On a property of random variables and Borel sets

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Proposition: Let $A$ be a Borel set of $\mathbb{R}$. We wish to show that for any two random variables, $X,Y:\Omega\to \mathbb{R}$, then we have the following inequality: $$ |\operatorname{P}(X\in A)-\operatorname{P}(Y\in A)|\leq \operatorname{P}(X\neq Y)$$

Here's where I'm at right now:

I tried showing it for an open interval $(-\varepsilon,\varepsilon)$ but I could not find the property to use. I was trying to start with $\operatorname{P}(X\neq Y)$ and using the fact that if $X\neq Y$, we have $|X-Y|>0$. If you have any hints, I would be grateful.


Edit: Thanks for the answers using expected value and caracteristic functions. It's a pretty elegant way to do things. Unfortunatly, I would prefer a solution without using theses methods since the book I'm reading did not introduce the expected value of a random variable yet.

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Hint: The left side is \begin{align*} |P(X \in A) - P(Y \in A)| &= \left|E \left[1_{X \in A} - 1_{Y \in A} \right] \right| \\ & \leq E \left|1_{X \in A} - 1_{Y \in A}\right| \\ & = E \left[1_{X \in A, Y \not \in A} + 1_{X \not \in A, Y \in A} \right] \end{align*}

EDIT, in response to edited original post: We can still play this game without resorting to expectations. Note that $$\{X \in A\} = \{X \in A, Y \not \in A\} \cup \{X \in A, Y \in A\}$$ and $$\{Y \in A\} = \{X \not \in A, Y \in A\} \cup \{X \in A, Y \in A\}$$ and that these are disjoint unions. Hence, \begin{align*} \left| P(X \in A) - P(Y \in A) \right| &= \left|P(X \in A, Y \not \in A) - P(X \not \in A, Y \in A) \right| \\ &\leq P(X \in A, Y \not \in A) + P(X \not \in A, Y \in A). \end{align*}

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\begin{align} &|\mathsf{E}[1\{X\in A\}-1\{Y\in A\}]|=|\mathsf{E}[(1\{X\in A\}-1\{Y\in A\})1\{X\ne Y\}]| \\[8pt] &\qquad\le\mathsf{E}[|1\{X\in A\}-1\{Y\in A\}|1\{X\ne Y\}]\le \mathsf{E}1\{X\ne Y\}. \end{align}