Valuing a Reverse Knock-Out Call with a Recombining Tree
Summary
The document explains how to value a reverse knock-out call in a finite, discrete price tree. The option expires worthless if the underlying has crossed an upper barrier at any point before maturity. The suggested procedure is to assign zero payoff to paths that breach the barrier, calculate the terminal call payoff for surviving paths, and use risk-neutral dynamic programming to work backward through the tree.
The example lists eight price paths, a strike of three, an upper barrier of nine, and zero interest. The accepted response gives terminal payoffs for each path and notes two reasons the barrier option is worth less than a vanilla call: its payoff is capped by the barrier minus strike, and paths may be knocked out before maturity even if they finish below the barrier. The example illustrates payoff construction, but it does not show the node probabilities or complete backward valuation, so it is not a full numerical price derivation.
Key ideas
- A reverse knock-out call pays zero if the asset crosses its upper barrier before expiry.
- Set breached paths to zero and calculate ordinary call payoffs on paths that survive.
- Risk-neutral dynamic programming can value the resulting path-dependent payoff.
- The barrier limits possible payoff and creates additional ways for the option to expire worthless.
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Full text
# Barrier Option Valuation
# Barrier Option Valuation
Good day,
> A reverse knock-out barrier call option expires worthless if the asset price ever goes above a given barrier level. Calculate the value of this barrier option struck at $K = 3$ with barrier level $B = 9$. Also, explain why the barrier call option worth less than the vanilla call? $r=0$
\begin{array}{|c|c|c|c|} \hline & S(t=0,\omega) & S(t=1,\omega)^* & S(t=2,\omega)^* &S(t=3,\omega)^* \\ \hline \omega_1 & 5& 8& 11 &15\\ \hline \omega_2 & 5& 8& 11 &10\\ \hline \omega_3 & 5& 8& 7 &10\\ \hline \omega_4 & 5& 8& 7 &5\\ \hline \omega_5 & 5& 4& 7 &10\\ \hline \omega_6 & 5& 4& 7 &5\\ \hline \omega_7 & 5& 4& 2 &5\\ \hline \omega_8 & 5& 4& 2 &1\\ \hline \end{array}
I found risk neutral probabilities for each path and at each node and I think I am correct (hard to go wrong as $r=0$ and no dividends are paid.I calculated the value of a vanilla call option using dynamic programming but Im not quite sure how to approach the Barrier option valuation. Do I simply put the paths with over $9$ to be equal to $0$ and apply dynamic programming again?
The additional question; its worth less as there is a range of values for which the option is worth anything whereas a vanilla option has only a minimum.
## Answer by Sanjay (score 2, accepted)
https://quant.stackexchange.com/a/44393
In this answer I assume that $K$ is the strike and the payoff at time t=3 is $X$:
$X= \begin{array}{cc} \{ & \begin{array}{cc} (S_3-K)^+ & \text{if }S_1,S_2,S_3\leq9 \\ 0 & \text{ else } \\ \end{array} \\ \end{array}$
Please correct me if I am mistaking.
The answer to your question is yes! The pay-offs are given as \begin{array}{|c|c|} \hline & \text{Pay-Off} \\ \hline \omega_1 & 0 \\ \hline \omega_2 & 0 \\ \hline \omega_3 & 0 \\ \hline \omega_4 & 2 \\ \hline \omega_5 & 0 \\ \hline \omega_6 & 2 \\ \hline \omega_7 & 2 \\ \hline \omega_8 & 0 \\ \hline \end{array}
Naturally the the payoff of the option has an upper bound $B-K$ and the vanilla call does not have this bound, so the the vanilla is worth more. This is pretty trivial. Even if $S_3<B$ then there is a probability that the $S_t>B$ for $t<3$. This fact will also push the price at $t=0$ down compared to vanilla call.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.