# How to prove stochastic dominance of negative binomial random variable

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. $$F_X(x)\le F_Y(x)\quad\forall\:x\in\Bbb R$$)?

My try

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, $$0=F_X(x)\le F_Y(x)\quad\forall\:x. However, I am finding it hard to prove it for $$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?

• Hint: construct an experiment where $X$ and $Y$ have the desired negative binomial distributions, but $Y \le X$ with probability 1. It would immediately follow from this that $F_X(x) \le F_Y(x)$ because $[X \le x] \subseteq [Y \le x]$. – guy May 26 at 2:36
• @guy, Tossing a coin until we get $n_1$ heads will do? Then let $X$ be the number of tosses and $Y$ be the number of tosses until $n_2$'th head. – Martund May 26 at 2:41
• Seems good to me! – guy May 26 at 2:48
• Post it as an answer @guy, I'll accept. – Martund May 26 at 2:53

It is possible to show that $$F_X$$ stochastically dominates $$F_Y$$ if and only if there exists a probability space $$(\Omega, \mathcal F, P)$$ such that there are random variables $$X \sim F_X$$ and $$Y \sim F_Y$$ defined on this space such that $$Y \le X$$ holds everywhere. The if part of the implication is trivial, since $$[X \le x] \subseteq [Y \le x]$$ so that $$F_X(x) = \Pr(X \le x) \le \Pr(Y \le x) = F_Y(x)$$. While you don't need the second part, note that the only if part is also easy, as we can take $$U \sim \text{Uniform}(0,1)$$ and apply the probability integral transform with $$X = F_X^{-}(U)$$ and $$Y = F_Y^{-}(U)$$ where $$F^-$$ denotes taking the generalized inverse of the cdf.
In the case of negative binomial random variables, consider infinitely flipping a coin with heads probability $$p$$ and let $$X$$ the flip at which we hit heads $$n_1$$ heads and let $$Y$$ be the flip at which we hit $$n_2$$ heads. Then $$Y \le X$$, so we can apply the above observation.