Answer to Question #204966 in Statistics and Probability for Sajid

Question

Let Y = |Z|, where Z ∼ N (0, 1) be a discrete random variable with the following PMF: (i) Find E(Y ) and V ar(Y ). (ii) Find V ar(Y ). (iii) Find the CDF and PDF of Y . 

Answer 1

The folded normal distribution is a probability distribution related to the normal distribution. Given a normally distributed random variable $X$  with mean $μ$  and variance $σ^2,$ the random variable $Y = |X|$ has a folded normal distribution. 

The mean of the folded distribution is

$$\mu_Y=\sigma\sqrt{\dfrac{2}{\pi}}e^{-\mu^2/(2\sigma^2)}+\mu\big[1-2\Phi(-\dfrac{\mu}{\sigma})\big]$$

Given $Z\sim N(0, 1), Y=|Z|.$ Then $\mu=0, \sigma=1$

$$\mu_Y=1\sqrt{\dfrac{2}{\pi}}e^{-0}+0\big[1-2\Phi(-0)\big]=\sqrt{\dfrac{2}{\pi}}$$

(i)

$$E(Y)=\mu_Y=\sqrt{\dfrac{2}{\pi}}$$

(ii)

The variance is expressed in terms of the mean:

$$\sigma_Y^2=\mu^2+\sigma^2-\mu_Y^2$$

Given $Z\sim N(0, 1), Y=|Z|.$ Then $\mu=0, \sigma=1$

$$\sigma_Y^2=0^2+1^2-(\sqrt{\dfrac{2}{\pi}})^2=1-\dfrac{2}{\pi}=\dfrac{\pi-2}{\pi}$$

(iii)

The probability density function (PDF) is given by

$$f_Y(y; \mu, \sigma^2)=\dfrac{1}{\sqrt{2\pi \sigma^2}}e^{-(y^2-\mu^2)/(2\sigma^2)}$$

$$+\dfrac{1}{\sqrt{2\pi \sigma^2}}e^{-(y^2+\mu^2)/(2\sigma^2)}$$

for $y ≥ 0,$ and everywhere else.

Given $Z\sim N(0, 1), Y=|Z|.$ Then

$$f_Y(y; 0, 1)=\dfrac{1}{\sqrt{2\pi(1)^2}}e^{-(y^2-0^2)/(2(1)^2)}$$

$$+\dfrac{1}{\sqrt{2\pi(1)^2}}e^{-(y^2+0^2)/(2(1)^2)}$$

$$f_Y(y; 0, 1)=\sqrt{\dfrac{2}{\pi}}e^{-y^2/2}$$

for $y ≥ 0,$ and everywhere else.

In our case $Y=|Z|$  follows a half-normal distribution. 

The cumulative distribution function (CDF) is given by

$$F_Y(y;0,1)=\displaystyle\int_{0}^y\dfrac{1}{1}\sqrt{\dfrac{2}{\pi}}e^{-x^2/(2\cdot1)}dx$$

$$F_Y(y;0,1)=\dfrac{2}{\sqrt{\pi}}\displaystyle\int_{0}^{y/\sqrt{2}}e^{-z^2}dz=\text{erf}(\dfrac{y}{\sqrt{2}})$$

where erf is the error function.