 Solution

Solving the geometric Brownian motion equation over $[t,T]$ givesThe Brownian increment is independent of ${\mathcal{F}}_t$ and has a normal distribution with variance $T-t$. Its exponential moment-generating function therefore givesThe required deterministic price function isThis includes $s=0$, $v=0$ and $t=T$. For $s>0$ and $t<T$, it is continuous and strictly decreasing in $v\ge 0$, with range $(0,\sqrt{s}]$.