An issue of dependent and independent random variables involving geometric Brownian motion.

Question

Let $X(t)=X(0)e^{\mu t + \sigma Z(t)}$ be a geometric Brownian motion (GBM) where $Z(t)$ is the standard Brownian motion with drift $0$ and the variance rate per unit of time is $1$. Now, let $s<t$ then I found in some of the text books that $X(t)$ and $X(s)$ are dependent random variables but $X(s)/X(0)$ and $X(t)/X(s)$ are independent random variables, but I don't understand why this is the case.

Answer 1

If $0<s<t$ then $Z(s)-Z(0)$ and $Z(t)-Z(s)$ are independent because the intervals $(0,s)$ and $(s,t)$ do not overlap.

Call these $A$ and $B$: \begin{align} A & = Z(s)-Z(0), \\ B & = Z(t)-Z(s). \end{align} Then $A,B$ are independent random variables.

But \begin{align} A & = Z(s) - Z(0) \\ A+B & = Z(t) - Z(0) \end{align} are not independent.

And the same applies to functions of $A$ and $B$, such as an exponential function.