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 $X_i$, $\operatorname{P}(X_m<X_k, \forall k\neq m)$, for each $m$?

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

  • $\begingroup$ How do you interpret $\operatorname{P}(X_m<X1,\dots,X_{m-1},X_{m+1},\dots,X_{k})$? $\endgroup$ Nov 19, 2020 at 6:22
  • $\begingroup$ The probability that $X_m<X_i$ for all $i\neq m$. $\endgroup$ Nov 19, 2020 at 6:27
  • $\begingroup$ That was my idea as well, but I wasn´t sure. $\endgroup$ Nov 19, 2020 at 6:29
  • $\begingroup$ 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 $\endgroup$
    – Henry
    Nov 19, 2020 at 12:27
  • $\begingroup$ @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. $\endgroup$ Nov 19, 2020 at 12:43

1 Answer 1


I don't think there can be a closed-form solution for $\operatorname{P}(X_m<X_k, \forall k\neq m)$ 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.

Once we move to approximations, moreover, we are inevitably trading off accuracy for computational ease. So, without knowing how you would weigh these, it's hard to know what kinds of approximations you would or would not find acceptable.


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