We say that a random variable $X$ stochastically dominates a random variable $Y$ if $P(X\ge t)\ge P(Y\ge t).$

I wonder if it is true that for two binomial random variables $X=Bin(n_1,p), Y=Bin(n_2,p),$ $X$ dominates $Y$ if $n_1\ge n_2?$ Actually it is known that one binomial random variable dominates another if the total trials are same but success probabilities are different. How to show a binomial random variable dominates another binomial random variable with a smaller success value?

It seems we only need to verify all natural numbers $t.$ For $t\in [n_2,n_1],$ $P(X\ge t)\ge P(Y\ge t)=0$ is ture. For smaller $t,$ it involves many binomial coefficients. I don't find an easy way to show it.

  • $\begingroup$ Yes: Let $Z$ denote some Bin$(n_1-n_2,p)$ random variable independent of $Y$ then $X$ is distributed like $Y+Z$ and $Y+Z\geqslant Y$ almost surely hence $P(X\geqslant x)=P(Y+Z\geqslant x)\geqslant P(Y\geqslant x)$ for every $x$. $\endgroup$
    – Did
    May 17, 2017 at 9:46

1 Answer 1


Let $P(n,k)$ be the probability we get at least $k$ successes out of $n$ trials.

Then clearly $P(n+1,k)=P(n,k)(1-p)+P(n,k-1)p\geq P(n,k)(1-p)+P(n,k)p = P(n,k)$


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