 Solution

A standard Brownian motion starts at zero, has continuous sample paths and independent increments, with $W_t-W_s\sim N(0,t-s)$ for $0\le s<t$.

Assume $S_0,K,T>0$ and $\sigma >0$; changing the sign of $\sigma$ just changes the sign of Brownian motion. Itô formula gives $d{S}_t=S_t[(\mu +{\sigma }^2/2)dt+\sigma d{W}_t]$. In the complete Black-Scholes model, the risk-neutral pricing theorem prices an integrable payoff $Y$ by $e^{-rT}{\mathbb{E}}_{Q}Y$, where the discounted asset is a martingale. Girsanov theorem therefore replaces the logarithmic drift by $r-{\sigma }^2/2$:Let $d_2=[\log (S_0/K)+(r-{\sigma }^2/2)T]/(\sigma \sqrt{T})$ and $d_1=d_2+\sigma \sqrt{T}$. The exercise event is $Z>-d_2$. Completing the square in the Gaussian integral yieldsConsequently ${\mathbb{P}}_Q(S_T>K)=\Phi (d_2)$ and ${\mathbb{E}}_Q[S_T{1}_{\{S_T>K\}}]=S_0{e}^{rT}\Phi (d_1)$. Thus the European call option price isThe physical drift $\mu$ disappears because the price is obtained under the martingale measure. If $\sigma =0$ in an arbitrage-free deterministic model, $S_t=S_0{e}^{rt}$ and the formula is interpreted by its zero-volatility limit.