Proof that $(X_n \Rightarrow\mathcal{N}(0,1)) \implies (-X_n \Rightarrow \mathcal{N}(0,1))$, where $"\Rightarrow"$ convergence in distribution

Question

Assume that a sequence of random variables $(X_n)$ converges in distribution to $\mathcal{N}(0,1)$, i.e. that for every $x \in \mathbb {R}$ we have $$\lim_{n \to \infty}\mathbb{P}(X_n \leq x) = \Phi(x).$$

The question is does that imply that the sequence $(-X_n)$ also converges in distribution to $\mathcal{N}(0,1)$. I'm not sure how to prove that it does (?). This is my attempt so far.

For any $x \in \mathbb{R}$, \begin{split} & \lim_{n \to \infty}\mathbb{P}(X_n \leq- x)=\Phi(-x) \\ \implies & \lim_{n \to \infty}\mathbb{P}(X_n \leq -x)=1-\Phi(x) \\ \implies & \lim_{n \to \infty}\left[\mathbb{P}(X_n \leq -x)-1\right]=-\Phi(x) \\ \implies & \lim_{n \to \infty}\left[1-\mathbb{P}(X_n \leq -x)\right]=\Phi(x) \\ \implies & \lim_{n \to \infty}\mathbb{P}(X_n > -x)=\Phi(x) \\ \implies & \lim_{n \to \infty}\mathbb{P}(-X_n < x)=\Phi(x), \\ \end{split} so we yet have to prove that for every $x \in \mathbb{R}$ $$ \lim_{n \to \infty}\mathbb{P}(-X_n < x)= \lim_{n \to \infty}\mathbb{P}(-X_n \leq x).$$ Let's assume opposite, i.e. that there's an $x \in \mathbb{R}$ such that $$\lim_{n \to \infty}\left[\mathbb{P}(-X_n \leq x)-\mathbb{P}(-X_n < x)\right] \neq 0.$$ Now, we have $$\lim_{n \to \infty} \mathbb{P}(-X_n=x)\neq 0,$$ i.e. $$\lim_{n \to \infty} \mathbb{P}(X_n=-x)\neq 0,$$ which means that the limiting cumulative distribution has a discontinuity at the point $-x$. But I'm not sure what to do with that fact - how to get a contradiction with my starting assumption. Thanks for any help, I appreciate it.

Answer 1

You have $\lim_{n \to \infty}\mathbb{P}(-X_n < x)=\Phi(x)$ and want to prove that $$\lim_{n \to \infty}\mathbb{P}(-X_n < x)= \lim_{n \to \infty}\mathbb{P}(-X_n \leq x).$$ You cannon get the contradiction without using continuity of limiting function.

Choose small $\delta>0$ and bound the r.h.s.: $$ \mathbb{P}(-X_n < x)\leq\mathbb{P}(-X_n \leq x) \leq \mathbb{P}(-X_n < x+\delta). $$ Take limits of all sides and get $$ \Phi(x)=\lim_{n\to\infty}\mathbb{P}(-X_n < x)\leq\liminf\mathbb{P}(-X_n \leq x)\leq\limsup\mathbb{P}(-X_n \leq x) \leq $$ $$ \leq\lim_{n\to\infty}\mathbb{P}(-X_n < x+\delta)=\Phi(x+\delta).$$ Since the above inequalities hold for any $\delta>0$, one can consider $\delta\to0$ and use the continuity $\Phi(x+\delta)\to\Phi(x)$ as $\delta\to0$.

Answer 2

It is sufficient to show that: $Z \sim N(0,1) \Rightarrow -Z \sim N(0,1)$.

\begin{align} \Phi(z) &= P(Z < z) \\ &= P(-Z > -z) \\ &= 1 - P(-Z < -z) \\ &= 1- F_{-Z}(-z)\\ &= 1 - \Phi(-z) \end{align}

This implies that $Z$ and $-Z$ have the same CDF. So if $F_n \Rightarrow F_Z$, then clearly $F_n \Rightarrow F_{-Z}$.