Auto-correlation of GBM

Question

The GBM is defined by
$ dS(t) = \mu S(t)dt + \sigma S(t) dW_t, $
with analytical solution
$ S(t^\prime) = S(t) exp\left[\left(\mu-\frac{\sigma^2}{2}\right)\left(t^\prime-t\right)+\sigma\left(W(t^\prime)-W(t)\right)\right]. $
What is the analytical expression for the covariance, $ Cov[S(t),S(t^\prime)] $ ?

Answer 1

W.l.o.g we use the initial condition $S(0)=1$ and define $\gamma:=\mu-\frac{\sigma^2}{2}$. Hence we have the dynamics

$$S_t=e^{\gamma t +\sigma W_t}$$

By definition $Cov(X,Y)=E((X-E(X))(Y-E(Y))=E(XY)-E(X)E(Y)$, where the last equality follows from the linearity of the expectation. Note $\gamma t+\sigma W_t$ is normal distributed with mean $\gamma t$ and variance $\sigma^2t$. Hence $S_t$ is lognormal with mean $\phi_t:=E[S_t]=e^{\gamma t+\frac{\sigma^2t}{2}}$. For $t>r$ this yields,

$$Cov(S_t,S_r) =E((S_t-\phi_t)(S_r-\phi_r))=E(S_rS_t)-\phi_r\phi_t$$

The last step is to calculate $E(S_tS_r)$.

$$E(S_rS_t)=e^{\gamma(t+r)}E(e^{\sigma(W_t+W_r)})=e^{\gamma(t+r)}E(e^{\sigma(W_t-W_r)}e^{2\sigma(W_r)})=e^{\gamma(t+r)}E(e^{\sigma(W_t-W_r})E(e^{2\sigma W_r})$$

where we have used that $W_t-W_r$ and $W_r$ are independent. Again, using the formula for the mean of a lognormal distribution we get $$E(e^{\sigma(W_t-W_r)})=e^{\frac{\sigma^2(t-r)}{2}}$$ and $$E(e^{2\sigma W_r})=e^{2\sigma^2 r}$$

Hence

$$Cov(S_t,S_r)=e^{\gamma(t+r)}e^{\frac{\sigma^2(t-r)}{2}}e^{2\sigma^2 r}-e^{\gamma t+\frac{\sigma^2t}{2}}e^{\gamma r+\frac{\sigma^2r}{2}}$$

I leave it to you to check if this can be further simplified.