod_{q=1}^Q \Gamma(u_q)} \prod_{q \neq j}^{Q} \theta_q^{u_q - 1} \mathbb{I}[\theta_q \geq 0] \int \delta\left(\theta_j - (1 - \sum_{i\neq j}\theta_i )\right) \theta_j^{u_j - 1} \mathbb{I}[\theta_j \geq 0] $$ With the definition of Dirac's delta, I got $$ \frac{\Gamma(\sum_{q=1}^Q u_q)}{\prod_{q=1}^Q \Gamma(u_q)} \left(\prod_{q \neq j}^{Q} \theta_q^{u_q - 1} \mathbb{I}[\theta_q \geq 0] \right) (1-\sum_{i\neq j} \theta_i)_j^{u_j - 1} \mathbb{I}\left[1-\sum_{i\neq j}\theta_i \geq 0\right] $$ But I don't know how it can be $\displaystyle\textrm{Dirichlet}(\theta_{\backslash j} | u_{\backslash j}) = \frac{\Gamma(\sum_{q\neq j}^Q u_q)}{\prod_{q \neq j}^Q \Gamma(u_q)} \delta \left( \sum_{q\neq j}^{Q} \theta_q - 1 \right) \prod_{q\neq j}^{Q} \theta_q^{u_q - 1} \mathbb{I}[\theta_q \geq 0]$

I searched internet a couple of minutes, then I found some relevant answers and further ones like this one and that one

But I cannot apply them to my problem due to different notations and definition. Especially, I want to know how to derive dirac delta term.

Xi'an
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yeomjy
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    Forget about Dirac's deltas, which are needlessly confusing, and consider this density to be defined over the simplex of $\mathbb R^{Q-1}$, that is, in terms of the vector $(\alpha_1,\ldots, \alpha_{Q-1})$, replacing $\alpha_Q$ by$$1-\sum_{i<Q}\alpha_i$$ – Xi'an Sep 18 '22 at 14:03
  • This is the representation adopted on Wikipedia and it is consistent with the $Q=2$ case of the Beta distribution. – Xi'an Sep 18 '22 at 15:04
  • Thanks for answering. Obviously, your notation seems more standard than dirac delta notation. – yeomjy Sep 19 '22 at 07:23

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