Deriving the Normal Moment-Generating Function by Completing the Square

For

X\sim N(\mu,\sigma^2)

, the moment generating function is

M_X(t)=\exp(\mu t+\sigma^2t^2/2)

. Complete the square inside the normal-density integral to derive it. Differentiate at zero to obtain ordinary moments, or substitute a nonzero argument to evaluate an expectation such as

E[e^X]

Complete the square around a shifted normal mean

Let

\mu

be the mean,

\sigma^2

the variance, and

t

the argument of the moment-generating function. The variable of integration

x

represents a possible value of

X

. For

\sigma>0

, write

E[e^{tX}]

using the normal density.

M_X(t)=\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{\infty} \exp\left(tx-\frac{(x-\mu)^2}{2\sigma^2}\right)\,dx.

Only the linear coefficient in $x$ changes when you multiply the density by $e^{tx}$ . The coefficient is now $\mu/\sigma^2+t$ . A normal kernel with mean $\mu+\sigma^2t$ has that coefficient, and its quadratic coefficient remains $-1/(2\sigma^2)$ . You can keep track of the constant by writing $tx=t(x-\mu)+t\mu$ .

\begin{aligned} tx-\frac{(x-\mu)^2}{2\sigma^2} &=-\frac{(x-\mu)^2-2\sigma^2t(x-\mu)}{2\sigma^2}+\mu t\\ &=-\frac{(x-\mu-\sigma^2t)^2}{2\sigma^2}+\mu t+\frac{\sigma^2t^2}{2}. \end{aligned}

The constant $\mu t+\sigma^2t^2/2$ is independent of $x$ , so its exponential comes outside the integral. What remains is the density of a normal with mean $\mu+\sigma^2t$ and variance $\sigma^2$ , integrated over the whole real line. Its integral is one. When $\sigma=0$ , $X=\mu$ almost surely and the formula reduces directly to $e^{\mu t}$ .

Use the formula to get moments

The normal tails allow differentiation under the expectation near zero. Each derivative brings down one factor of $X$ , so $M'_X(0)=E[X]$ and $M''_X(0)=E[X^2]$ . Differentiating the closed form gives

M_X'(0)=\mu,\qquad M_X''(0)=\mu^2+\sigma^2,\qquad \operatorname{Var}(X)=M_X''(0)-M_X'(0)^2=\sigma^2.

For an exponential moment such as $E[e^X]$ , evaluate $M_X(1)$ directly. It is a different calculation from taking a derivative at zero.

Three moments from one formula

Let $X\sim N(2,9)$ . Its first moment is 2, its second raw moment is $2^2+9=13$ , and its exponential moment is $E[e^X]=M_X(1)=e^{6.5}$ . Exponentiating the mean alone would give $e^2$ , which misses the contribution from dispersion.

In a stochastic calculus exercise, suppose

B_T\sim N(0,T)

and define

Z_T=e^{\sigma B_T-\sigma^2T/2}

. Using the normal moment generating function at argument

\sigma

gives

E[Z_T]=e^{-\sigma^2T/2}E[e^{\sigma B_T}] =e^{-\sigma^2T/2}e^{\sigma^2T/2}=1.

The same idea applied to a Brownian increment gives the corresponding conditional expectation in a martingale proof.

Verify the exponential moment numerically

This example computes the three quantities for the normal variable in the exercise and checks the first derivative with a centred difference.

from math import exp

mu, variance = 2.0, 9.0

def mgf(t):
    return exp(mu * t + variance * t**2 / 2)

h = 1e-5
print("Mean by centred difference:", (mgf(h) - mgf(-h)) / (2 * h))
print("Second raw moment:", mu**2 + variance)
print("Exponential moment:", mgf(1))

Numerical differentiation is a useful check on an analytic result. If you try the second derivative numerically, a very small step can cause cancellation because both nearby function values are close to one.

Complete the square around a shifted normal mean

Use the formula to get moments

Three moments from one formula

Verify the exponential moment numerically

References and next steps