Mean of exponential Brownian motion

Question

I am new to stochastics and I am trying to compute the expectation of $S_t = e^{\sigma W_t}$, where $W_t$ is a standard Brownian motion and $\sigma>0$. My attempt (using the log-normal PDF here and scaling the variance of $W_t$) is as follows:

$\mathbb{E}[S_t] = \int_\mathbb{R} e^{\sigma x}\frac1{\sqrt{2\pi t\sigma^2}e^{\sigma x}} e^{-\frac{(\sigma x)^2}{2t\sigma^2}}dx = \int_\mathbb{R} \frac1{\sqrt{2\pi t}\sigma} e^{-\frac{x^2}{2t}}dx$,

but I have been told the correct answer is

$\mathbb{E}[S_t] = \int_\mathbb{R} \frac1{\sqrt{2\pi t}} e^{\sigma x} e^{-\frac{x^2}{2t}}dx$.

Can someone please point out why my attempt is incorrect?

Answer 1

It is not clear how you are using or you would use the fact that $S_t$ is lognormal in the integral. What you have is that $\displaystyle S_t = e^{\sigma\sqrt{t}Z}$, where $Z$ is an standard normal, therefore its mean is $\displaystyle e^{\sigma^2t/2}$, which is the same result you get with the second integral.

To clarify, you have to possible paths for computing $E[S_t]$ by using the integral that involves the PDF. We make $X = W_t$.

1. A direct application of the definition of expected value, using the PDF of $S_t$:

\begin{align} E[S_t] &= \int_\mathbb{R} s_tf_{s_t}(s_t)ds_t\tag{1}\\ \end{align}

2. Using the PDF of $X$, noting that $S_t = g(X) = e^{\sigma X}$:

\begin{align} E[S_t] = E[g(X)]&= \int_\mathbb{R} g(x)f_{X}(x)dx\\ &= \int_\mathbb{R} e^{\sigma x}f_{X}(x)dx\tag{2}\\ \end{align}

The second approach is the one used in your second integral. As I said, it is not clear how you came up with the integrand in the first integral. A direct use of the fact that $S_t$ is lognormal would be that if $S_t = e^{\mu + \sigma Z}$, where $Z$ is standard normal, then its mean is $e^{\mu + \sigma^2/2}$, which is a result that can be derived using $(2)$.

EDIT: If $Y = e^{\mu + \sigma Z}$, with $Z$ the standard normal, then $Y$ is a lognormal random variable, and its PDF is given by

$$f_Y(y) = \frac{1}{y\sqrt{2\pi}\sigma}e^{-(\text{ln}y-\mu)^2/(2\sigma^2)},\qquad y>0$$

In your case $\displaystyle S_t = e^{\sigma X} = e^{\sigma\sqrt{t}Z}$, therefore

$$f_{S_t}(s_t) = \frac{1}{s_t\sqrt{2\pi}\sigma\sqrt{t}}e^{-(\text{ln}s_t)^2/(2\sigma^2t)},\qquad s_t>0$$

and $(1)$ becomes

\begin{align} E[S_t] &= \int_0^\infty s_t\frac{1}{s_t\sqrt{2\pi}\sigma\sqrt{t}}e^{-(\text{ln}s_t)^2/(2\sigma^2t)}ds_t\\ &= \int_0^\infty \frac{1}{\sqrt{2\pi}\sigma\sqrt{t}}e^{-(\text{ln}s_t)^2/(2\sigma^2t)}ds_t\\ \end{align}

If we let $u = \text{ln}s_t$,

\begin{align} E[S_t] &= \int_\mathbb{R} \frac{1}{\sqrt{2\pi}\sigma\sqrt{t}}e^ue^{-u^2/(2\sigma^2t)}du\tag{3}\\ \end{align}

But since $\displaystyle S_t = e^{\sigma X}$, $u = \text{ln}s_t = \sigma x$, and replacing this in $(3)$, we get $(2)$.