Probability & Statistics

Moments of the Normal

For a standard normal, symmetry zeroes the odd moments and integration by parts yields the ladder E[X^n] = (n-1) E[X^(n-2)], so the fourth moment is 3, the benchmark for zero excess kurtosis.

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Let XX be a standard normal random variable, with density φ(x)=12πe−x2/2\varphi(x) = \tfrac{1}{\sqrt{2\pi}}e^{-x^2/2}. Find the first four moments E[X]\E[X], E[X2]\E[X^2], E[X3]\E[X^3], and E[X4]\E[X^4].

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#Odd moments vanish by symmetry

The density φ\varphi is even, so for odd nn the integrand xnφ(x)x^n \varphi(x) is odd and its integral over the whole line cancels term for term,

E[X]=E[X3]=0.(1)\E[X] = \E[X^3] = 0. \tag{1}
The density is even, so the integrand for the mean splits into mirror lobes of equal area on opposite sides of the axis and cancels. Every odd moment vanishes the same way, while the even moments climb the ladder E[X^n] = (n-1) E[X^(n-2)].

#Even moments climb a ladder

The density satisfies φ′(x)=−x φ(x)\varphi'(x) = -x\,\varphi(x), which turns one factor of xx into a derivative. Integrating by parts for n≥2n \ge 2,

E[Xn]=∫−∞∞xn−1 (x φ(x)) dx=∫−∞∞xn−1 (−φ′(x)) dx=(n−1) E[Xn−2],(2)\E[X^n] = \int_{-\infty}^{\infty} x^{n-1}\,\big(x\,\varphi(x)\big)\,dx = \int_{-\infty}^{\infty} x^{n-1}\,\big(-\varphi'(x)\big)\,dx = (n-1)\,\E[X^{n-2}], \tag{2}

since the boundary term [−xn−1φ(x)]−∞∞\big[-x^{n-1}\varphi(x)\big]_{-\infty}^{\infty} vanishes. Starting from E[X0]=1\E[X^0] = 1,

E[X2]=1⋅E[X0]=1,E[X4]=3⋅E[X2]=3.(3)\E[X^2] = 1 \cdot \E[X^0] = 1, \qquad \E[X^4] = 3 \cdot \E[X^2] = 3. \tag{3}

#Collect them

E[X]=0,E[X2]=1,E[X3]=0,E[X4]=3.(4)\E[X] = 0, \qquad \E[X^2] = 1, \qquad \E[X^3] = 0, \qquad \E[X^4] = 3. \tag{4}

The second moment 11 is just the unit variance, and the fourth moment 33 is the reason the normal has zero excess kurtosis, the benchmark every other distribution's tail is measured against.