# Maximum of a probability vector distributed as a Dirichlet variate

Let $p_1, p_2, \ldots \sim \text{Dirichlet}(\alpha_1, \alpha_2, \ldots)$. What is the distribution of $\max(p_1, p_2, \ldots)$?

I have searched for the order statistics of the Dirichlet distribution without success.

## 1 Answer

I am not sure there is a closed-form solution for the distribution of $p_{(k)}$ when $(p_1,\ldots,p_k)\sim\text{Dir}(\alpha_1,\ldots,\alpha_k)$ and the $\alpha_i$'s are different. At least, it seems to be the case when following the natural derivation of the density of $p_{(k)}$.

Indeed, if we start with the Gamma representation of $\text{Dir}(\alpha_1,\ldots,\alpha_k)$ random variables, namely that $$p_i=\xi_i\Big/\sum_{i=1}^k \xi_i,$$where $$\xi_i\sim\text{Ga}(\alpha_i,1)\,,$$ the index of the maximum in $(p_1,\ldots,p_k)$ is the same as the index of the maximum in $(\xi_1,\ldots,\xi_k)$. Considering the distribution of the maximum of the $\xi_i$'s, namely $$\xi_{(k)}=\arg\max\{\xi_1,\ldots,\xi_k\},$$ the same argument as the one used for order statistics leads to a marginal distribution with density \begin{align*}f_k(\xi)&=\sum_{i=1}^k \left\{\prod_{j\ne i} F(\xi;\alpha_j)f(\xi;\alpha_i)\right\}\\ &=\left\{\prod_{i=1}^kF(\xi;\alpha_i)\right\}\sum_{i=1}^k\dfrac{f(\xi;\alpha_i)}{F(\xi;\alpha_i)}\end{align*} where $f(\cdot;\alpha)$ and $F(\cdot;\alpha)$ are the Ga$(\alpha,1)$ pdf and cdf, respectively. (The later is not available in closed form.) This means in particular that the probability that the maximum is associated with index $\iota$ is given by $$\pi_\iota=\int_0^\infty \left\{\prod_{i\ne\iota}F(\xi;\alpha_i)\right\}f(\xi;\alpha_\iota)\,\text{d}\xi$$ (which does not seem to enjoy a closed-formed expression in the general case).

If we now move to the Dirichlet case, we have $$p_{(k)}=\xi_{(k)}\Big/\sum_{i=1}^k \xi_i=\xi_{(k)}\Big/\{\xi_{(k)}+\sigma\}\,,$$where $\sigma$ denotes the sum of the remaining $\xi_i$'s. Conditional on $\xi_{(k)}$, those $\xi_i$'s are distributed as $$\sum_{i=1}^k \pi_i \prod_{j\ne i} \left\{\dfrac{f(\xi_j;\alpha_j) }{F(\xi_{(k)};\alpha_j)}\mathbb{I}_{(0,\xi_{(k)})}(\xi_j)\right\}$$modulo a permutation that does not matter for the sum. Therefore the sum $\sigma$ is distributed from the convolution of this density, namely $$g(\sigma|\xi_{(k)})=\int_{[0,\xi_{(k)}]^{k-2}} \sum_{i=1}^k \pi_i \left\{\prod_{j\ne i,i+1} \dfrac{f(\xi_j;\alpha_j)}{F(\xi_{(k)};\alpha_j)}\right\} \dfrac{f(\sigma_{-i+1};\alpha_{i+1})}{F(\xi_{(k)};\alpha_{i+1})}\,\mathbb{I}_{(0,\xi_{(k)})}(\sigma_{-i+1})\,\prod_{j\ne i,i+1}\text{d}\xi_j$$where $$\sigma_{-i+1}=\sigma-\sum_{j\ne i,i+1}\xi_j$$denotes the remaining $\xi_{i+1}$ determined by the sum $\sigma$ and the other $\xi_j$, and where $i+1$ is understood modulo $k$, meaning that $k+1\equiv 1$. From there, we can derive the joint density of $(\sigma,\xi_{(k)})$ as $f_k(\xi_{(k)})g(\sigma|\xi_{(k)})$, hence the density of $$\dfrac{\xi_{(k)}}{\sigma+\xi_{(k)}}$$but there is no hope (?) this approach leads to a closed-form density.