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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?

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    $\begingroup$ 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? $\endgroup$ – Michael Nov 9 '17 at 7:52
  • $\begingroup$ PS: No need to worry about distributions or $\epsilon$-thingys here. The random variables $X, Y$ can have different distributions and can be dependent. $\endgroup$ – Michael Nov 9 '17 at 7:52
  • $\begingroup$ 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. $\endgroup$ – pkenneth81 Nov 9 '17 at 8:12
  • $\begingroup$ 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. $\endgroup$ – Michael Nov 9 '17 at 8:16
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    $\begingroup$ 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 $P[E \cup F] \leq P[E] + P[F]$. $\endgroup$ – Michael Nov 9 '17 at 17:56
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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 $$\begin{align} P[X\in A] &= P[X\in A \;\cap\; X \neq Y] + P[X\in A \;\cap\; X = Y]\\ &= P[X\in A \;\cap\; X \neq Y] + P[Y\in A \;\cap\; X = Y]\\ &\leq P[ X \neq Y] + P[Y\in A ] \end{align}$$

and we're done.

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  • $\begingroup$ Yeah. I found out that there something called the coupling Lemma, well documented on books of Advanced Probability theory. This answer seems to be the standard proof of such lemma. $\endgroup$ – pkenneth81 Nov 15 '17 at 6:37

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