how they went from $E(1,|X−Y|>1)=0$ to $P(|X−Y|>1)=0$

Question

Consider the following mathematical statements involving random variables $X$ and $Y$:

The notation $|X-Y|\wedge 1$ denotes the minimum of $|X-Y|$ and 1.

$ \begin{align*} E(|X-Y|\wedge 1)=0 &\implies E(|X-Y|, |X-Y|\leq 1) + E(1, |X-Y| > 1) = 0 \\ &\implies E(|X-Y|, |X-Y|\leq 1) = 0 \text{ and } P(|X-Y| > 1) = 0 \\ &\implies E(|X-Y|) = E(|X-Y|, |X-Y|\leq 1) = 0 \\ &\implies X=Y \text{ almost surely.} \end{align*} $

I have two specific questions regarding the above statements:

1. how they went from $E(1,|X−Y|>1)=0$ to $P(|X−Y|>1)=0$

2. Similarly, why is it true that $E(|X-Y|) = E(|X-Y|, |X-Y|\leq 1)$?

I would appreciate any explanations or insights into these equalities. Thank you!

Answer 1

The steps in your post seem to be a little out of order.

We have the following two equalities. \begin{align} E[|X-Y|] &= E[|X-Y|, |X-Y| \le 1] + E[|X-Y|, |X-Y| > 1] \tag{a}\\ E[|X-Y| \land 1] &= E[|X-Y|, |X-Y| \le 1] + E[1, |X-Y| > 1]\tag{b} \end{align}

If $E[|X-Y| \land 1] = 0$, then both terms on the right-hand side of (b) must be zero. Note that $E[1, |X-Y|>1]$ can be rewritten as $P(|X-Y| > 1)$.

Since $P(|X-Y|>1)=0$, we have $E[|X-Y|, |X-Y| > 1] = 0$.

We have shown that the two terms on the right-hand side of (a) are both zero, so we have $E[|X-Y|]=0$.