Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X \sim Binomial(100, 0.9)$. Apparently, it is possible to have another random variable $Y\sim Binomial(100, 0.5)$ for which $P(X \ge Y) = 1$.

How is that possible? I don't feel that it is because $0.5 < 0.9$.

share|improve this question
    
To be clear, how can one guarantee that $X \ge Y$? A binomial distribution spans 0 to 100, right? –  Minden Petrofsky Oct 15 '12 at 18:32
    
I do not see any such way to do so. I mean, you make two Bernoulli experiments 100 times each, independently of the probability parameter, and there's still the chance you get no 'wins' in the one with highest probability and one 'win' in the one with less chances. Little, it is, but still non-zero. If the random variables were continuous, then some thing could be argued, but in this case I'd say the statement is wrong. Where did you get it? –  busman Oct 15 '12 at 18:40

2 Answers 2

up vote 3 down vote accepted

It is indeed possible that $Y\leqslant X$ almost surely, and this fact stems from what is called the coupling method.

Obviously, not every pair $(X,Y)$ of random variables with respective distributions Bin$(n,x)$ and Bin$(n,y)$ with $y\leqslant x$ in $(0,1)$ is such that $\mathbb P(Y\leqslant X)=1$. Hence, what this result means is that one can construct some pair $(X,Y)$ on some common probability space, with the desired marginal distributions, and such that $\mathbb P(Y\leqslant X)=1$.

To do so, let us start from the favorite building block of anyone interested in the simulation of random variables, namely a sequence $(U_k)_{1\leqslant k\leqslant n}$ of i.i.d. random variables with uniform distribution on $[0,1]$. As is well known, each random variable $\mathbf 1_{U_k\leqslant x}$ is Bernoulli$(x)$, for every $x$ in $(0,1)$. Let us consider, for some $y\leqslant x$, $$ X=\sum_{k=1}^n\mathbf 1_{U_k\leqslant x} \qquad\text{and}\qquad Y=\sum_{k=1}^n\mathbf 1_{U_k\leqslant y}. $$ The following points should be obvious: the distribution of $Y$ is Bin$(n,y)$; the distribution of $X$ is Bin$(n,x)$; and $Y\leqslant X$ almost surely. QED.

This yields a coupling of the family of distributions $($Bin$(n,x))_{x\in[0,1]}$, for each fixed $n$. Your case is when $n=100$, $x=0.9$ and $y=0.5$.

share|improve this answer

If $X \sim Binomial(n_X,p_X)$ and $Y \sim Binomial(n_Y,p_Y)$, then the only ways to get $P(X \geq Y) = 1$ are

  1. $p_X=1$ and $n_X \geq n_y$, because you then have $X = n_X$ and $P(Y \leq n_x) = 1$ (the latter because $n_Y \leq n_X$).
  2. $p_Y=0$ or $n_Y = 0$, because you then have $Y = 0$.

This assumes that the random variable are independent, of course.

share|improve this answer
1  
But other cases could get the probability so close to 1 that software, like R, get the numerical answer 1. Then you must remember that numerical 1 is not the same as 1. –  kjetil b halvorsen Oct 15 '12 at 18:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.