Pricing a Put-on-Call with a Binomial Tree

# Put-on-call option confusion


# Put-on-call option confusion












So the question asks: Given a 3-steps Binomial Tree model with $S(0) = 50$, $U = 20%,D = 􀀀20%$, and $R = 5%$. A European call option has the strike price $X = 40$ and maturity time $T = 3$. Also, a Put-on-Call option is written on this European call option with maturity time T = 2, i.e, the final payoff (at T = 2) of PoC option is given by H(T) = max ($K_p 􀀀-C_E(2), 0)$, where $C_E(2)$ is the value (price) of the underlying European call option at T = 2. The strike price of the outside put option $K_p = 12$. Find out the initial price of this Put-on-Call option.

So so far I have: Where I constructed the binomial tree model for the European call option only ( without the put-on-call option). So I got the European call option price is $16.4669042$.

But what exactly is the $C_E(2)$, the value (price) of the underlying European call option at T = 2?

Is it C(2) = max{0, 0.5(S(1)+S(2)) - X}? If so I then have three C(2)?

Also, what is the final payoff H(T) at T=2? Where should it go on the graph?

## Answer by Quantuple (score 2)

https://quant.stackexchange.com/a/25108

Your confusion certainly comes from the fact the underlying on which the put option is written is not the stock anymore but rather another option. We call that a compound option in derivatives lingo.

How would you price the option if it were a simple put expiring at $T=2$ and not an obscure put-on-call? Well, you would have started at the end of the period representing the maturity of your option (here the put expiries at $T=2$, so this is the end of the second period) and for each tree node you found there, call them $S_{uu}=S(0)(1+u)^2, S_{ud}=S(0)(1+u)(1+d)$ and $S_{dd}=S(0)(1+d)^2$ for up-up, up-down and down-down states, you would have set the option value as equal to its payoff. For a standard put option, because you know the value which the stock takes on the 3 different tree nodes, this would have given: $P_{uu}=\max(K_p - S_{uu},0)$, $P_{ud}=\max(K_p - S_{ud},0)$ and $P_{dd}=\max(K_p - S_{dd},0)$. Once done, you would have then worked your way up to the root by backwards induction, computing $P_{u}$ and $P_{d}$ and finally inferring $P$.

The only thing that changes here, is that the underlying consists of a call option rather than the stock. At the end of the second period, your put-on-call option is therefore worth $P_{uu}=\max(K_p - C_{uu},0)$, $P_{ud}=\max(K_p - C_{ud},0)$ and $P_{dd}=\max(K_p - C_{dd},0)$. This is precisely what is meant by the formula $$H(T=2) = \max(K_p - C_E(2), 0)$$ where $C_E(2)$ represents the value at $T=2$ of a European call option with maturity $T=3$ and strike $X$ as given in your exercise.

Now finding $C_E(2)$ (in other words the values $C_{uu}, C_{ud}$ and $C_{dd}$) is a standard binomial tree pricing problem and is exactly what you already did above:

- Use a 3-period binomial tree. Start at the end of the tree and set the option value as equal to its payoff, that is: $C_{uuu}=\max(S_{uuu}-X,0)$, $C_{uud}=\max(S_{uud}-X,0)$, $C_{ddu}=\max(S_{ddu}-X,0)$ and $C_{ddd}=\max(S_{ddd}-X,0)$). Then move backwards by one period to obtain the desired values $C_{uu}, C_{ud}$ and $C_{dd}$.

- Now that you understood how it works, you see that you could also have simply 'extended' the tree on which you were working to price the put(-on-call) option and use the extra part to find the value at $T=2$ of the call option.

Anyway, once you have determined the values $P_{uu}$, $P_{ud}$ and $P_{dd}$ of the put-on-call option at $T=2$ (the values you came up with for $C_{uu}$, $C_{ud}$ and $C_{dd}$ in your tree above i.e. $33.905$, $10.476$ and $0$ are perfectly fine), you can go back to using the standard approach. Simply move backwards computing $P_{u}$ and $P_{d}$ and finally $P$ and you're done!