I'm currently struggling with the derivation of a formula to price the following exotic option with Black-Scholes.

The option has the maximum payoff of $(S_T-z)$ and $(y - S_T)$, where $S_T$ is the price of the underlying at maturity $T$, $X$ the strike price, and $z$ and $y$ are constants. E.g. $z = 20$ and $y = 40$.

My first approach was to build a formula where I combine the payoff of $max\{(S_T-z);0\}$ and $max\{(y-S_T);0\}$.

$ d_1 =\frac{\ln(\frac{X}{S_0})+(r_f-0.5*\sigma^2)*T}{\sigma*\sqrt{T}} $ and $d_2 = d_1 - \sigma*\sqrt{T}$

$Price = (S_0*\phi{(d_1)}-a*e^{-r_fT}*\phi{(d_2)})+(b*e^{-r_fT}*\phi{(-d_2)}-S_0*\phi{(-d_1)}) $

EDIT: Thanks to the notes I came up with a second approach on how to price this option:

Payoff structure $max(S_T-z;y-S_T) = y - 2*max(S_T-(y+z)/2,0)$, where the price is represented by

$y-S_0*e^{-r_fT} + 2*(S_0*\phi(d_1)-0.5*(y+z)*e^{-r_fT}*\phi(d_2))$

Is this correct and do I have to adjust $d_1$ as well?

So that it would be $d_1 = \frac{\ln(\frac{0.5*(z+y)}{S_0})+(r_f-0.5*\sigma^2)*T}{\sigma*\sqrt{T}}$

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    $\begingroup$ Note that $\max(S_T-z, y-S_T) = y-S_T+2\max(S_T- (y+z)/2, \, 0)$. You can now use Black-Scholes. $\endgroup$ – Gordon Jul 21 at 12:26
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    $\begingroup$ You are trying to price a chooser option, which a type of compound option. Your approach of summing the payoffs is not correct: if $z<S_T<y$, then per your approach you will receive a payoff from both trades, whereas in reality you just receive $\max\{S_T-z,y-S_T\}$. $\endgroup$ – Daneel Olivaw Jul 21 at 12:38
  • $\begingroup$ You can also represent it as combination of options. Notice that $max(S_T-z, y-S_T)=max(y-S_T, 0)+2*max(S_T-(y+z)/2, 0)-max(S_T - y, 0)$ $\endgroup$ – emot Jul 21 at 19:11
  • $\begingroup$ Peet: regarding your edit: your formula is incorrect, it should be $ye^{-rT}-S_0+2*(N(d_1)S_0-N(d_2)*(y+z)*0.5e^{-rT})$ where $d1$ and $d2$ is formula of 0.5*(z+y), i.e. standard d1 where strike $K=0.5*(y+z)$ then: $d_1=\frac{1}{\sigma \sqrt T} * [ ln( \frac{S_0}{0.5*(z+y)})+(r+0.5\sigma^2)*(T-t)]$ and $d_2=d_1-\sigma \sqrt T$ $\endgroup$ – emot Jul 22 at 12:22

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