Clue for solving problem about Coupling of Random Variables.

Just have been trying to approach this problem from Resnick's book on probability but have got no clue so far.

The problem is like this:

We are giving two random variables X, Y on the same space $(\Omega, \mathcal{B})$, and we are asked to show: $\sup_{A \in \mathcal{B} } | P[X\in A] - P[Y\in A] | \leq P[X \neq Y]$.

What I have thought:

1. My intuition is that maybe X and Y have the same distribution, although I don't see how the distribution of the two RVs plays a role here.

2. For the RHS I can say that if we set $P[X \neq Y] = \epsilon$, and we can check that the LHS $< \epsilon$ we may be able to get this done.

3. I realized that, if X, Y are random variables, then both satisfy the mapping:

$X:(\Omega, \mathcal{B}) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$

Then, $A \in \mathcal{B}(\mathbb{R})$, so I'm confused why the problem states that $A \in \mathcal{B}$.

Any solid hint please?

• You want to fix $A \in \mathcal{B}$ and then show $P[X \in A] \leq P[Y\in A] + P[X\neq Y]$, then a similar inequality. Can you show that? – Michael Nov 9 '17 at 7:52
• PS: No need to worry about distributions or $\epsilon$-thingys here. The random variables $X, Y$ can have different distributions and can be dependent. – Michael Nov 9 '17 at 7:52
• Thanks for you reply. I kinda got the intuition of your hint, but stuck on how to implement it. Moreover, I'm still confused on why $A\in \mathcal{B}$? since I'm thinking of: $X^{-1}(A)= \{ \omega \in \Omega: X(\omega) \in A\}$ and $A \in \mathcal{B}(\mathbb{R})$. I'm confused about measures in this space. Any help will be helpful. Thank you all. – pkenneth81 Nov 9 '17 at 8:12
• You are right. I stated you should fix "$A \in \mathcal{B}$" only because that was the notation in the sup inequality in your question. But indeed the notation of the inequality of your question is not correct: $A$ is a subset of the reals, not necessarily a subset of $\Omega$. So indeed that should be changed to "fix $A \in \mathcal{B}(\mathbb{R})$" (using that to denote the collection of measurable subsets of $\mathbb{R}$). Overall, you just want to show $|P[X \in A] - P[Y \in A]| \leq P[X \neq Y]$ for all measurable subsets $A$ of the reals. – Michael Nov 9 '17 at 8:16
• I'm not sure I follow. The sample space is not necessarily discrete, so summing is not appropriate here. No need to use $\epsilon$ either. The approach of the first comment I gave is likely better. Also note that $|x-y|\leq c$ if and only if $-c \leq x-y \leq c$, which is a useful fact for getting rid of pesky absolute value bars. And, recall the union bound says $P[E \cup F] \leq P[E] + P[F]$. – Michael Nov 9 '17 at 17:56

It suffices to prove that for any $A\in \mathcal B$, $$| P[X\in A] - P[Y\in A] | \leq P[X \neq Y]$$
By symmetry, it suffices to prove that $$P[X\in A] - P[Y\in A] \leq P[X \neq Y]$$
which rewrites as $P[X\in A] \leq P[Y\in A] + P[X \neq Y]$.
The last inequality follows from \begin{align} P[X\in A] &= P[X\in A \;\cap\; X \neq Y] + P[X\in A \;\cap\; X = Y]\\ &= P[X\in A \;\cap\; X \neq Y] + P[Y\in A \;\cap\; X = Y]\\ &\leq P[ X \neq Y] + P[Y\in A ] \end{align}