Stochastic domination and sum of random variables 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.
 A: $$\begin{align}
P(X_1+X_2\geq t)&= E(1_{X_1+X_2\geq t})\\
&= \int 1_{x_1+x_2\geq t} dP_{(X_1,X_2)}(x_1,x_2) \tag{1}\\
&= \int \int 1_{x_1+x_2\geq t} dP_{X_1}(x_1) dP_{X_2}(x_2)\tag{2}\\
&= \int \left(\int 1_{x_1\geq t-x_2} dP_{X_1}(x_1)\right) dP_{X_2}(x_2)\\
&\geq \int \left(\int 1_{y_1\geq t-x_2} dP_{Y_1}(y_1)\right) dP_{X_2}(x_2)\tag{3}\\
&= \int \left(\int 1_{x_2\geq t-y_1} dP_{X_2}(x_2)\right) dP_{Y_1}(y_1)\\
&\geq \int \left(\int 1_{y_2\geq t-y_1} dP_{Y_2}(y_2)\right) dP_{Y_1}(y_1)\tag{4}\\
&= \int \int 1_{y_1+y_2\geq t} dP_{Y_1}(y_1) dP_{Y_2}(y_2)\\
&= P(Y_1+Y_2\geq t)
\end{align}$$
$(1)$: Law of the unconscious statistician
$(2)$: Independence of $X_1$ and $X_2$
$(3)$: $X_1$ dominates $Y_1$ hence for fixed $x_2$, $P(X_1\geq t-x_2)\geq P(Y_1\geq t-x_2)$
$(4)$: $X_2$ dominates $Y_2$
Note that the independence of $(X_1,X_2)$ with $(Y_1,Y_2)$ is not needed.
A: 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 $Y_1+Y_2$.
