Binomial Distribution: Stochastic Dominance Suppose

*

*$X_1 \sim \operatorname{Bin}(N_1,p)$ and $X_2 \sim \operatorname{Bin}(N_2,p)$

*$N_2>N_1$
Does $X_2$ first-order stochastically dominate $X_1$?
 A: Yes! One proof I know is via coupling. Let $X_{1},\ldots,X_{n}$ be an IID sequence of Bernoulli random variables
with $P\left[X_{i}=1\right]=p$. Also, we have $S_{n}:=\sum_{i=1}^{n}X_{i}\sim\text{Bin}\left(n,p\right)$.
Then $S_{n+1}\geq S_{n}$. Hence, $P\left[S_{n+1}>s\right]=P\left[\sum_{i=1}^{n+1}X_{i}>s\right]\geq P\left[\sum_{i=1}^{n}X_{i}>s\right]=P\left[S_{n}>s\right]$
which establishes the claim.
See Theorem 4.23 and example 4.24 in these notes. Altenatively, you may relate the Binomial distribution with the order statistics distribution of a distribution and use the known stochastic ordering relation wrt $n$ (see e.g. this book)
A: 
Lemma: If $X,Y$ are random variables on a common probability space where $X\ge Y$ holds always, then $X$ scholastically dominates $Y$.

Proof: $\{Y>c\}\subseteq \{X>c\}$, so $P(Y>c)<P(X<c)$. $\square$
Now, flip $N_2$ independent coins where each has a probability $p$ of heads. Let $X_2$ be the total number of heads, and let $X_1$ be the total number of heads in a particular subset of $N_1$ coins. Clearly $X_2\ge X_1$, so $X_2$ stochastically  dominates $X_1$. 
