How to show that $X_2$ is also a standard normal variable?

Question

Suppose $X_1$ is a standard normal variable. Define, $$X_2=\begin{cases}-X_1, & \text{if } |X_1|<1,\\X_1, & \text{otherwise}\end{cases}$$

Show that $X_2$ is also a standard normal random variable.

My approach:

Let,$$F(X_2)=P[X_2\leq x_2]\\=P[X_2\leq x_2\mid|X_1|\leq 1].P[|X_1|< 1]+P[X_2\leq x_2\mid|X_1|\geq 1].P[|X_1|\geq 1]\\=P[-X_1\leq x_2].P[|X_1|<1]+P[X_1<x_2].P[|X_1|>1]$$

But I cannot compute the probabilities.May be,my approach is not right.

If you were to draw a picture of the bivariate distribution of $(X_1, X_2)$ it would likely suggest a solution. — whuber, Jun 10 '16 at 16:03
Possible duplicate of [Normal Distribution Existence Non-affine Invariant Transformation?](http://stats.stackexchange.com/questions/200380/normal-distribution-existence-non-affine-invariant-transformation) — Xi'an, Jun 10 '16 at 17:40

Greenparker · Accepted Answer · 2016-06-10T16:14:14.070

2

You have the correct idea so far, but are missing a key trick. Recall that if $X_1$ is a standard normal distribution, then $X_1$ is symmetric around 0. This means that

$$P(X_1 < x_2) = P(X_1 > -x_2) = P(-X_1 < x_2) = P(-X_1 \leq x_2). $$

Using this, the result will follow directly from your last step.

edited Jun 10 '16 at 16:14

answered Jun 10 '16 at 16:01

Greenparker

14,131
3
36
80

score 2 · Answer 2 · answered Jun 10 '16 at 16:12

I would divide it into 3 cases:

$X_2 \le 1$

$-1 \lt X_2 \le 1$

$1 < X_2$

The first is trivial, since $X_1$ is standard normal. For the second,

$ \begin{align} F(x_2) &= F(-1) + P(-1 \lt X_2 \le x_2) \\ &= F(-1) + P(-x_2 \le X_1 \lt 1) \\ &= F(-1) + P(-1 \lt X_1 \le x_2) ~by~symmetry \end{align} $

which is standard normal since $X_1$ is.

The third case is similar.

How to show that $X_2$ is also a standard normal variable?

2 Answers2