Let $X,Y$ be two random variables taking real values. Suppose that $P(Y\leq t) \leq P(X\leq t)$ for any $t\in{\mathbb R}$. Is there always a random variable $Y^{*}$ with the same distribution as $Y$, such that $P(X \leq Y^{*})=1$ ?


1 Answer 1


The question you asked is about Coupling of two independent random variable. Coupling of $X,Y$ are random variables $X',Y'$ such that they have the same marginal as $X$ and $Y$ but different joint distribution.

To see an example of coupling, let's start with $X$ and $Y$ two independent Bernoulli random variable such that: $$ p=P(Y=0)\leq q=P(X=0) $$ Suppose that $X$ and $Y$ are related to two distinct events and we want to compare some probabilistic features of these events. But to do that, it is interesting to see if we can somehow couple the behavior of these two events, which is whether we can find $X^*$ and $Y^*$ such that they are dependent but still have the same marginal. If we do that it seems that we have found a hidden random variable that connects these two events.

Now consider a uniform random variable $U$ over $[0,1]$ and define: $$ X^*=\mathbb{1}(U>q), Y^*=\mathbb{1}(U>p) $$ These two random variables have the same marginal as the original $X,Y$, as it can be seen here: $$ P(X^*=0)=P(U<q)=q ,P(Y^*=0)=P(U<p)=p $$ But they are not independent and they are related by $U$. Moreover we have $X^* \leq Y^*$.

The previous example is not the same as your question. In your example, $X$ should remain untouched and in previous example, $X^*$ is not equal to $X$.

The solution is to construct $U$ based on $X$. Define $U$ based on $X$ as follows. If $X=0$, then $U$ is drawn uniformly from $[0,q]$ and if $X=1$, then $U$ is drawn uniformly from $[1-q,1]$. Therefore $U$ is dependent on $X$ however it can be seen that $U$ is uniformly distributed over $[0,1]$. Now we can define $Y^*=\mathbb{1}(U>p)$. $Y^*$ has the same distribution as $Y$ and for each $U=u$, it can be seen that $Y^*\geq X$.

The question is whether we can generalize this approach for random variables on $\mathbb R$. The answer is yes.

Let's start with finding a coupling of $X$ and $Y$. First we define the generalized inverse function of $F_X(t)=P(X\leq t)$ and $F_Y(t)=P(Y\leq t)$ as $G_X(t)$ and $G_Y(t)$. Of course $F_Y(t)$ and $F_X(t)$ are not in general invertible and hence we have to define the inverse as follows: $$ G_X(t)=\inf\{x\in\mathbb R; F_X(x)\geq t\}\\ G_Y(t)=\inf\{y\in\mathbb R; F_Y(y)\geq t\} $$ Note that $F_Y(t)\leq F_X(t)$ and therefore $G_Y(t)\geq G_X(t)$.

Suppose that $X^{-}=G_X(F_X(X))$. Take $Y^*$ and $U$ random variables defined as follows: $$ X^{-}=G_X(U),\\ Y^*=G_Y(U) $$

First, it can be seen that $U$ has uniform distribution on $[0,1]$. Then it is easy to see that $Y^*$ have the same marginal as $Y$ and of course $X\leq Y^*$.

  • $\begingroup$ +1. The last part (X and Y*) assumes that the probability space X and Y are defined on is large enough so that one can define U on it as well. $\endgroup$
    – Did
    Sep 25, 2013 at 5:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .