# Convergence In Distribution vs Probability Explanation

Having trouble following lecture notes:

Sequence of random variables, $$X_1, ..., X_n,...$$ with corresponding c.d.f's $$F_1,...,F_n,...$$ converges in distribution to a random variable X with c.d.f $$F$$ if: $$\lim _{n \rightarrow \infty} F_{n}(x)=F(x)$$ for all $$x$$ where $$F$$ is continuous.

To see that convergence in distribution in general does not imply convergence in probability let $$X_1, X_2,...$$ be a sequence of i.i.d. random variables all with cdf $$F=\Phi$$. Also let $$X$$ be a random variable that is independent of these and has the same standard normal distribution.
$$X_{n} \stackrel{d}{\longrightarrow} X$$ but for all $$\epsilon \gt 0$$, $$\operatorname{Pr}\left(\left|X_{n}-X\right|<\varepsilon\right)=2 \Phi\left(\frac{\varepsilon}{\sqrt{2}}\right)-1$$ and this does not converge to 1.

I'm very rusty in probability theory and rather confused by this. Where did the expression, $$2 \Phi\left(\frac{\varepsilon}{\sqrt{2}}\right)-1$$, come from? And why is he testing for convergence at 1? From previous notes, the convergence is explained to be 0: $$\lim _{n \rightarrow \infty} \operatorname{Pr}\left(\left|X_{n}-X\right|>\varepsilon\right)=0$$

1. Let $$Z_n \equiv X_n - X \sim N(0,2)$$, hence \begin{align} P(|X_n - X| < \epsilon)& = P(|Z_n|< \epsilon)= P( - \epsilon < Z_n < \epsilon)\\ & = P(Z_n < \epsilon) - ( 1 - P(Z_n < \epsilon) )\\ &=2\Phi(\epsilon/\sqrt{2})-1 \end{align} .
2. And why is he testing for convergence at 1? $$P(|X_n - X| < \epsilon)=1 \iff P(|X_n - X| \ge \epsilon) = 0$$
$$1 = P(|X_n - X| < \epsilon) + P(|X_n - X| = \epsilon) + P(|X_n - X| > \epsilon)$$ you can drop the strict equality since it is continues r.v