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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.

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

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