Pricing a call option with payoff function max{$S_T - S_{T/2}, 0$}$C=\max\{S_T - S_{T/2}, 0\}$, where $S_T$ is geometric brownian motion. AppreciateI appreciate any help! Please close this question if this is a duplicated question. Thanks all!
My approach is to take out $S_{T/2}$,:
C = $S_{T/2}$ max{$\frac{S_T}{S_{T/2}} - 1, 0$}$$C = S_{T/2} \max\left\{\frac{S_T}{S_{T/2}} - 1, 0\right\}$$
thenThen we can define a risk neuralneutral measure:
E[C] = E[$S_{T/2}$ max{$\frac{S_T}{S_{T/2}} - 1, 0$}] = $\tilde{E}$[max{$\frac{S_T}{S_{T/2}} - 1, 0$}]$$\begin{align} E[C] & = E\left[S_{T/2} \max\left\{\frac{S_T}{S_{T/2}} - 1, 0\right\}\right] \\[3pt] & = \tilde{E}\left[\max\left\{\frac{S_T}{S_{T/2}} - 1, 0\right\}\right] \end{align}$$
where we trickuse $S_{T/2}$ as the numeraire. Then plug into the call option BSBlack-Scholes formula with $S=1$, $K=1$ and $r=0$.
Is this approach correct? My main concern is that, can Iwe use $S_{T/2}$ as the numeraire?