Geometric Brownian motion This question is related to conditional expectation of a geometric Brownian motion. 

The price of a stock is $10$ times a Geometric Brownian Motion with drift $\mu  = 0.05$ and $\sigma  = 0.2$.
  Assume the stock price is $30$ at time $16$. What is the expected value of the stock price at time $25$?

The answer is $56.3283$
What formula should I use to get this answer?  
According to the question, the stock price is $S(t) = 10{e^{x(t)}}$
Should I use $E[Z(t)] = {e^{\mu t + {{\sigma {t^2}} \over 2}}}$?
Thanks in advance.
 A: Take $S(t)$ as the price of the stock at time $t$.
We can write $S(t)$ as $S(t)=Ce^{0.2B(t)+0.05t}$ where $B(t)$ is your standard Brownian motion and $C$ is a constant. You're asked to compute $\mathbb{E}\left(S(25)\big|S(16)=30\right)$.
Since $S(16),\frac{S(25)}{S(16)}$ are independent and $B(25)-B(16)\sim \mathcal{N}(0,9)$ we can say $$\begin{eqnarray*}\mathbb{E}\left(S(25)\big|S(16)=30\right)&=&\mathbb{E}\left(\frac{S(25)}{S(16)}\times S(16)\Bigg|S(16)=30\right)\\&=& 30 \mathbb{E}\left(\frac{S(25)}{S(16)}\right) \\ &=& 30e^{0.05\times 9}\mathbb{E}\left(e^{0.02(B(25)-B(16))}\right) \\ &=& 30e^{0.05\times 9} \int_{-\infty}^{\infty}e^{0.02x}\times \frac{1}{3\sqrt{2\pi}}e^{-\frac{x^2}{18}}\mathrm{d}x \\ & \approx & 56.3283 \end{eqnarray*}$$
I know I'm way late to the party, but I saw this question unanswered while searching through some resources on geometric Brownian motion and thought I'd add my two cents.
A: Under a geometric Brownian motion $$ dS_t = \mu S_t {dt} + \sigma S_t {dW_t} $$
where $\mu = 0.05$ and $\sigma = 0.2$,
the stock price is written $$ P_t = cS_t $$ where $c=10$.
Since $ \mathbb{E}[S_t] = S_0 \exp( \mu t ) $, the expected stock value is
$$ \mathbb{E}[P_t] = P_0 \exp( \mu t ) = 30 \exp(0.05 t) $$
where we have shifted $t$ to "start" at $16$.
So at time $25$, we have $t=9$, so:
$$ \mathbb{E}[P_t] = 30 \exp(0.05 \times 9) = 47.05 $$
which is wrong, apparently.
Either the question or I have made a mistake? Perhaps you can figure out which one?
