Let $W(t)$ be a standard Brownian motion. Define $X(t)=e^{W(t)}$. Let $0<s<t$. Give the expression for $Cov(X(s),X(t))$

Question

Let $W(t)$ be a standard Brownian motion. Define $X(t)=e^{W(t)}$. Let $0<s<t$. Give the expression for $Cov(X(s),X(t))$.

Here is my work:

$$Cov(X(s),X(t))=E[X(s)X(t)]-E[X(s)]E[X(t)]$$

Then since $W(s)$ and $W(t)-W(s)$ are independent, so are $X(s)$ and $X(t)-X(s)$ $$E[X(s)X(t)]=E[X(s)(X(s)+X(t)-X(s))]=E[X(s)^2]+E[X(s)]E[X(t)-X(s)]$$

Overall, $$Cov(X(s),X(t))=E[X(s)^2]+E[X(s)]E[X(t)-X(s)]-E[X(s)]E[X(t)]$$

I got this question wrong in my assignment but I couldn't figure out where.

Thanks in advance.

Answer 1

This is incorrect:

Then since $W(s)$ and $W(t)-W(s)$ are independent, so are $X(s)$ and $X(t)-X(s)$

In fact, $X(s)$ and $X(t)-X(s)$ are very much positively correlated. Instead, you need to go back to the definition see that $X(s) = e^{W(s)}$ and $$e^{W(t) - W(s)} = \frac{e^{W(t)}}{e^{W(s)}} = \frac{X(t)}{X(s)}$$ are independent. This allows you to write

$$\begin{align*}E\left[X(s)X(t)\right] &= E\left[X(s)^2\cdot \frac{X(t)}{X(s)}\right] \\ &= E\left[X(s)^2\right]E\left[ \frac{X(t)}{X(s)}\right] \\ &= E\left[X(s)^2\right]E\left[X(t-s)\right]\end{align*}$$ or for the covariance: $$\text{Cov}(X(s),X(t)) = E\left[X(s)^2\right]\cdot E\left[X(t-s)\right] - E[X(s)]\cdot E[X(t)]$$ To evaluate these, it's useful to know that, for a standard normal $Z,$ $$E\left[e^{\sigma Z}\right] = e^{\sigma^2/2}$$