# Dirichlet-Multinomial probability that a specific category has the lowest count

Say we have $$X=(X_1,X_2,\dots,X_n)\overset{d}{=}\operatorname{Dirichlet-Multinomial}(\boldsymbol{\alpha},N)$$. How can we find the probability that $$X_m$$ is the smallest of the $$X_i$$, $$\operatorname{P}(X_m, for each $$m$$?

Edit: The marginal distributions for $$X_i$$ are $$\operatorname{Beta-Binomal(\alpha_i,\sum_{j\neq i}\alpha_j, N)}$$. Hence:$$\operatorname{P}(X_i=x_i)=\binom{N}{x_i}\frac{\operatorname{B}(x_i + \alpha_i, N-x_i + \sum_{j\neq i}\alpha_j)}{\operatorname{B}(\alpha_i,\sum_{j\neq i}\alpha_j)}.$$

• How do you interpret $\operatorname{P}(X_m<X1,\dots,X_{m-1},X_{m+1},\dots,X_{k})$? – callculus Nov 19 '20 at 6:22
• The probability that $X_m<X_i$ for all $i\neq m$. – Floyd Everest Nov 19 '20 at 6:27
• That was my idea as well, but I wasn´t sure. – callculus Nov 19 '20 at 6:29
• So $X_m$ being equal-lowest presumably is not good enough. Are the $p_m$s different for each $m$? You may find simulation the easiest approach for particular cases – Henry Nov 19 '20 at 12:27
• @Henry I think $X_m$ being equal lowest could be good enough, and I was hoping a closed form would offer some improvement over simulation in my case. And yes, the $p_i$ are not always equal. – Floyd Everest Nov 19 '20 at 12:43

I don't think there can be a closed-form solution for $$\operatorname{P}(X_m because for any particular $$x_m$$, we would need to work out partitions with a minimum size for each part for the remaining $$x_k$$ in order to ensure that each was at least $$x_m+1$$. If, for example, $$x_m$$ was $$2$$, we would need to partition the other $$N-2$$ samples into $$i-1$$ parts of at least $$3$$ each. If, as contemplated in your comment, equally low values were included, the minimum would decrease to $$2$$, but there would still be some minimum for each non-$$m$$ $$x_k$$ to satisfy.
Each acceptable partition might, in turn, represent multiple combinations, depending on the exchangeability of the remaining $$x_k$$ and $$\boldsymbol{\alpha}_k$$. If all of the remaining $$x_k$$ values were identical and all of the associated $$\boldsymbol{\alpha}_k$$ values were identical, then a single calculation would suffice to cover every instantiation of that partition. At the other extreme, if no $$x_k$$ equalled another $$x_k$$ and no $$\boldsymbol{\alpha}_k$$ equalled another $$\boldsymbol{\alpha}_k$$, then each acceptable partition could be realized in $$k!$$ unique ways (mappings of partition values to $$i$$'s), each of which would require a separate calculation. In any event, the impossibility of counting in closed form even the number of acceptable partitions for a given $$x_m$$ (let alone summing the probabilities associated with all the different realizations of that partition) dooms, it seems to me, the possibility of a closed-form solution.