Expected Absolute Value of a Normal Random Variable
Let $X \sim N(0, \sigma^2)$. Compute $E[|X|]$ in terms of $\sigma$, then evaluate at $\sigma^2 = 2\pi$.
Hints
- Write $X = \sigma Z$ where $Z \sim N(0,1)$. This lets you factor out $\sigma$ and work with the standard normal instead.
- Apply LOTUS: $E[|Z|] = \int_{-\infty}^{\infty} |x| \phi(x)\, dx$. The integrand is even, so fold the integral onto $[0, \infty)$ and multiply by $2$.
- After folding, evaluate $\frac{2}{\sqrt{2\pi}} \int_0^{\infty} x e^{-x^2/2}\, dx$ using the substitution $u = x^2/2$, so $du = x\, dx$.
Worked Solution
How to Think About It: The absolute value kills the sign of $X$ and folds the distribution onto the positive real line. Since $X$ is symmetric around $0$, $E[|X|] = 2 \cdot E[X \cdot \mathbf{1}_{X > 0}]$. The quickest path is to factor out $\sigma$ using the scaling property of Gaussians, reduce to the standard normal case, then evaluate the resulting integral via a $u$-substitution.
Quick Estimate: A rough bound: $E[|X|] \leq \sqrt{E[X^2]} = \sigma$ by Jensen's inequality (since $|\cdot|$ is concave on the positive reals... actually convex, so Jensen goes the other way -- but $\sqrt{E[X^2]} = \sigma$ gives the right order). For $\sigma^2 = 2\pi$, we get $\sigma = \sqrt{2\pi} \approx 2.51$, so the answer should be somewhat less than $2.51$. A rough mental model: the mean of the half-normal is about $80\%$ of $\sigma$, so roughly $0.8 \times 2.51 \approx 2$. Spoiler: the exact answer is $2$.
Approach: Factor using $X = \sigma Z$ where $Z \sim N(0,1)$, then apply LOTUS to compute $E[|Z|]$.
Formal Solution:
Since $X = \sigma Z$ with $Z \sim N(0,1)$:
$$E[|X|] = E[|\sigma Z|] = \sigma \, E[|Z|]$$
Apply LOTUS with the standard normal density $\phi(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$:
$$E[|Z|] = \int_{-\infty}^{\infty} |x| \cdot \frac{1}{\sqrt{2\pi}} e^{-x^2/2}\, dx$$
Both $|x|$ and $\phi(x)$ are even functions, so the integrand is even. Fold the integral onto $[0, \infty)$:
$$E[|Z|] = \frac{2}{\sqrt{2\pi}} \int_0^{\infty} x \, e^{-x^2/2}\, dx$$
Substitute $u = x^2/2$, so $du = x\, dx$:
$$E[|Z|] = \frac{2}{\sqrt{2\pi}} \int_0^{\infty} e^{-u}\, du = \frac{2}{\sqrt{2\pi}} \cdot 1 = \sqrt{\frac{2}{\pi}}$$
Therefore:
$$E[|X|] = \sigma \sqrt{\frac{2}{\pi}}$$
Answer: $E[|X|] = \sigma \sqrt{\dfrac{2}{\pi}}$. For $\sigma^2 = 2\pi$, we have $\sigma = \sqrt{2\pi}$, so:
$$E[|X|] = \sqrt{2\pi} \cdot \sqrt{\frac{2}{\pi}} = \sqrt{4} = 2$$
Intuition
The result $E[|X|] = \sigma \sqrt{2/\pi} \approx 0.798 \sigma$ is a number worth memorizing. It tells you the mean absolute deviation of a normal is about $80\%$ of the standard deviation. This is useful for quick sanity checks: if someone reports a daily P&L standard deviation of $\$1\text{M}$, the average absolute daily move is around $\$800\text{K}$.
The calculation itself illustrates two techniques that come up constantly: (1) the scaling trick -- writing $X = \sigma Z$ to reduce to the standard case -- is used everywhere in derivatives (standardizing payoffs, scaling vol) and in statistics (pivoting test statistics); and (2) exploiting symmetry to fold integrals, which appears in computing option prices, moments of symmetric distributions, and Fourier transforms. The $u = x^2/2$ substitution is also the key step in computing the Gaussian normalization constant, so this integral is closely related to the reason $\int e^{-x^2/2} dx = \sqrt{2\pi}$.