Stochastic ordering Assume $Y$ is non negative random variable. Prove that $X+Y$ is stochastically greater than $X$ for any random variable $X$.
We have to prove there that $\Pr(X+Y > x) \geq \Pr(X>x) $ for all $x$
 A: Recalling that $A \subset B \Rightarrow P(A) \leq P(B)$, and that $B := \{ X > x -\epsilon \} \supset \{ X > x \} := A$  for any $\epsilon \geq 0$, and that by positivity of $Y$, $x - Y = x - \epsilon$ for some $\epsilon \geq 0$ (for every sample point $\omega$): 
$$P(X+Y > x) = P(X > x - Y) \geq P(X>x).$$
A: One approach:
Would be glad if anyone points out any mistake if there is one.(Or comment to say if there is no mistake).
$P(X+Y \leq x) = \int_{-\infty}^{\infty} F_X(x-y)dG_Y(y)dy$ where $F$ and $G$ are cdf of $X,Y$ respectively.
$\leq \int_{-\infty}^{\infty} F_X(x)dG_Y(y)dy$.     As $F(x-t) \leq F(x) \forall t\geq 0$ ($F$ is non decreasing.)
$= F_X(x) \int_{-\infty}^{\infty}dG_Y(y)$
$=F_X(x)$  as $\int_{-\infty}^{\infty}dG_Y(y)=1$
First step used convolution
A: Stochastic ordering is equivalent to the existence of a coupling which strictly preserves the order. I.e. there is a random variable $u$ study that $X(u) +Y(u) \geq X(u)$. The statements about probabilities fall out of this readily.
