Now that we have a function for the asset price, we can commence our foray into derivatives. Here we prove the function satisfied by a European call option. The function $$ w \rightarrow S(0)e^{(r \; – \sigma^2/2)T +\sigma \sqrt{T} w} \; – K \geq 0 $$ if and only if $w(K) \leq w$ as the function is strictly increasing. Careful observation of the exponent in the equation for $K$ compared to the exponent in $S(T) – K$ also reveals this. Hence
\begin{align}
\mathbb{E} f_c (S(T), T) &= \frac{1}{\sqrt{2 \pi}} \int_{-w(K)}^\infty \Big(S(0)e^{(r \; – \sigma^2/2)T + \sigma\sqrt{T}w} \; – K \Big)e^{-w^2/2} dw
\\ \\
&= S(0)e^{(r \; – \sigma^2/2)T} \frac{1}{\sqrt{2 \pi}} \int_{-w(K)}^\infty e^{-\frac{1}{2}(w^2 -2\sigma \sqrt{T} w)} dw \quad – \quad \frac{1}{\sqrt{2 \pi}} \int_{-w(K)}^\infty K e^{-w^2/2} dw \\ \\
&= S(0)e^{(r \; – \sigma^2/2)T} \frac{1}{\sqrt{2 \pi}} \int_{-w(K)}^\infty e^{-\frac{1}{2}(w^2 -2\sigma \sqrt{T} w)} dw \; – K \Phi (w (K))\\ \\
&= S(0)e^{rT}\underbrace{ \frac{1}{\sqrt{2 \pi}} \int_{-w(K)}^\infty e^{-\frac{1}{2}(w^2 -2w \cdot \sigma \sqrt{T} \; + \sigma^2 T)} dw}_{\Phi (w (K)\; – \; \sigma \sqrt{T})} \; – K \Phi (w (K))
\end{align} observe carefully that the exponent $e^{-\frac{1}{2}(w \; – \sigma \sqrt{T})^2}$ means the remaining integral is the distribution $\Phi (w (K))$ shifted by $\sigma \sqrt{T}$. Hence
\begin{align}
\mathbb{E} f_c (S(T), T) &= S(0)e^{rT}\Phi (w (K)\; – \; \sigma \sqrt{T}) \; – \; K \Phi (w (K)).
\end{align}