Suppose I have a Geometric Brownian Motion process, $$dX_t=\mu X_t dt + \sigma X_t dW_t$$

I'd like to find the covariance of $\log(X_t)$ and $\log(X_s)$ where $s<t$. We can write $\log(X_t)$ in differential form as $$d\log(X_t)=\sigma dW_t+\left(\mu-\frac{\sigma^2}{2}\right)dt$$

That's $$cov(\log(X_t),\log(X_s))=E[\log(X_t)\log(X_s)] - E[\log(X_t)]E[\log(X_s)]$$ $$=\sigma^2 s - ts\left(\mu-\frac{\sigma^2}{2}\right)^2$$

Is there anything wrong with my derivation? As my intuition tells my there shouldn't be any term associated with $\sigma^4$. Any help is appreciated!


1 Answer 1


Let $Y = \log X$, then:

$$\begin{align} Y &= Y_0 + (\mu-\frac{\sigma^2}{2})t + \sigma W_t \\ EY_t &=Y_0 + (\mu-\frac{\sigma^2}{2})t \\ EY_tEY_s &= Y_0^2 + Y_0 (\mu-\frac{\sigma^2}{2}) (t+s) + (\mu-\frac{\sigma^2}{2})^2 t s \\ E(Y_tY_s) &= E\left((Y_0 + (\mu-\frac{\sigma^2}{2})t + \sigma W_t) (Y_0 + (\mu-\frac{\sigma^2}{2})s + \sigma W_s)\right) \\ &= Y_0^2 + Y_0 (\mu-\frac{\sigma^2}{2}) (t+s) + (\mu-\frac{\sigma^2}{2})^2 t s + \dots + \sigma ^2 \min(t,s) \end{align}$$

What remains: $$C\text{ov}(Y_t, Y_s) = \sigma^2 \min(t,s)$$

  • $\begingroup$ thank you! my mistake was $E(Y_t Y_s)$. $\endgroup$ Commented Jan 17, 2020 at 13:30

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