I need to compute the probability of sampling a specific vector from any multinomial generated by a Dirichlet parametrization.
Let's call the vector $\mathbf{x} = \{x_i\}, i \in K$
I need to compute $$ p(\mathbf{x}|\alpha) = \int Mult_{PMF}(x, y) \, Dirichlet_{PDF}(y|\alpha)\;dy$$
Multinomial PMF is $$Mult_{PMF}(\mathbf{x}, \mathbf{p}) = B(\mathbf{x} +1) \prod_{i=1}^K p_i^{x_i} = \frac{\Gamma(\sum_{i=1}^K x_i +1)}{\prod_{i=1}^K \Gamma(x_i +1)} \prod_{i=1}^K p_i^{x_i}$$
Dirichlet PDF is $$Dirichlet_{PDF}(\mathbf{x}, \mathbf{\alpha}) = B(\mathbf{\alpha}) \prod_{i=1}^K x_i^{\alpha_i - 1} = \frac{\Gamma(\sum_{i=1}^K \alpha_i )}{\prod_{i=1}^K \Gamma(\alpha_i )} \prod_{i=1}^K x_i^{\alpha_i - 1}$$
This is what I derived
$$ \begin{align} p(\mathbf{x}|\alpha) & = \int Mult_{PMF}(x, y) \, Dirichlet_{PDF}(y|\alpha) \; dy \\ & = \int B(\mathbf{x} + 1) \prod_{i=1}^{K}y_i^{x_i} \; B(\alpha) \prod_{i=1}^K y_i^{\alpha_i-1} \; dy \\ & = B(\mathbf{x} + 1) B(\alpha) \int \prod_{i=1}^{K}y_i^{x_i} \; \prod_{i=1}^K y_i^{\alpha_i-1} \; dy \\ & = B(\mathbf{x} + 1) B(\alpha) \int \prod_{i=1}^{K}y_i^{x_i} y_i^{\alpha_i-1} \; dy \\ & = B(\mathbf{x} + 1) B(\alpha) \int \prod_{i=1}^{K}y_i^{x_i + \alpha_i-1} \; dy \\ \end{align} $$
At thi point I try to bring the part inside the integral in the form of the Dirichlet
$$ \begin{align} p(\mathbf{x}|\alpha) & = \frac{B(\mathbf{x} + 1) B(\alpha)}{B(\mathbf{x} + \alpha)} \int B(\mathbf{x} + \alpha) \prod_{i=1}^{K}y_i^{x_i + \alpha_i-1} \; dy \\ & = \frac{B(\mathbf{x} + 1) B(\alpha)}{B(\mathbf{x} + \alpha)} \int Dirichlet_{PDF}(y, \mathbf{x} + \mathbf{\alpha}) \; dy \\ & = \frac{B(\mathbf{x} + 1) B(\alpha)}{B(\mathbf{x} + \alpha)} \end{align}$$
I'm not sure I did all the steps correctly, as the result I obtained is pretty different from what I see online (more specifically at How to derive the Dirichlet-multinomial? )
In my result I have an additional $ B(x+1)$.
Is my result correct?
Why does it differ from the one showed in the other answer?