If we chop a standard normal distribution in half and use only the positive side (scaled up by a factor of $2$ to maintain a proper density), then we get the so-called ‘half normal’ density:


If $Z ∼ N(0, 1)$ then $|Z|$ has a half normal density.

How if $X$ is half normal and $S = ±1$ with probability half each, independently of $X$, then $Z = SX ∼ N(0, 1).$ $S ∼ U{(−1, +1)}$ to indicate that the distribution of $S$ is uniformly distributed over the finite set $\{−1, +1\}$?


$S$ is either $+1$ or $-1$ with equal probabilities. So $SX$ gives you back the negative side of the original normal you had. And you would scale it down (in your words) by a factor of two, regaining the original standard normal random variable.

  • $\begingroup$ Can't $S$ accept value between $-1$ and $1$ ? When we say $X\sim U(0,1)$, we mean $X$ can assume any value between $0$ and $1$ $\endgroup$ – ABC Sep 24 '13 at 1:11
  • $\begingroup$ From the problem description, $S$ is discrete and cannot take on any values except $-1$ and $+1.$ I agree the notation is confusing. Whoever wrote it should have been more careful. $\endgroup$ – soakley Sep 24 '13 at 3:10

I will derive the mean and variance, and leave the distribution to you. Define $I\{S=1\}\equiv I_1$ for simplicity, to be the indicator function taking the value $1$ when $S=1$ and the value $0$ when $S=-1$. Note also that when $S=1 \Rightarrow Z = X$ and when $S=-1 \Rightarrow Z = -X$. Combining we can re-write the functional form of $Z$ as

$$Z= XI_1 - X(1-I_1) = X(2I_1-1)$$

Then, given also independence between $X$ and $S$ $$E(Z) = E\left(X(2I_1-1)\right) = E(X)E(2I_1-1) = \sqrt{\frac{2}{\pi}}\cdot(2E(I_1)-1)$$

and by the properties of indicator function, $E(I_1) = P(S=1) = 1/2$ we get

$$E(Z) = \sqrt{\frac{2}{\pi}}\cdot(2\frac 12-1) = 0$$

For the variance we have, since the mean is zero,

$$ \text {Var}(Z) = E(Z^2) = E\left(X^2(2I_1-1)^2\right) = E\left(X^2\right)E\left(4I_1^2 - 4I_1+1\right)$$

Now $$\text {Var}(X) = E(X^2) - \left[E(X)\right]^2 \Rightarrow E(X^2) = \text {Var}(X) + \left[E(X)\right]^2 = 1-\frac {2}{\pi} + \frac {2}{\pi} = 1$$

So $$ \text {Var}(Z) = 1\cdot E\left(4I_1^2 - 4I_1+1\right)= 4E\left(I_1^2\right) - 4P(S=1)+1 = 4E\left(I_1^2\right) - 1$$

By the properties of the product of the indicator function, $I_1^2= I_1\cdot I_1 = \min \{I_1,I_1\} = I_1$


$$\text {Var}(Z) = 4E\left(I_1\right) - 1 = 4P(S=1) - 1 = 2 - 1 = 1$$


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