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Ok. This is bit confusing. Let $g$ be a Gaussian random variable (normalized, i.e. with mean 0 and standard deviation 1). Then, in the expression $$\|g\|_{L_1}=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}|x|\exp(\frac{-x^2}{2})dx =\sqrt{\frac{2}{\pi}},$$ shouldn't the term $|x|$ not appear in the integral?

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The term $|x|$ will not be there in the integral if you were computing the $L_1$ norm of the density function. Here you are computing the $L_1$ norm of the random variable. – Dilip Sarwate Nov 18 '12 at 23:00
Oh. I see. Silly me. – Nirakar Neo Nov 19 '12 at 2:10
up vote 4 down vote accepted

If $X$ is a random variable with density $f$, and $\phi$ is a measurable function, then $E[\phi(X)]=\int_{\Bbb R}\phi(t)f(t)dt$. As the $L^1$ norm of a random variable $X$ is $E[|X|]$, we have, when $X$ is normally distributed, the announced result.

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It should! The $L_1$-norm of a random variable $X$ with density $f(x)$ equals $$\int_{\Omega} |X(w)| d\mu(\omega) = \int_{-\infty}^{+\infty} |x| f(x) dx,$$ where $(\Omega, \mu)$ is the probability space $X$ defined on.

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