Let us say that $P$ is a normal random variable having expected value $\mu$ and variance $\sigma^2$. I am asked to compute the expected value of the variable $Y = |P|$.
Could someone explain?
Let us say that $P$ is a normal random variable having expected value $\mu$ and variance $\sigma^2$. I am asked to compute the expected value of the variable $Y = |P|$.
Could someone explain?
Hint:
By definition,
$E(|P|)=\int_{-\infty}^\infty |x| P(x)dx=\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^\infty |x| e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx$
You can divide the integral into $\int_{-\infty}^\infty=\int_{-\infty}^0+\int_0^\infty$ Now you can calculate these integrals (hint: what is $|x|$ for $x<0$? and for $x>0$?)
Y has folded normal distribution, in the following link you can find its expected value using the expected value and variance of P. http://en.wikipedia.org/wiki/Folded_normal_distribution