Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given the probability density function of $X$ (folded standard normal distributed) is:

$$f(x) = \frac{2}{\sqrt{2 \pi}} \exp\left(-\frac{x^2}{2}\right),\quad x \geqslant 0 $$

How can one show that $Z = |X|$ with probability $1/2$ and $Z=-|X|$ with probability $1/2$ has a standard normal distribution?

Please give me a hint only. Thanks!

(I tried to show that E(Z) = 0 and Var(Z) = 1, but it was not sufficient.)

share|cite|improve this question
ANYTHING works. What did you try? – Did Feb 28 '13 at 22:23
Could you please give me a hint? Thanks. – Guess Gucci Feb 28 '13 at 22:41
Which word do you fail to understand in "What did you try"? – Did Feb 28 '13 at 22:51
For example what happens when you take the absolute value of a random variable? – Seyhmus Güngören Feb 28 '13 at 23:19
I tried to show E(Z) = 0 and Var(Z) = 1 but realized that they together do not mean that Z has a standard normal distribution. Now I am looking for another hint to start with the proof. – Guess Gucci Mar 1 '13 at 3:30
up vote 0 down vote accepted


  • Since the density of $X$ has zero mass on the negative real line, we have, with probability 1, that $X \ge 0$. So you can discard the absolute value sign around $X$. That is, $Z = X$ or $-X$ with equal probability. More precisely, there is a symmetric Bernoulli variable $$ B \sim \begin{cases} 0 & \text{with probability}\; 1/2 \\ 1 & \text{with probability}\; 1/2 \end{cases} $$ which is independent of $X$ and $Z = B X + (1-B)(-X)$. (This a more detailed description of for how $Z$ is constructed from $X$.)

  • You can now use the law of total probability. For any (measurable!) subset of the real-line (say $A=(-\infty,x]$), we have $$ P (Z \in A) = P(X \in A|B = 0) P(B = 0) + P(-X \in A| B = 1) P(B = 1) $$ You should be able to figure out the rest.

share|cite|improve this answer
Thanks for your help! – Guess Gucci Mar 4 '13 at 21:51
You are welcome. – passerby51 Mar 5 '13 at 16:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.