First Four Moments of the Standard Normal
Let $X \sim N(0, 1)$ be a standard normal random variable. Compute
$$E[X], \qquad E[X^2], \qquad E[X^3], \qquad E[X^4].$$
Derive them (for example from the moment generating function or by integration by parts) rather than quoting them.
Hints
- The density is symmetric about 0. What does that immediately say about every odd moment?
- The MGF is $M(t) = E[e^{tX}] = e^{t^2/2}$. The $n$-th moment is $M^{(n)}(0)$; expand $e^{t^2/2}$ as a power series and read off coefficients.
- $e^{t^2/2} = 1 + \tfrac{t^2}{2} + \tfrac{t^4}{8} + \cdots$, and $E[X^n] = n! \times [\text{coefficient of } t^n]$, giving $E[X^4] = 4!/8 = 3$.
Worked Solution
How to Think About It: Symmetry kills the odd moments instantly. For the even moments, the MGF of a standard normal is $e^{t^2/2}$; differentiating it four times is tedious, but expanding it as a power series and matching against $M(t) = \sum E[X^n]\, t^n / n!$ takes one line. Integration by parts gives the same recursion $E[X^n] = (n-1)E[X^{n-2}]$.
Quick Estimate: $E[X^2] = 1$ by definition of a unit variance, and $E[X^4]$ must exceed $(E[X^2])^2 = 1$ by Jensen. The normal's fourth moment is the benchmark "kurtosis 3."
Formal Solution:
*Step 1 -- Odd moments by symmetry.* The density $\phi(x) = e^{-x^2/2}/\sqrt{2\pi}$ is even, so $x^n \phi(x)$ is odd for odd $n$ and integrates to zero over $\mathbb{R}$ (the integrals converge absolutely). Hence $E[X] = E[X^3] = 0$.
*Step 2 -- MGF.* Completing the square,
$$M(t) = E[e^{tX}] = \int \frac{e^{tx - x^2/2}}{\sqrt{2\pi}}\,dx = e^{t^2/2} \int \frac{e^{-(x-t)^2/2}}{\sqrt{2\pi}}\,dx = e^{t^2/2}.$$
*Step 3 -- Match coefficients.* Expanding both sides,
$$\sum_{n \ge 0} \frac{E[X^n]}{n!}\, t^n = e^{t^2/2} = \sum_{k \ge 0} \frac{t^{2k}}{2^k\, k!}.$$
So $E[X^{2k}] = \dfrac{(2k)!}{2^k k!} = (2k - 1)!!$ and the odd moments vanish. In particular $E[X^2] = 2!/2 = 1$ and $E[X^4] = 4!/(4 \cdot 2) = 24/8 = 3$.
*Step 4 -- Integration-by-parts check.* Since $\phi'(x) = -x\phi(x)$,
$$E[X^n] = \int x^{n-1} \cdot x\phi(x)\,dx = \left[-x^{n-1}\phi(x)\right]_{-\infty}^{\infty} + (n-1)\int x^{n-2}\phi(x)\,dx = (n-1)E[X^{n-2}],$$
giving $E[X^2] = 1 \cdot E[X^0] = 1$ and $E[X^4] = 3 \cdot E[X^2] = 3$.
Answer: $E[X] = 0$, $E[X^2] = 1$, $E[X^3] = 0$, $E[X^4] = 3$ (in general $E[X^{2k}] = (2k-1)!!$ and odd moments are zero).
Intuition
Odd moments vanish by symmetry, $E[X^2] = 1$ is the variance, and $E[X^4] = 3$ is the number behind "the normal has kurtosis 3." The fastest derivation is to expand the MGF $e^{t^2/2}$ and match coefficients, which also gives the general pattern $E[X^{2k}] = (2k-1)!!$. Knowing $E[X^4] = 3$ cold matters in practice: it is the baseline against which excess kurtosis of returns is measured, and it drives the variance of a sample variance and the vega/volga terms in option pricing.