Assume that the risk free rate is $0$ and that the stock price is given by the equation $S(t)=6e^{2t+2B(t)}$ where $B(t)$ is the standard Brownian motion. Determine the price at time $0$ of the European call option with strike $K=6e^{18}$ and expiration $9$.
The formula I have is $C_E(0)=S\Phi(d_+)-Ke^{-rt}\Phi(d_-)$
where $d_+=\frac{ln\frac{S(0)}{K}+\Big(r+\frac{\sigma^2}{2}\Big)T}{\sigma\sqrt{T}}$ and $d_-=\frac{ln\frac{S(0)}{K}+\Big(r-\frac{\sigma^2}{2}\Big)T}{\sigma\sqrt{T}}$
There are two things I don't understand:
1. The solution uses $\sigma=2$, how do you find the standard deviation from the information given in the question?
2. I need to find $S(0)$ so I plug in $0$ for $t$ in $S(t)=6e^{2t+2B(t)}$ and I get $S(t)=6e^{2B(0)}$, I know $B(0)$ is a normal distribution with mean $0$ and variance $0$ but does that mean that $B(0)=0$? Why?
Generally speaking, in the Black-Scholes model, the stock price process $\{S(t), t \geq 0\}$ is modelled as $$S(t) = S(0) e^{\displaystyle \left(r-\frac{1}{2}\sigma^2\right)t + \sigma W(t)}.$$ Now, comparing the above with $S(t) = 6e^{-2t+2W(t)}$ (note that I wrote $-2t$ instead of $2t$, otherwise it does not hold, so I assume it is a typo) and given that $r=0$, you immediately obtain $$\begin{cases} S(0) = 6, \\ \sigma = 2. \end{cases}$$ Note that $W(0) = 0$ by definition (in fact, to be precise you should write 'almost sure equal to zero'). You might recall that a standard Brownian motion always starts at zero.