Sufficient statistic for product of $\Gamma(x + a_i)$ Background
Suppose that the national mint mints coins with bias $p_i \sim \rm{Beta}(A,B)$ for some unknown constants $A$, $B$.
Given $n$ coins, you flip each coin a certain number of times.  Coin $i$ comes up heads $a_i$ times and tails $b_i$ times.  (The flips are clearly exchangeable, so we don't need the actual sequence of flips.)
If I did the math right, I figured out that the likelihood of $A, B$ given the statistics $(a_i, b_i)$ is:
$$L(A,B \mid a_i, b_i) = \frac{\prod_i\rm{Beta}(A + a_i, B + b_i)}{\rm{Beta}(A, B)^n}.$$
(Edit in 2012: This is the product of $n$ Pólya–Eggenberger urn schemes.)
I would like to summarize the statistics $(a_i, b_i)$ even further, into a finite set of numbers if possible.
Question
Can $\prod_i\Gamma(A + a_i)$ be written using sufficient statistics calculated from the $a_i$?
Example
For example, $\prod_i\exp(A + a_i)$ can be written $\exp(nA + S)$ where $n, S$ are sufficient statistics calculated from the $a_i$.
 A: I think I would agree with your expression for the likelihood, although this approach is not without problems (see below). 
Notice, that since $a_i$ and $b_i$ are integers, the likelihood is just a rational function in $A$ and $B$:
$$
   L(A,B \mid a,b) = \prod_{i=1}^n \left( \frac{ \prod\limits_{k=0}^{a_i-1} (A+k) \prod\limits_{m=0}^{b_i-1} (B+m) }{\prod\limits_{j=0}^{a_i+b_i-1} (A+B+j)}\right)
$$
The maximum likelihood equations are easily seen to be
$$
 \begin{eqnarray}
   \sum_{i=1}^n \sum_{k=0}^{a_i-1} \frac{1}{A+k} &=&  \sum_{i=1}^n \sum_{j=0}^{a_i+b_i-1} \frac{1}{A +B+j}  \\
   \sum_{i=1}^n \sum_{m=0}^{b_i-1} \frac{1}{B+m} &=&  \sum_{i=1}^n \sum_{j=0}^{a_i+b_i-1} \frac{1}{A +B+j}
  \end{eqnarray}
$$
Given that $A>0$ and $B>0$ these equations become polynomial equations in $A$ and $B$ with, hopefully, a unique positive root. However, checks in Mathematica for small values of $n$ and $a_i$, $b_i$ indicate that the resulting system admits no positive finite solutions.

It is easily seen, that infinite $A$ and $B$ solve MLE equations. Let's fix $r=A/B$. Then, in the limit of infinite $A$ and $B$, equations reduce to
$$
 \begin{eqnarray}
   \sum_{i=1}^n a_i &=&  \sum_{i=1}^n \frac{r}{r+1} (a_i+b_i)  \\
   \sum_{i=1}^n r b_i &=&  \sum_{i=1}^n \frac{r}{r +1} \cdot (a_i+b_i)
  \end{eqnarray}
$$
which gives 
$$
   r = \frac{\sum_{i=1}^n a_i}{\sum_{i=1}^n b_i}
$$
This result is confirmed numerically by the following heuristic. Use $a_i$ and $b_i$ to estimate head probability $p_i$ for each coin using ML estimator $p_i = \frac{a_i}{a_i+b_i}$. Then use ML estimation for beta distribution to determine $\alpha$ and $\beta$ from the data $\{p_1, \ldots, p_n \}$.

I do not have a good explanation as to why ML estimation can not seem to determine both $\alpha$ and $\beta$. This may be worth a separate question.
