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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?

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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 $P[X=j|Y_1\leqslant y_1, \ldots, Y_k\leqslant y_k]$.

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  • $\begingroup$ Could you say something for the advantage of using expectation rather than the integration of p.d.f? $\endgroup$ – fishiwhj Aug 9 '13 at 10:27
  • $\begingroup$ Not much (basically because I do not understand your comment)--except that the manipulations in your post (factoring P(X=j), reversing Bayes formula (apparently), simplifying by P(X=j)) are somewhat odd. $\endgroup$ – Did Aug 9 '13 at 10:44
  • $\begingroup$ OK...Please forgive my poor English, by the way, is there something wrong(or non-standard) with my solution? $\endgroup$ – fishiwhj Aug 10 '13 at 4:22

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