Expected value of exponential Brownian motion

Question

I want to prove that

$$E[e^{2B_t}] = e^{2t}$$

where $B_t$ is a Brownian motion.

I have been reading up on Mean of exponential Brownian motion but it does not show how the rest of the log-normal function just disappears from the answer?

Answer 1

I would say that the first thing you should check out (if you haven't already) is the Change of Variables Theorem (see for instance "Probability: Theory and Examples" by Durrent - Theorem 1.6.9). This theorem essentially says how you can compute expected values:

\begin{gather*} \int_\Omega g(X(\omega)) dP = \int_\mathbb{R} g(x) dP_X \end{gather*}

In your case, $g(x) = e^{2x}$, so:

\begin{gather*} E[e^{2B_t}] = \int_\Omega e^{2B_t} dP = \int_{\mathbb{R}} e^{2x} \frac{1}{\sqrt{2\pi t}} e^{-x^2/2t} dx = \int_{\mathbb{R}} \frac{1}{\sqrt{2\pi t}} e^{\frac{-x^2 + 4xt}{2t}} dx = \int_{\mathbb{R}} \frac{1}{\sqrt{2\pi t}} e^{\frac{-x^2 + 4xt - 4t^2 + 4t^2}{2t}} dx \end{gather*} \begin{gather*} \int_{\mathbb{R}} \frac{1}{\sqrt{2\pi t}} e^{\frac{-(x - 2t)^2 + 4t^2}{2t}} dx = e^{2t} \int_{\mathbb{R}} \frac{1}{\sqrt{2\pi t}} e^{\frac{-(x - 2t)^2}{2t}} dx = e^{2t} \end{gather*}

Answer 2

$E[e^{2B_t}]=\int e^{2x} \frac 1 {\sqrt {2\pi t}}e^{-x^{2}/2t}dx$. Put $y=x/\sqrt t$ to get $E[e^{2B_t}]=\int e^{2y\sqrt t} \phi (y)dy$ where $\phi$ is the standard normal pdf. Now use the following:

$$\int e^{ay} e^{-y^{2}/2} dy$$ $$=e^{a^{2}/2} \int e^{-(y-a)^{2}/2} dy$$ $$=e^{a^{2}/2} \int e^{-z^{2}/2} dz$$ $$ =\sqrt {2 \pi} e^{a^{2}/2}.$$ You get $E[e^{2B_t}]=e^{2t}$