0
$\begingroup$

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

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.