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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.)

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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

1 Answer 1

up vote 0 down vote accepted

Hints

  • 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.

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

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