For the standard Dirichlet, the expectation of $X_i$ is $\alpha_i/\alpha_0$, where $\alpha_0 = \sum_i \alpha_i$ [http://en.wikipedia.org/wiki/Dirichlet_distribution].

I am considering the following generalization. Suppose we are playing a simple poker game as follows. We are player 1 and observing player 2's plays. Player 2 can be dealt one of two hands with probability 1/2 (or more generally probability p) -- K or Q (player 2 sees this and player 1 does not). Then player 2 selects one of i actions. Let $X_i$ denote the rv for the probability player 2 plays action i with a K, and let $Y_i$ denote the rv for Q. Suppose that he is following a static probability distribution for all rounds, and that the $X_i$'s and $Y_i$'s are independent (his strategy for a K is independent of his strategy for a Q). Player 1 only observes the action i of P2, and not his card.

I am trying to find a closed form for $E[X_i]$ and $E[Y_i]$.

This generalization differs from the "Generalized Dirichlet distribution," which has a relatively simple closed-form solution for the expectations (http://en.wikipedia.org/wiki/Generalized_Dirichlet_distribution).

I think the pdf of this distribution is the following, where p denotes the probability P2 is dealt a K, $\alpha_i$ is the observed number of times he has taken action $i$ so far, and $N(\alpha)$ is some normalization constant I'm not sure how to compute:

$$f(x,y;\alpha) = N(\alpha) * \prod_i [p x_i + (1-p)y_i]^{\alpha_i}$$ where $x_i, y_i \geq 0, \sum_i x_i = 1, \sum_i y_i = 1.$

One can apply the binomial theorem here, but I am not sure how to proceed and if that will even help. Somehow the gamma/beta function will need to come in to play.

Thanks a lot.