E[exp(Zt-Zu)] is not exp(E[Zt-Zu]).
You can approximately visualize this from a simple Taylor series expansion:
LHS: E[1 + (Zt-Zu)/1! + (Zt-Zu)^2/2! + ....]
= E[1] + E[Zt-Zu] + (1/2!)*E[(Zt-Zu)^2] + (1/3!)*E[(Zt-Zu)^3] + .....
= 1 + 0 + .5*(t-u) + 0 + 0 + 0 + ..... = 1 + .5*(t-u)
RHS: exp(E[Zt-Zu]) = exp(0) = 1
Last edited: Sep 17, 2015