Problem:
In the Black & Scholes model, the price $S_t$ of a risky asset at time $t$ is given by the formula
$$S_t = \text{exp}\{(r-\frac{\sigma^2}{2})t + \sigma B_t \}$$
The value of a European call option, with maturity $T$ and strike price $K$, is $(S_T − K)^+$ at time $T$. If $T > t$, compute explicitly
$$E[(S_T − K)^+|\mathcal{F}_t ].$$
In the given solution, it reads:
Note: \begin{equation} \begin{split}E[(S_T − K)^+|\mathcal{F}_t] &= E[(S_0\text{exp}\{(r−\frac{\sigma^2}{2})T +σB_T\} - K)^+ |\mathcal{F}_t] \\ &= E[(S_0\text{exp}\{(r−\frac{\sigma^2}{2})(T-t+t) +σ(B_T-B_t+B_t)\} - K)^+ |\mathcal{F}_t] \\ & = E[(S_t\text{exp}\{(r−\frac{\sigma^2}{2})(T-t) +σ(B_T-B_t)\} - K)^+ |\mathcal{F}_t] \end{split} \end{equation}
I fail to see why
\begin{equation} \begin{split} S_T &= \text{exp}\{(r-\frac{\sigma^2}{2})t + \sigma B_t \} \\ &= (S_0\text{exp}\{(r−\frac{\sigma^2}{2})T +σB_T\} - K) \\ & = (S_0\text{exp}\{(r−\frac{\sigma^2}{2})(T-t+t) +σ(B_T-B_t+B_t)\} \\ &= S_t\text{exp}\{(r−\frac{\sigma^2}{2})(T-t) +σ(B_T-B_t)\} \end{split} \end{equation}
holds. Any help is very appreciated.
The solution of the price process in the risk neutral framework is $S_t=S_0e^{(r-\sigma^2/2)t+\sigma W_t}$ for $S_0>0$ where $W_t$ is the driving Brownian motion. So if $T>t$ we have ($S_t$ is always strictly positive) $$S_T=S_t\frac{S_T}{S_t}=S_t\frac{S_0e^{(r-\sigma^2/2)T+\sigma W_T}}{S_0e^{(r-\sigma^2/2)t+\sigma W_t}}=S_te^{(r-\sigma^2/2)(T-t)+\sigma (W_T-W_t)}$$ The solution you are provided with says, equivalently: $$\begin{aligned}S_T&=S_0e^{(r-\sigma^2/2)(T-t+t)+\sigma (W_T-W_t+W_t)}=\\ &=S_0e^{(r-\sigma^2/2)(T-t)+\sigma (W_T-W_t)}\underbrace{e^{(r-\sigma^2/2)t+\sigma W_t}}_{=S_t/S_0}=\\ &=S_te^{(r-\sigma^2/2)(T-t)+\sigma (W_T-W_t)}\end{aligned}$$