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.
Best Answer
I think altering the notation a little bit can be useful here. Let's start with the hand that player 2 has been given. This is a binary variable ($H$) with probability $Pr(H=K)=p$ of event K. In code, $H \sim Bern(p)$.
Second, the actions that can be taken by player 2. Let $n$ be the number of possible actions he can choose. I am going to assume that whatever hand he has been dealt, he can choose from the same set of actions. Some of those actions might not be smart to choose, but he has that option.
Third, the probability of player 2 taking action $i$ for $i \in \{1, ..., n\}$. Let us call this probability $X_i$ and let $X=(X_1, ...X_n)$ the vector with probabilities of all actions. Whatever hand player 2 has been dealt. So get rid of $Y_i$.
You assume that $X$ has the Dirichlet distribution. That can be correct, but it should be noted that $X$ can only have this distribution conditional on $H$. Thus, $X|\{H=h\} \sim Dir(n,\alpha)$ ($h$ can be a number representing $K$ or $Q$). Only then the condition that their sum equals 1 is met.
So the probability of player 2 taking action $i$ depends in some way on the cards dealt, but I did not specify how. I think the most reasonable choice is to let this happen via $\alpha$. Thus, your model would look something like $X|\{H=h\} \sim Dir(n,\alpha(h))$.
The expectation of $X_i$ can then be calculated using $E(X_i) = E(E(X_i|H)) = E(\frac{\alpha_i(H)}{\sum_i \alpha_{i}(H)})$. The exact expression and whether it can be obtained in close form depends on how you let $\alpha$ depend on $h$.
I hope this helps.