Raw Moments of a Standard Normal Random Variable

Let \(X \sim N(0,1)\). We want to compute the raw moments

\[ E[X^n] \quad \text{for } n=1,2,3,4 \]

The standard normal distribution is symmetric around \(0\). This symmetry is the key idea:

If \(n\) is odd, then \(x^n\) is an odd function, so the expectation is \(0\).
If \(n\) is even, then the moment is positive and can be computed directly.

Step-by-Step Moments

\[ E[X] = 0 \]

Since the standard normal distribution is symmetric about \(0\), the positive and negative parts cancel out.

\[ E[X^2] = 1 \]

For a standard normal random variable, the variance is

\[ \mathrm{Var}(X) = E[X^2] - (E[X])^2 \]

Since \(E[X]=0\) and \(\mathrm{Var}(X)=1\), we get

\[ E[X^2] = 1 \]

\[ E[X^3] = 0 \]

Because \(x^3\) is an odd function and the normal density is symmetric, the expectation is zero.

\[ E[X^4] = 3 \]

A standard result for the standard normal distribution is

\[ E[X^4] = 3 \]

This is also consistent with the general even-moment formula for a standard normal:

\[ E[X^{2k}] = (2k-1)!! \]

For \(k=2\):

\[ E[X^4] = 3!! = 3 \]

\[ E[X] = 0,\qquad E[X^2] = 1,\qquad E[X^3] = 0,\qquad E[X^4] = 3 \]

Therefore, the first four raw moments of \(X \sim N(0,1)\) are: