0
$\begingroup$

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

New contributor
Peet is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
4
  • 5
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

Peet is a new contributor. Be nice, and check out our Code of Conduct.

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.