Take the tour ×  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)] $ ?  share|improve this question add
comment  1 Answer  up vote 4 down vote accepted  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^