Let the probability density function of $(Y_1, \ldots, Y_k)$ be a Dirichlet distribution parameterized by $\alpha_1 , \ldots, \alpha_k$, that is
$$
f(y_1, \ldots, y_k|\alpha_1, \ldots, \alpha_k) = \text{Dir}(\alpha_1, \ldots, \alpha_k)
$$
Let $D(y_1, \ldots, y_k|\alpha_1, \ldots, \alpha_k)$ denote the distribution function of $(Y_1, \ldots, Y_k)$,
and $X$ is a discrete random variable with p.f.
$$
P(X=j|Y_1, \ldots, Y_k) = Y_j \ \text{ a.s } \text{ for } j = 1, \ldots, k
$$
What I want to do is to calculate the following joint distribution of $X$ and $Y_1, \ldots, Y_k$:
$$
P(X=j, Y_1 \leq z_1, \ldots, Y_k \leq z_k)
$$
My way is using p.d.f. of the joint distribution above:
\begin{align}
&P(X=j, Y_1 \leq z_1, \ldots, Y_k \leq z_k) =\\
&\ \ \ \ \ \ \ P(X=j)P(Y_1 \leq z_1, \ldots, Y_k \leq z_k|X=j) \\
&\ \ \ \ \ \ = P(X=j)\int_0^{z_1}\cdots\int_0^{z_k}\frac{P(X=j|y_1, \ldots, y_k)f(y_1, \ldots, y_k)}{P(X=j)}dy_1 \cdots dy_k \\
&\ \ \ \ \ \ = \int_0^{z_1}\cdots\int_0^{z_k}y_jdD(y_1, \ldots, y_k|\alpha_1, \ldots, \alpha_k)
\end{align}
However, I'm looking for another solution by factorizing the joint distribution as
\begin{align}
&P(X=j, Y_1 \leq z_1, \ldots, Y_k \leq z_k) =\\
&\ \ \ \ \ \ \ P(X=j|Y_1 \leq z_1, \ldots, Y_k \leq z_k)P(Y_1 \leq z_1, \ldots, Y_k \leq z_k)
\end{align}
How can I deal with this factorization?
Best Answer
Let $A=P[X=j,Y_1\leqslant y_1, \ldots, Y_k\leqslant y_k]$. Then $$ A=E[P[X=j|Y_1, \ldots, Y_k]; Y_1\leqslant y_1, \ldots, Y_k\leqslant y_k]=E[Y_j; Y_1\leqslant y_1, \ldots, Y_k\leqslant y_k], $$ hence $$ A=\int z_j\,\mathbf 1_{z_1\leqslant y_1,\ldots,z_k\leqslant y_k}\,\mathrm{Dir}(z_1,\ldots,z_k)\,\mathrm dz. $$ Note that there is no simple formula for $P[X=j|Y_1\leqslant y_1, \ldots, Y_k\leqslant y_k]$.