# Exotics - Combination of different payoffs using Black-Scholes

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

• 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. Commented Jul 21, 2021 at 12:26
• 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 $\max\{S_T-z,y-S_T\}$. Commented Jul 21, 2021 at 12:38
• You can also represent it as combination of options. Notice that $max(S_T-z, y-S_T)=max(y-S_T, 0)+2*max(S_T-(y+z)/2, 0)-max(S_T - y, 0)$
– emot
Commented Jul 21, 2021 at 19:11
• 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 $d_2=d_1-\sigma \sqrt T$
– emot
Commented Jul 22, 2021 at 12:22