Answer in Statistics and Probability for Sajid #204966
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 .
The folded normal distribution is a probability distribution related to the normal distribution. Given a normally distributed random variable “X” with mean “u03bc” and variance “u03c3^2,” the random variable “Y = |X|” has a folded normal distribution.
The mean of the folded distribution is
“mu_Y=sigmasqrt{dfrac{2}{pi}}e^{-mu^2/(2sigma^2)}+mubig[1-2Phi(-dfrac{mu}{sigma})big]”
Given “Zsim N(0, 1), Y=|Z|.” Then “mu=0, sigma=1”
“mu_Y=1sqrt{dfrac{2}{pi}}e^{-0}+0big[1-2Phi(-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 “Zsim 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{2pi sigma^2}}e^{-(y^2-mu^2)/(2sigma^2)}”
“+dfrac{1}{sqrt{2pi sigma^2}}e^{-(y^2+mu^2)/(2sigma^2)}”
for “y u2265 0,” and everywhere else.
Given “Zsim N(0, 1), Y=|Z|.” Then
“f_Y(y; 0, 1)=dfrac{1}{sqrt{2pi(1)^2}}e^{-(y^2-0^2)/(2(1)^2)}”
“+dfrac{1}{sqrt{2pi(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 u2265 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)=displaystyleint_{0}^ydfrac{1}{1}sqrt{dfrac{2}{pi}}e^{-x^2/(2cdot1)}dx”
“F_Y(y;0,1)=dfrac{2}{sqrt{pi}}displaystyleint_{0}^{y/sqrt{2}}e^{-z^2}dz=text{erf}(dfrac{y}{sqrt{2}})”
where erf is the error function.
