First order stochastic dominance and summation of random variables Suppose $X$ first-order stochastically dominates $Y$.
Moreover, $W_i$ is I.I.D. and follows Bernoulli distribution with success probability of $\alpha$
Then, does $X +\sum_{i=1}^{f(X)} W_i$ first-order stochastically dominate $Y +\sum_{i=1}^{f(Y)} W_i$? 
 A: Yes. 
Claim:
Suppose that


*

*We have random variables $X, Y, \{W_i\}_{i=1}^{\infty}$.

*$W_i\geq 0$ for all $i \in \{1, 2, 3, ...\}$. 

*$(X,Y)$ is independent of $\{W_i\}_{i=1}^{\infty}$. 

*$P[X>x] \geq P[Y>x]$ for all $x \in \mathbb{R}$. 

*$f:\mathbb{R}\rightarrow \{0, 1, 2, 3, \ldots\}$ is a nondecreasing function. 
Then $P[X + \sum_{i=1}^{f(X)} W_i > x] \geq P[Y + \sum_{i=1}^{f(Y)} W_i > x]$ for all $x \in \mathbb{R}$.
Proof:
Extend the probability space to include a random variable $U$ that is uniform over $(0,1)$ and is independent of $(X, Y, \{W_i\}_{i=1}^{\infty})$. Define random variables $A$ and $B$ (based entirely on $U$) by:  
\begin{align}
A &= \inf \{x \in \mathbb{R} : P[X\leq x]\geq U\}\\
B &= \inf \{ x \in \mathbb{R} : P[Y\leq x]\geq U\}
\end{align} 
By standard theory for generating random variables according to a given distribution, we know that $A$ has the same distribution as $X$, and $B$ has the same distribution as $Y$.  Further, since $A$ and $B$ are based entirely on $U$, they are both independent of $(X,Y, \{W_i\})$. Finally, we see that $A\geq B$ always (since $P[X\leq x]\leq P[Y\leq x]$ for all $x$). So we have: 
\begin{align}
A &\geq B \\
f(A) &\geq f(B)\\
\sum_{i=1}^{f(A)} W_i &\geq \sum_{i=1}^{f(B)} W_i
\end{align}
Thus 
$$ A + \sum_{i=1}^{f(A)} W_i \geq B + \sum_{i=1}^{f(B)} W_i$$
But the left-hand-side has the same distribution as $X + \sum_{i=1}^{f(X)}W_i$ while the right-hand-side has the same distribution as $Y+\sum_{i=1}^{f(Y)}W_i$. 
$\Box$
