# Help understanding a probability inequality

I'm working throught Wasserman's "All of Statistics" book. When proving convergence of random variables/distributions in chapter 5, he lists the following inequality:

$$F_n(x) = \mathbb{P}(X_n\le x)=\mathbb{P}(X_n\le x, X\le x+\epsilon) + \mathbb{P}(X_n\le x, X\gt x+\epsilon)\\ \le \mathbb{P}(X \le x + \epsilon) + \mathbb{P}(\vert X_n - X \vert \gt \epsilon)\\ = F(x + \epsilon) + \mathbb{P}(\vert X_n - X \vert \gt \epsilon)$$

Where $$\mathbb{P}$$ is the probability that the random and $$F$$ is the cdf of the random variable $$X$$

Here, $$X$$ and $$X_n$$ are different random variables (it's part of a proof showing $$X_n$$ converges to $$X$$).

What I don't understand is how he arrives at the inequality on the second line. Could someone help me understand this. I feel like my lack of some essential fundamental knowledge of statistics is being exposed by not being able to figure this out.

Thanks!

• If two numbers are separated by at least $\epsilon,$ then the absolute value of their difference is at least $\epsilon.$
– whuber
Oct 25 '18 at 23:34

Note that $$X_n \le x$$ and $$X \le x+\epsilon$$ implies $$X \le x+ \epsilon$$, that is

$$\{\omega \in \Omega| X_n (\omega) \le x, X(\omega) \le x+ \epsilon \} \subseteq \{\omega \in \Omega| X(\omega) \le x+ \epsilon \}$$

Hence we have $$\mathbb{P}(X_n \le x, X \le x+\epsilon)\le \mathbb{P}(X \le x+\epsilon).$$

Also if $$X_n \le x, X > x+\epsilon$$ then we have

$$X_n \le x \le x+\epsilon < X$$

and $$|X_n - X|=X-X_n > x+\epsilon - x=\epsilon.$$

That is $$\{\omega \in \Omega| X_n (\omega) \le x, X(\omega) > x+ \epsilon \} \subseteq \{\omega \in \Omega| X_n(\omega)-X(\omega) > \epsilon \}$$

Hence $$\mathbb{P}(X_n \le x, X > x+\epsilon) \le \mathbb{P}(X_n -X > \epsilon).$$

Summing up the two inequalities would give us the result.