How would you price the following option on underlying $S$ without dividends?

Time to maturity of option $\tau = 12$ months

Option has a strike $K > 0$ and constant barrier $B > 0$.

$t_0$ is the current point in time, while ${t_1, t_2,...t_{12}}$ are observation (potential trigger) dates and $t_{12}$ is the expiration date.

When the underlying is above the barrier at any of the observation dates, the option is called (similar to up and in call) and the difference between underlying at the observation date $S_{t_i}$ and strike $K$ is payed out.

Assume further for our example:

$ 0 < S_{t_0} < B $

$ K = S_{t_0}$ (ATM)

At each observation date, i.e. ${t_1, t_2,...t_{11}}$, excluding the expiration date, the payoff looks as follows:

$$ \begin{align*} if \quad S_{t_i} \geq B: \quad payoff = max(S_{t_i} - K, 0)\\ elif \quad S_{t_i} < B: \qquad \qquad \qquad payoff = 0 \end{align*}$$

The first case in our setting is per definition positive and implies that option is triggered / redeemed

At maturity / expiration if option hasn't been triggered on one of the observation dates, the payoff is independent of the barrier:

$$ \quad payoff = S_{t_i} - K $$

i.e. you get back the payoff of holding the stock over that the whole maturity when $K = S_{t_0}$

In other words, over the whole maturity the option bears the entire downside potential, but each month the option can be triggered if the stock price level is above the barrier $B$ and is then also redeemed at that observation date $t_i$. At maturity you essentially get back the payoff of holding purely the stock if we assume that $ K = S_{t_0}$.

I´d argue that the profile can be replicated dynamically by shorting a 12 month put and being long 1 month calls at the beginning of each month conditional on being knocked out in the prior period.

Simple two step binomial tree

  • $\begingroup$ Aren't there some contradictory statements in the problem description ? You start with saying that the difference between S and B is payed out if S > B, but then describe payoff=max(S-K,0) when S > B at each observation date. At first I would not think that you could replicate it exactly with a series of vanilla options as the next observation values are dependent on what happen previously: has it been KO or not. It may be replicated with a series of discrete barrier options. $\endgroup$ – jherek Mar 20 '19 at 11:31
  • $\begingroup$ Yes, sorry my mistake it should say difference between S and K. I edited it. Also I think you´re right about the path-dependence...so would you say there is no closed form solution, rather do it with simulation? $\endgroup$ – Alfi Mar 20 '19 at 11:37
  • $\begingroup$ There exists a "closed-form" solution for these kinds of options in a Black-Scholes setting - see the paper "The Quintessential Option Pricing Formula" by Skipper and Buchen. I write "closed-form" because it will involve a 12-dimensional cumulative normal distribution. $\endgroup$ – LocalVolatility Mar 21 '19 at 8:57
  • $\begingroup$ I see, thanks @LocalVolatility. Unfortunately I´m helplessly impractical & only know how to approach these type of Qs from the BS-type of setting. And actually I´m not looking for a neat formula that tells me the theoretical premium I should expect under some model. Rather how would you price/replicate this payoff dynamically in practice? $\endgroup$ – Alfi Mar 21 '19 at 12:07
  • $\begingroup$ Hmm why isn’t this replicable by just holding a 12m forward on the stock struck at K? If it gets triggered early, say at t, just borrow the stock in (t,12m) and invest the premium K for that period. If interest rates are zero, it replicates the structure doesn’t it ? Another way of aging it is: the payoff is always S-K, so we’re just talking about timing. $\endgroup$ – dm63 Mar 22 '19 at 3:31

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