Expectation of absolute random variables with mean 1 and standard deviation 1

Question

For a random variable $\gamma \sim \mathcal{N}(\mu,\sigma)$ , were is $ \mathcal{N}$ is the normal distribution.

What is the way to calculate the following:

$ \mathbb{E}[|\gamma|] = ? $

And speciely for $ \mu =\sigma =1 $

Answer 1

The expected value of a normal distribution is the mean, in your case $\mu$, because the normal is symmetrical from let to right.

For any probability density function $f(x)$ this can be calculated by integrating $x\times f(x)$ across its domain.

The normal is defined as having the pdf: $$f(x)=\frac{1}{\sqrt{2\pi}}\exp{(\frac{-x^2}{2})}$$

The mean of that is always zero and its standard deviation is $1$. To adjust for $\mu= 1$ it is easiest to say:

$$f(x)=\frac{1}{\sqrt{2\pi}}\exp{(\frac{-(x-1)^2}{2})}$$

To calculate the expected value of $\gamma$ then:

$$E(\gamma)=\int_{-\infty}^{\infty}{x\times\frac{1}{\sqrt{2\pi}}\exp{(\frac{-(x-1)^2}{2})}} dx = 1$$

To calculate the expected value of $\lvert\gamma\rvert$ you need to subtract the area to the left of zero and then add it back in negated, effectively reflecting that part of the p.d.f in the y-axis and adding it to the part to the right. This is equivalent to integrating from zero to infinity the sum of the normal with mean $\mu$ and the normal with mean $-\mu$:

$$E(\lvert\gamma\rvert)=\int_{0}^{\infty}{x\times\frac{1}{\sqrt{2\pi}}\exp{(\frac{-(x-1)^2}{2})}} dx + \int_{0}^{\infty}{x\times\frac{1}{\sqrt{2\pi}}\exp{(\frac{-(x+1)^2}{2})}} dx$$ $$=1.08332+0.08332=1.166$$