The Mean of e to the X

Suppose $X$ is distributed normal with mean $0$ and variance $\sigma^2$ . What is $\E[e^X]$ ?

Reveal solutionHide solution

#Write the integral

\E[e^X] = \int_{-\infty}^{\infty} e^{x} \cdot \frac{1}{\sigma\sqrt{2\pi}}\, e^{-x^2 / (2\sigma^2)}\, dx. \tag{1}

#Complete the square

Merge the two exponents and group the powers of $x$ ,

x - \frac{x^2}{2\sigma^2} = -\frac{1}{2\sigma^2}\big(x^2 - 2\sigma^2 x\big) = -\frac{(x - \sigma^2)^2}{2\sigma^2} + \frac{\sigma^2}{2}. \tag{2}

#Integrate

The constant $e^{\sigma^2/2}$ pulls out, and what remains is a normal density centred at $\sigma^2$ , whose integral is $1$ ,

\E[e^X] = e^{\sigma^2/2} \int_{-\infty}^{\infty} \frac{1}{\sigma\sqrt{2\pi}}\, e^{-(x - \sigma^2)^2 / (2\sigma^2)}\, dx = e^{\sigma^2/2}. \tag{3}

Exponentiating a symmetric normal produces a right-skewed lognormal. The long upper tail drags the mean up to e to the one half, strictly above the median of 1, so averaging an exponential is not the exponential of the average.

#Result

\E[e^X] = e^{\sigma^2/2}. \tag{4}

This is the mean of the lognormal variable $e^X$ , which exceeds its median $e^0 = 1$ because the exponential stretches the upper tail far more than the lower. It is also the normal moment generating function $\E[e^{tX}] = e^{\sigma^2 t^2 / 2}$ read off at $t = 1$ .