How to calculate the expectation of the exponetial of stochastic process?

Question

Let X be stochastic process defined by $X_t = exp(\sigma W_t)$, where W is a Wiener process and $\sigma$ is a constant. Then we have

$$E(X_T)=E(e^{\sigma W_T})=\exp(\frac{1}{2}\sigma^2 T)$$.

I know this is the correct result but I don't know how to get this?

Answer 1

We know that $E(W_T)=0$ and $E(g(W_t))=\int_\mathbb{R}g(w)\frac{1}{\sqrt{2T\pi}}e^{-\frac{1}{2T}{w^2}}dw$.

$g(x)=e^{\sigma x} $

And finally we get: $$\frac{1}{\sqrt{2T\pi}}\int_{\mathbb{R}}e^{\sigma w}e^{-\frac{w^2}{2T}}dw =\frac{1}{\sqrt{2T\pi}}\int e^{-\frac{w^2-2T\sigma w +T^2\sigma^2}{2T}}e^{\frac{T^2\sigma^2}{2T}}dw=e^{\frac{T\sigma^2}{2}}\int \frac{1}{\sqrt{2T\pi}}e^{-\frac{w^2-2T\sigma w +T^2\sigma^2}{2T}}dw$$

$$E(e^{\sigma W_T})=e^{\frac{T\sigma^2}{2}}$$

Answer 2

Let $Y_t = \exp \left( -\frac{\sigma^2}{2}t \right) X_t = \exp \left( \sigma W_t - \frac{\sigma^2}{2}t \right)$. Ito's lemma yields $$ dY_t = \left( \sigma Y_t dW_t - \frac{\sigma^2}{2} Y_t dt \right) + \frac 12 \cdot \sigma^2 dt = \sigma Y_t dW_t, $$ hence $Y_t$ is a martingale. It follows that $$ \exp \left( -\frac{\sigma^2}{2}t \right) \mathbb{E} X_t = \mathbb{E} Y_t = \mathbb{E} Y_0 = \mathbb{E} 1 = 1. $$