For uni I've got the following exercise:
and I cant get to an answer, what I've tried is the following:
\begin{align*} E[e^{-rt}e^{W(t) + c(t)}|F(s)] &= e^{c(t)-rt}E[e^{W(t)}|F(s)]\\ &= e^{c(t)-rt}E[e^{W(t)-W(s)}e^{W(s)}|F(s)] \\ &= e^{W(s) + c(t)-rt}E[e^{W(t)-W(s)}] \\ &= e^{W(s) + c(t)-rt+\frac{t-s}{2}} \end{align*} this implies $W(s) + c(t)-rt+\frac{t-s}{2} = W(s) + c(s) - rs$, meaning $c(t)-c(s) + \frac{t-s}{2} = r(t - s)$, how do I obtain a formula for $c(t)$ from this?Any help is appreciated!!
I think I have found quite a beautiful proof for this (correct me if I'm wrong):
Because $c(t)-c(s) = (t-s)(r+\frac{1}{2})$, we can say that $\frac{c(t)-c(s)}{t-s} = (r+\frac{1}{2})$ for all $t> s \geq 0$, so $\lim_{t\rightarrow s}\frac{c(t)-c(s)}{t-s} = (r+\frac{1}{2}) = c'(s)$. so $c(x) = rx + \frac{x}{2} + k$. As all equations mentioned before are valid for all $k$ we choose $k=0$. So $c(x) = rx + \frac{x}{2}$ is valid!