Wiener process analytic expression from geometric brownian motion

Question

The solution to the SDE

$dx= -kx\ dt + cx \ dW$

is

$x(t) = x_0 e^{(c - k^2/2)t}e^{-k W}$

with mean

$\langle x(t) \rangle = x_0 e^{(c - k^2/2)t}$

where $W(t)$ is the Wiener process.

Im looking to find the expression

$\langle [W(s) + W(t) - 2W(0)]^2 \rangle$

but am unsure of how to proceed.

Help would be appreciated.

Answer 1

There is a serious problem in your post, namely the assertion

$x(t) = x_0 e^{(c - k^2/2)t}e^{-k W}$ with mean $\langle x(t) \rangle = x_0 e^{(c - k^2/2)t}$

cannot hold. Actually, $\langle x(t) \rangle = x_0 e^{(c - k^2/2)t}\langle e^{-k W} \rangle$ and $\langle e^{-k W} \rangle\ne1$ hence you have to rethink this part.

Later on, you seem to switch to an entirely different question, which is:

to find the expression $\langle [W(s)+W(t)−2W(0)]^2\rangle$.

This is so unrelated to the rest that one gets the impression the preceding is just some noise... but anyway, to compute this expression, assume without loss of generality that $s\leqslant t$, then $W(t)-W(s)$ is independent of $W(s)$ and $W(0)=0$ hence $$\langle [W(s)+W(t)−2W(0)]^2\rangle=\langle [2W(s)+W(t)−W(s)]^2\rangle=4\langle W(s)^2\rangle+\langle [W(t)−W(s)]^2\rangle, $$ thus, $$ \langle [W(s)+W(t)−2W(0)]^2\rangle=4s+t-s=t+3s.$$