# 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.

• What is the payout if this option if the value hasn't gone above the barrier level? – Sanjay Mar 3 '19 at 21:45
• Consider accepting the answer if your question has been answered. – Sanjay Mar 6 '19 at 16:08

In this answer I assume that $$K$$ is the strike and the payoff at time t=3 is $$X$$:
$$X= \begin{array}{cc} \{ & \begin{array}{cc} (S_3-K)^+ & \text{if }S_1,S_2,S_3\leq9 \\ 0 & \text{ else } \\ \end{array} \\ \end{array}$$
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 $$S_3 then there is a probability that the $$S_t>B$$ for $$t<3$$. This fact will also push the price at $$t=0$$ down compared to vanilla call.