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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)] $ ?

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1 Answer 1

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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.

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  • $\begingroup$ You are here calculating the covariance between two independent GBMs. Indeed, substituting back $\gamma=\mu-\sigma^2/2$ into your formula gives $Cov(S_t,S_r) = e^{\mu(t+r)}\left(e^{\sigma^2r}-1\right)$. In particular without drift, $\mu=0$, we have $Cov_{\mu=0}(S_t,S_r) = e^{\sigma^2r}-1$, which is independent on $t$ and is just the variance of a GBM. What I am looking for is the auto-correlation of a single GBM trajectory: $S_0\rightarrow S_t \rightarrow S_{t^\prime}$. $\endgroup$
    – RRG
    Apr 11, 2013 at 1:46
  • $\begingroup$ @RPG Sorry, but I do not get your point. In general $S_t$ and $S_r$ are not independent. Furthermore, you ask what $Cov(S_t,S_r)$ is. I'm sorry if I misunderstood your question. $\endgroup$
    – math
    Apr 11, 2013 at 7:16
  • $\begingroup$ It seems your solution is correct. I will accept it as an answer to my question. Thanks. $\endgroup$
    – RRG
    Apr 23, 2013 at 8:09
  • $\begingroup$ You're welcome. I'm glad if I could help. $\endgroup$
    – math
    Apr 23, 2013 at 8:23
  • $\begingroup$ I think the exponent in the last expectation shouldn't have the 2. Therefore, if you simplify, the covariance becomes 0 $\endgroup$
    – chuse
    Apr 21, 2022 at 9:53

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