Use Ito's lemma to find the process's mean and variance

Question

The process is defined as $$x_t = \exp{(kt+\sigma W_t)}$$ where $k,\sigma\in\mathbb{R}$ and $W_t$ is Brownian motion.

I want to find $Ex_t, Dx_t$.

To find $Ex_t$, I try to solve the equation: \begin{align*} &dx_t=k\exp{(kt+\sigma W_t)}dt+\sigma\exp{(kt+\sigma W_t)}dW_t+\frac{1}{2}\sigma^2\exp{(kt+\sigma W_t)}dt=\newline &=kx_tdt+\frac{1}{2}\sigma^2\exp{(kt+\sigma W_t)}dt+\sigma\exp{(kt+\sigma W_t)}dt=\newline &=x_tdt(k+\frac{1}{2}\sigma^2)+\sigma x_tdW_t \end{align*} and can't seem to arrive at the final solution from here.

Speaking of variance, I could use \begin{align*} &Dx_t = Ex^2_t-(Ex_t)^2\newline &d(x_t^2)=2dx_t \end{align*} but to do so, I need that $Ex_t$.

Any advice on finding these moments could be helpful. Am I missing something?

Answer 1

The simplest solution is to use that for a Gaussian $Z$ with mean zero and variance $v$ $$ E\big[e^{Z-v/2}\big]=1\,. $$ Proof.

$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{\sqrt{v}\,x-v/2}e^\frac{-x^2}{2}\,dx=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{-\frac{(x-\sqrt{v})^2}{2}}\,dx=1\,.$$

This yields $$ \boxed{\quad\phantom{\Bigg|}E\big[e^{\alpha W_t-\alpha^2t/2}\big]=1\,.\quad} $$ which could also be shown using Ito's formula:

$$M_t:=e^{\alpha W_t-\alpha^2t/2}\;\Longrightarrow\; dM_t=\alpha M_t\,dW_t\;\Longrightarrow\; M_t=1+\textstyle\int_0^t\alpha M_s\,dW_s\,.$$ The integral in the last term has expectation zero.

Then \begin{align} E\big[e^{k\,t\,+\,\sigma\, W_t}\big]&=e^{\,k\,t\,+\,\sigma^2\,t/2}\,,\\[2mm] E\big[e^{2\,k\,t\,+\,2\,\sigma\, W_t}\big]&=e^{\,2\,k\,t\,+\,2\,\sigma^2\,t} \end{align} so that the variance of $e^{k\,t\,+\,\sigma\, W_t}$ is \begin{align} {\rm Var}\big[e^{k\,t\,+\,\sigma\, W_t}\big]&=e^{\,2\,k\,t\,+\,2\,\sigma^2\,t}\,-\, e^{\,2\,k\,t\,+\,\sigma^2\,t}\\ &=e^{\,2\,k\,t\,+\,\sigma^2\,t}\,(e^{\sigma^2t}-1)\,. \end{align}

Answer 2

The exponential of a sum is the product, always

$$ \sum _{n=0}^{\infty } \frac{1}{n!} (k t + \sigma W_t)^n = \sum _{n=0}^{\infty } \frac{1}{n!}\sum _{m=0}^n \frac{n!}{m! (n-m)!}\ (k t )^m \ \left(\sigma W_t\right)^{n-m} = \ \sum _{l=0}^\infty \frac{1}{l!}\ (k t )^l \ \sum _{n=0}^{\infty } \frac{1}{n!} \left(\sigma W_t\right)^n = e^{k \ t} e^{\sigma W_t} $$

Since $W_t$ is normal distributed with $E(W_t^{2n+1})=0$, all one needs are the expectations of the even powers

$$\frac{1}{\sqrt{2 \pi } \sigma } \ \int_{-\infty }^{\infty } x^{2 n} e^{-\frac{x^2}{2 \sigma ^2}} \, dx = \frac{2^n \sigma ^{2 n} \Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi }}$$

Sum the even terms of the exponential series

$$ e^{k \ t} \sum _{n=0}^{\infty } \frac{t^n \left(2^n \sigma ^{2 n} \Gamma \left(n+\frac{1}{2}\right)\right)}{\sqrt{\pi } (2 n)!} = e^{k t + \frac{\sigma ^2 t}{2}}$$