2 typo

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 is in probability.

1

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 is probability.