TITLE: Stochastic domination and sum of random variables
QUESTION [10 upvotes]: I'm dealing with a statement that seems trivial but I couldn't prove it rigorously.
Suppose that I have two i.i.d. random variables $X_1$ and $X_2$ and another couple of i.i.d. variables $Y_1$ and $Y_2$ ($X$ and $Y$ are independent on each other and do not need to have the same law). Suppose that $X$ dominates $Y$ in the sense that 
$$\mathbb{P}(X \ge t) \ge \mathbb{P}(Y \ge t), \quad \forall \: t \in \mathbb{R}  $$
How can I show that 
$$\mathbb{P}(X_1+X_2 \ge t) \ge \mathbb{P}(Y_1+Y_2 \ge t), \quad \forall \: t \in \mathbb{R}  $$
This seems pretty intuitive to me but I didn't get far.

REPLY [2 votes]: A coupling based argument:
Since $X_1$ dominates $Y_1$ and $X_2$ dominates $Y_2$ we know there are couplings $(X_1', Y_1')$ and $(X_2', Y_2')$ such that $X_1'\ge Y_1'$ and $X_2'\ge Y_2'$ almost surely.
That implies $X_1'+X_2' \le Y_1'+Y_2'$ which means $X_1+X_2$ dominates $Y_1+Y_2$.