I am in doubt with expected value of stochastic process :
(*) As prof. desmond higham says in his article http://www.caam.rice.edu/~cox/stoch/dhigham.pdf
$$u_t=e^{t+\frac{B_t}{2}} \to mean=average=e^{\frac{9}{8}t}$$ I understand this ,because (by Ito formula)$$du=1.u.dt+\frac{1}{2}.u.dB_t+\frac{1}{2}.\frac{1}{4}u.dt \to du=\frac{9}{8}u . dt +\frac{1}{2}.u.dB_t$$
this implies that $u_0 .e^{\frac{9}{8}t}$ is $E[u]$
(**) when we have $u_t=e^{(\mu-\frac{1}{2} \sigma ^2).t+\sigma .B_t}$ by Ito formula we obtain $$du=\frac{\partial u}{\partial t}.dt+\frac{\partial u}{\partial x}.dB_t+\frac{\partial^2 u}{2\partial x^2}=\\ (\mu).udt+\sigma.u.dB_t \to \\ E[u_t]=u_0.e^\mu$$
(***)when we have $du_t=\mu .u.dt+\sigma.u.dt $ and $\mu , \sigma$ are constant $E[u_t]=e^\mu$
now
Am I right in $(*) ,(**),(***)$ ? thanks in advanced Can you show me more example like those , on this site ?
For any lognormal process u, such as $du_t=u_t(\mu dt+\sigma dB_t)$, we have $E[u_t]=e^{\mu t}$.
By using Ito's lemma on the process $y_t=log(u_t)$, and using the fact that for any random variable normally distrubuted X, with mean m , and variance $\sigma^2$, we have $E[e^X]=e^{m+\frac{\sigma^2}{2}}$, you can conclude the result.