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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, $0=F_X(x)\le F_Y(x)\quad\forall\:x<n_1$. 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?

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  • $\begingroup$ 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 $[X \le x] \subseteq [Y \le x]$. $\endgroup$ – guy May 26 at 2:36
  • $\begingroup$ @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. $\endgroup$ – Martund May 26 at 2:41
  • $\begingroup$ Seems good to me! $\endgroup$ – guy May 26 at 2:48
  • $\begingroup$ Post it as an answer @guy, I'll accept. $\endgroup$ – Martund May 26 at 2:53
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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.

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