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

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.

• Maybe use MGFs? Commented Jun 10, 2016 at 14:49
• If you were to draw a picture of the bivariate distribution of $(X_1, X_2)$ it would likely suggest a solution.
– whuber
Commented Jun 10, 2016 at 16:03
• Possible duplicate of Normal Distribution Existence Non-affine Invariant Transformation? Commented Jun 10, 2016 at 17:40

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

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

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.
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