Concept:
A Gaussian random variable X with mean μ and variance σ2 can be denoted as X N (μ, σ2) and its probability density function (pdf) is given by:
\({f_x}\left( x \right) = \frac{1}{{\sqrt {2\pi {\sigma ^2}} }}{e^{ - {{\left( {x - \mu } \right)}^2}/2{\sigma ^2}}}\)
The expectation of any function of x is calculated as:
\(E\left[ {g\left( x \right)} \right] = \mathop \smallint \limits_{ - \infty }^\infty g\left( x \right) \cdot {f_x}\left( x \right)dx\)
If f(-x) = f(x) then f(x) is said to be even function of x.
And, an even function satisfies:
\(\mathop \smallint \limits_{ - \infty }^\infty f\left( x \right)dx = 2\mathop \smallint \limits_0^\infty f\left( x \right)d\tau \)
Calculation:
Given:
X ∼ N (0, 1)
\({f_X}\left( x \right) = \frac{1}{{\sqrt {2\pi } }}{e^{ - {x^2}/2}}\)
\(E\left[ {\left| X \right|} \right] = \mathop \smallint \limits_{ - \infty }^\infty \left| x \right|\;{f_X}\left( x \right)dx\)
\( = \mathop \smallint \limits_{ - \infty }^\infty \left| x \right| \cdot \frac{1}{{\sqrt {2\pi } }}{e^{ - {x^2}/2}}dx\;\)
\(= \frac{1}{{\sqrt {2\pi } }}\mathop \smallint \limits_{ - \infty }^\infty \left| x \right| \cdot {e^{ - {x^2}/2}}dx\)
Since:
\(\left| x \right| \cdot {e^{ - {x^2}/2}}\) is an even function of ‘x’, we can write:
\(E\left[ {\left| x \right|} \right] = \frac{1}{{\sqrt {2\pi } }} \times 2\mathop \smallint \limits_0^\infty x \cdot {e^{ - {x^2}/2}}dx\)
Using substitution of variables:
\(Let\;\frac{{{x^2}}}{2} = y\)
\(x\;dx = dy\)
\(E\left[ {\left| x \right|} \right] = \frac{1}{{\sqrt {2\pi } }} \times 2\mathop \smallint \limits_0^\infty {e^{ - y}}dy\)
\( = \sqrt {\frac{2}{\pi }} \;\{ - {e^{ - y}}\left. \right|_0^\infty = \sqrt {\frac{2}{\pi }} \;\left\{ {1 - 0} \right\}\)
\(E\left[ {\left| x \right|} \right] = \sqrt {\frac{2}{\pi }} \simeq 0.798\)
E[|x|] ≃ 0.8
