I'm trying to solve it for a European call option. Basically, I have to solve the following \begin{align} e^{{-r(T-t)}}{\textrm {E}}_s [(S_{{T}}-X){\mathbf{1}}_{{S_{{T}}>X}}]=e^{{-r(T-t)}}{\textrm {E}}_s[S_{{T}}{\mathbf{1}}_{{S_{{T}}>X}}]-Xe^{{-r(T-t)}}\underbrace{{\textrm{E}}_s [{\mathbf{1}}_{{S_{{T}}>X}}]}_{\Pr_s[S_{{T}}>X]} \end{align} where the subscript $s$ means the operator is conditioned on $S_t=s$, with $t<T$. The process $S_t$ is a geometric Brownian motion, and $X$ is a constant (the strike price).
\begin{equation} S_t=S_0e^{\left(r-\frac{1}{2}\sigma^2\right)t}e^{\sigma W_t}\leftrightarrow W_t = \frac{\ln\frac{S_t}{S_0}-\left(r-\frac{1}{2}\sigma^2\right)t}{\sigma}\sim\mathcal{N}(0,t)\tag{1} \end{equation} $\leftrightarrow$ is a bijection, so the probability can be shifted over $W_t$, but... how? \begin{equation} {\Pr}_s(S_T>X)=:{\Pr}(S_T>X|S_t=s)\overset{?}{=}{\Pr}\left(W_T-W_t>-{\frac {\ln{\frac {S_t}{X}}+\left(r-{\frac {1}{2}}\sigma ^{{2}}\right)(T-t)}{\sigma}}\right) \end{equation}
I don't understand how to get to the last fraction on the right from $(1)$.