1
$\begingroup$

An exotic option is described as follows:

Let $S_t$ be the underlying at $t$. The holder has the option to lock in the current price during the lifetime of the option, which he does for $S_{t}=50$. The strike price is $K=40$. At maturity, the option returns the payoff of a traditional European call or the intrinsic value at time $t$, whichever is greater. For example, if $S_T<50$ then the payoff is $10$. If $S_T>50$ then the payoff is the excess of the asset price over $50$.

I am quite confused, since by the description (if we ignore the last sentence) I understand the payoff to be simply the max of two calls: $\max(S_t-K,0)\textbf{1}_{S_T\le S_t}+\max(S_T-K,0)\textbf{1}_{S_T>S_s}$, where $\textbf{1}_{(\cdot)}$ is the indicator function.

After the last sentence, the payoff seems to be: $\max(\underbrace{S_t-K}_{=10},0) \textbf{1}_{S_T\le S_t} + \max(S_T-S_t,0) \textbf{1}_{S_T \gt S_t}$,

which is weird because e.g. for $S_T=55$ the payoff is $5$, i.e. less than $10$. If so, then the term "intrinsic value" refers to $S_T-S_t$ and not $S_t-K$.

Does this type of exotic option have a name? I tried searching "lock-in" option with no result so far. Is anyone familiar with what the text might be referring to and what the correct payoff is? Is there an error (contradiction) in the description?

$\endgroup$

1 Answer 1

9
$\begingroup$

The option described is called a "Shout" option.

$\endgroup$
1
  • $\begingroup$ Thanks a lot! So according to wikipedia my initial payoff is correct and is equivalent to $\max(S_T-K,S_t-K,0)$, meaning that the description above was poorly formulated at the end. $\endgroup$
    – PaulG
    Mar 21, 2022 at 10:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.