Let $X\sim NegBin(n_1,p)$ and $Y\sim NegBin(n_2,p)$ with $n_1>n_2$. How do I show that $X$ stochastically dominates $Y$ (i.e. $F_X(x)\le F_Y(x)\quad\forall\:x\in\Bbb R$)?
Since a Negative binomial random variables is the number of tosses required before $r$ heads occur, where $r$ is the first parameter, we know that the minimum value it takes is at least $r$. Therefore, $0=F_X(x)\le F_Y(x)\quad\forall\:x<n_1$. However, I am finding it hard to prove it for $x\ge n_1$.
Also, intuitively it's clear. If we want to get more heads, then we have to perform more tosses, that's what I believe it says intuitively. But is there any way to show it algebraically?