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


$$Y\_i = \frac{G\_{a\_i}+G\_{b\_i}\mathbb{I}(i\in\{1,2\})}{G\_{a\_1}+G\_{b\_1}+G\_{a\_2}+G\_{b\_2}+G\_{a\_3}}. $$



Is there a monotone coupling of Dirichlet random variables?
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.


