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Let $X=(X_1,X_2,X_3)\sim \text{Dirichlet}(a_1,a_2,a_3)$ and $Y=(Y_1,Y_2,Y_3)\sim \text{Dirichlet}(a_1+b_1,a_2+b_2,a_3)$, where all $a_i$ and $b_i$ are positive. Is there a natural coupling between $X$ and $Y$ such that $X_1\geq Y_1$ and $X_2\geq Y_2$ with probability 1?

The following coupling does not guarantee this property: Define independent random variables $G_{c}\sim \text{Gamma}(c)$ for $c\in\{a_1,a_2,a_3,b_1,b_2\}$, and let

$$X_i = \frac{G_{a_i}}{G_{a_1}+G_{a_2}+G_{a_3}} $$

and

$$Y_i = \frac{G_{a_i}+G_{b_i}\mathbb{I}(i\in\{1,2\})}{G_{a_1}+G_{b_1}+G_{a_2}+G_{b_2}+G_{a_3}}. $$

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1 Answer 1

up vote 1 down vote accepted

This is not always possible.

Fix $a_1,a_2,a_3,b_1$. As $b_2\to\infty$, we have $Y_1\to0$ in probability, so it is not stochastically larger than $X_1$.

A necessary condition is domination of the expectations, namely $\frac{a_i+b_i}{\sum a_j+b_j} \ge \frac{a_i}{\sum a_j}$ for $i=1,2$, but this is not sufficient either. If $b_1/a_1=b_2/a_2$ and $b_i\to\infty$ then the $Y_i$'s converge to constants in probability.

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