# Black Scholes Geometric Brownian Motion Option Pricing

I'm doing a past paper for one of my masters modules and I'm stuck on this and I have no idea how to tackle such a thing. It's worth 30% of the exam so would be great if anyone here has any suggestions.

Thanks

## 1 Answer

For Question (1), the payoff is given by \begin{align*} (S_T-S_{T_0})^+. \end{align*} Note that \begin{align*} E\left(e^{-r(T-T_0)} (S_T-S_{T_0})^+ \mid \mathscr{F}_{T_0}\right) &=S_{T_0}N(d_1)-e^{-r(T-T_0)}S_{T_0}N(d_2), \end{align*} where \begin{align*} d_1 &= \frac{(r+\frac{1}{2}\sigma^2)(T-T_0)}{\sigma \sqrt{T-T_0}}\\ &=(r+\frac{1}{2}\sigma^2)\sqrt{T-T_0}/\sigma, \end{align*} and \begin{align*} d_2 = d_1 - \sigma \sqrt{T-T_0}. \end{align*} The option price is then given by \begin{align*} C_0 &\equiv e^{-rT}E\left( (S_T-S_{T_0})^+ \right)\\ &=e^{-rT_0}E\left(E\left(e^{-r(T-T_0)} (S_T-S_{T_0})^+ \mid \mathscr{F}_{T_0}\right)\right)\\ &=e^{-rT_0}E\left(S_{T_0}N(d_1)-e^{-r(T-T_0)}S_{T_0}N(d_2) \right)\\ &=S_0 \big[N(d_1) - e^{-r(T-T_0)} N(d_2)\big].\tag{*} \end{align*}



For Question (2), note that \begin{align*} \max(S_T,\, S_{T_0}) = S_{T_0} + (S_T-S_{T_0})^+. \end{align*} Then, from (*), the contingent price is given by \begin{align*} P_0 &\equiv e^{-rT}E\left( S_{T_0} + (S_T-S_{T_0})^+ \right)\\ &=e^{-r(T-T_0)}S_0 + S_0 \big[N(d_1) - e^{-r(T-T_0)} N(d_2)\big]\\ &=S_0 \big[N(d_1) + e^{-r(T-T_0)} (1-N(d_2))\big]\\ &=S_0 \big[N(d_1) + e^{-r(T-T_0)} N(-d_2)\big]. \end{align*}