# Convergence in distribution absolute value of random variables.

How to proof from definition of convergence in distribution that fact:

Let $$X_{1}, X_{2}, . . .$$ and $$X$$ be real random variables with respective distribution functions $$F_{X_1}, F_{X_2}, . . .$$ and $$F_{X}$$. If $$X_{n}$$ converges in distribution to random variable $$X$$ then $$|X_{n}|$$ converge in distribution to $$|X|$$.

We say that sequence of random variable $$X_{1}, X_{2}, . . .$$ converges in distribution to $$X$$ if and only if $$\displaystyle \lim_{n \rightarrow \infty} F_{X_n}(x)=F_{X}(x)$$ for every $$x \in \mathbb{R}$$ at which $$F_X$$ is continuous.

I will be very grateful for the tips.

• What have you tried ? Also your definition of convergence in distribution is wrong as stated. It holds for continuous variables, but you need to account for continuity points of $F_X$ otherwise. Commented Nov 2, 2017 at 19:50
• $P(|X_n| \leq x)=P(X_n \leq x, X_n \geq 0)+P(-X_n \leq x, X_n <0)$ but here I can't use the assumption Commented Nov 2, 2017 at 19:54
• Yes, I have corrected the definition. Commented Nov 2, 2017 at 19:59
• Hint: For $x\ge 0$ we have $\Pr\{|X_n| \le x\} = \Pr\{ X_n \le x \} - \Pr\{ X_n < -x \}$. Commented Nov 2, 2017 at 22:07

Going by the comment, \begin{align*} \displaystyle \lim_{n \rightarrow \infty} F_{\vert X_n \vert } (x) &= \displaystyle \lim_{n \rightarrow \infty} P(\vert X_n \vert \leq x) \\ &= \displaystyle \lim_{n \rightarrow \infty} P(-x \leq X_n \leq x) \\ &= \displaystyle \lim_{n \rightarrow \infty} \{ P(X_n \leq x) - P(X_n < -x) \} \\ &= \displaystyle \lim_{n \rightarrow \infty} P(X_n \leq x) - \displaystyle \lim_{n \rightarrow \infty} P(X_n < -x) \\ &= P(X \leq x) - P(X < -x) = P(\vert X \vert \leq x), \end{align*} where the fourth inequality follows by the algebraic properties of limits.