Covariance function exponential of brownian motion with drift

Question

Let $X_t = \exp\{\mu t + \sigma B_t\}$ where $B_t$ is a standard brownian motion and $\mu, \sigma >0$. I need to compute $m(t)=\mathbb{E}[X_t]$ and $c(t,s)=\text{Cov}(X_t,X_s),\ t,s \geq 0$. Using the moment generating function of $B_t \sim \mathcal{N}(0,t)$, I have found that $m(t)=\exp\{\mu t + 0.5 t \sigma^2\}$. Now I compute \begin{align*} c(t,s)&=\text{Cov}(X_t, X_s)\\ &=\mathbb{E}[X_tX_s]-\mathbb{E}[X_t]\mathbb{E}[X_s]\\ &=\mathbb{E}[\exp\{\mu t+\sigma B_t+\mu s +\sigma B_s\}]-\exp\{\mu t + 0.5 t \sigma^2 + \mu s + 0.5 s \sigma^2\}\\ &=\mathbb{E}[\exp\{\mu t\}\exp\{\sigma B_t\}\exp\{\mu s\}\exp\{\sigma B_s\}]-\exp\{\mu t + 0.5 t \sigma^2 + \mu s + 0.5 s \sigma^2\}\\ &=\exp\{\mu(t+s)\}\underbrace{\mathbb{E}[\exp\{\sigma (B_t + B_s)\}]}_{B_t, B_s\text{ are not independent}} - \exp\{0.5 \sigma^2 (t+s) + \mu(t+s)\} \end{align*} How can I compute this expectation? Does a MGF of the sum of normal random variables intervene here? How does one handle the correlation between $B_t$ and $B_s$? Thanks!

Answer 1

Following @angryavian's answer, \begin{align*} c(t,s)&=\text{Cov}(X_t, X_s)\\ &=\mathbb{E}[X_tX_s]-\mathbb{E}[X_t]\mathbb{E}[X_s]\\ &=\mathbb{E}[\exp\{\mu t+\sigma B_t+\mu s +\sigma B_s\}]-\exp\{\mu t + 0.5 t \sigma^2 + \mu s + 0.5 s \sigma^2\}\\ &=\mathbb{E}[\exp\{\mu t\}\exp\{\sigma B_t\}\exp\{\mu s\}\exp\{\sigma B_s\}]-\exp\{\mu t + 0.5 t \sigma^2 + \mu s + 0.5 s \sigma^2\}\\ &=\exp\{\mu(t+s)\}\mathbb{E}[\exp\{\sigma (B_t + B_s)\}] - \exp\{0.5 \sigma^2 (t+s) + \mu(t+s)\}\\ &=\exp\{\mu(t+s)\}\mathbb{E}[\exp\{\sigma (B_s + (B_t-B_s) + B_s)\}] - \exp\{0.5 \sigma^2 (t+s) + \mu(t+s)\}\\ &=\exp\{\mu(t+s)\}\mathbb{E}[\exp\{2\sigma B_s\}] \mathbb{E}[\exp\{\sigma(B_t-B_s)\}] - \exp\{0.5 \sigma^2 (t+s) + \mu(t+s)\}\\ &=\exp\{\mu(t+s)\}\exp\{2\sigma^2s\}\exp\{0.5\sigma^2(t-s)\}-\exp\{0.5 \sigma^2 (t+s) + \mu(t+s)\}\\ &=\exp\{\mu(t+s)+2\sigma^2s +0.5\sigma^2(t-s)\}-\exp\{\mu(t+s) + 0.5 \sigma^2 (t+s)\} \end{align*}