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 $0.5 < 0.9$.
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 $0.5 < 0.9$.
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 $y=0.5$.
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
This assumes that the random variable are independent, of course.