# Introduction

Let $$X$$ and $$Y$$ be random varaibles not necessarily defined on the same probability space, thus they may not even have a joint distribution, and we can not make sense of an expression like $$X+Y$$. Let $$X'$$ and $$Y'$$ be random variables defined on the same probability space. We say that $$(X', Y')$$ is coupling of $$X$$ and $$Y$$ if $$X$$ and $$X'$$ have the same law and $$Y$$ and $$Y'$$ have the same law. There is always at least one coupling given by the case where we take $$X'$$ to be independent of $$Y'$$; usually, this is not the coupling we are interested in. Let $$U$$ be uniformly distributed in $$[0,1]$$. In the case where $$X$$ and $$Y$$ are real-valued random variables, with (increasing) cdfs $$F_X$$ and $$F_Y$$, the quantile coupling is given by taking $$X'=F_X^{-1}(U)$$ and $$Y'=F_X^{-1}(U)$$. Notice that $$X'$$ and $$Y'$$ are coupled since they were generated with the same randomization $$U$$. We will have more to say about this coupling in a bit. First, I want to convince you that you already know how to do coupling: any procedure you have for generating a random variable from a uniform, or a sequence of uniforms, is a coupling of the random variable you generated and the uniforms! Let’s go back to the beginning.

# Bernoulli’s and Binomials

Let $$n \geq 1$$ be an integer and $$p \in (0,1)$$. Recall that we say that $$X \sim Bin(n,p)$$ if

$\mathbb{P}(X=k) = { n \choose k} p^k (1-p)^{n-k}$ and $$X$$ is integer-valued on $$\{0,1,\ldots, n\}$$. If I told you to compute the mean and variance of $$X$$ from the definition, then what happens? It is quite brutal, even the mean: most of the proof is applying the definition, and manipulating the sigma notation. We have

$\begin{eqnarray*} \mathbb{E} X &=& \sum_{k=0}^n k {n \choose k}p^k(1-p)^{n-k} \\ &=& \sum_{k=1}^n k {n \choose k}p^k(1-p)^{n-k} \\ &=& p\sum_{k=1}^n k {n \choose k}p^{k-1}(1-p)^{n-k} \\ &=& p\sum_{k=1}^n k \frac{n!}{k!(n-k)!}p^{k-1}(1-p)^{n-k} \\ &=& pn\sum_{k=1}^n \frac{(n-1)!}{(k-1)!(n-k)!}p^{k-1}(1-p)^{n-k} \\ &=& pn\sum_{k=0}^{n-1} \frac{(n-1)!}{k!(n-k-1)!}p^{k}(1-p)^{n-k-1} \\ &=& pn\sum_{k=0}^{n-1} \frac{(n-1)!}{k!(n-1 - k)!}p^{k}(1-p)^{n-1-k} \\ &=& pn \sum_{k=0}^{n-1} {n-1 \choose k} p^{k}(1-p)^{n-1-k} \\ &=& pn(1) = pn; \end{eqnarray*}$ the most interesting part of this proof is the in the second to last line, where we recognize that the sum is the sum of the pmf of a binomial random variable with parameter $$(n-1, p)$$.

It is possible you never even saw this proof. A much easier proof is given by the fact that, if $$Y_1, \ldots, Y_n$$ are independent Bernoulli random variables with paramter $$p$$, then $X ' = Y_1 + \cdots + Y_n$ has the same law as $$X$$, and thus the same expectation and variance! The upshot is the linearity of expecation comes to the rescue. $\mathbb{E} X = \mathbb{E} X' = \sum_{i=1} ^n \mathbb{E} Y_i = np.$ Moreover, the independence of the Bernoulli random variables give $var(X) = var(X') = \sum_{i=1} ^n var(Y_i) = n(1-p)p.$

Thus by finding this coupling of a binomial random variable with a (sequence) of Bernoulli random variables, the computation becomes much easier. This coupling did not need to be found of course, since this is how the binomial random variable arises naturally!

Next, we consider another standard application of coupling.

# Stochastic domination

Let $$X$$ and $$Y$$ are real-valued random variables. We say that $$X$$ stochastically dominates $$Y$$ if for all $$z \in \mathbb{R}$$, we have $F_X(z)= \mathbb{P}(X \leq z) \leq F_Y(z) =\mathbb{P}(Y \leq z).$ Let us say that a coupling $$(X', Y')$$ of $$X$$ and $$Y$$ is monotone if $$\mathbb{P}(X' \geq Y') =1$$.

Notice that if $$X$$ stochastically dominates $$Y$$, then the quantile coupling produces a monotone coupling! We have

$X'= F_X^{-1}(U) \geq F_{Y}^{-1}(U) = Y'$ In particular, if $$X$$ stochastically dominates $$Y$$, we have $\mathbb{E} X = \mathbb{E} X' \geq \mathbb{E} Y' = \mathbb{E} Y.$