Here is a problem in Hull's book and the given solution:

enter image description here

  1. My approach was to compute the profit $\pi = \pi_{SP} + \pi_{LC}$ (short put, long call).

One can show that $\pi = \pi_{SP} + \pi_{LC} = S_T - K + p_{SP} - p_{LC}$, where $p_{SP}, p_{LC}$ are the prices of the options.

So if we want $p_{SP} = p_{LC}$, then we must have $\pi = S_T - K$

Is that right?

  1. I'm not quite sure I understand the last two sentences of the answer in the solutions manual. What is the difference between forward price and delivery price? What I think the answer means:

Because the payoff from LC and SP (long call and short put) is $S_T - K$, because the payoff from a fwd contract is $S_T - F$ and because a fwd contract is LC and SP combined, $p_{SP} = p_{LC}$ when $K = F$?

Am I understanding right? How exactly does the conclusion follow if so? If not, what exactly does the solutions manual mean to say?


You are mostly right, I don't really get what you don't understand. The answer in the book is quite clear, but let me put it that way :

Selling a put and buying a call on the same underlying $S$ with same maturity and same stike $K$ is always equivalent to a long position in a forward contract on $S$ with delivery price $K$. The easiest way to see that is to draw the payoff of such a strategy in a simple graph, you should get a 45° line crossing the $x$-axis at $K$. Then we know that a forward contract with delivery price $F$ (the forward rate) costs nothing. Then if you want your strategy to cost nothing, you should set $K=F$.

It seems that it's the beginning of the book, but you may want to look at the Put-Call parity. Without details, it is a relation that links the price of a call with the price of a put on the same underlying, same maturity and same strike. The Put-Call parity can be expressed as follows (with $r$ the risk-free rate) : \begin{equation} C-P=S-Ke^{-rt} \end{equation}

With $F = Se^{rt}$ you can see that : \begin{equation} C-P=e^{-rt}(F-K) \end{equation}

So we have $F=K$ $\iff$ $C=P$. The demonstration should be in the Hull's book.

  • $\begingroup$ Thanks Louis. B. I was thinking of doing something like the latter part you mentioned, but that's in a later part of the book, as you pointed out. As for the former part.... $\endgroup$ – BCLC Nov 3 '15 at 20:09
  • $\begingroup$ ...the thing is, the original problem doesn't mention any forward contracts. So what is 'F' ? So the solutions manual is trying to say that the trader's position is equivalent to a forward contract w/ delivery price F that should be equal to K? How does that answer the question? The solutions manual seems to be creating another scenario and then saying that the new scenario is equivalent to the first scenario iff the 'K' in the first scenario equals the 'F' in the second scenario... $\endgroup$ – BCLC Nov 3 '15 at 20:15
  • $\begingroup$ ...anyway, I think I understand a little better: The payoff turns out to be $S_T - K$, the same as the payoff of a forward contract and since it costs nothing to enter a forward contract (what the trader's strategy practically is), it follows that the cost needed to get the payoff $S_T - K$ should be 0. Hence $p_{SP} - p_{LC} = 0$...this seems to follow from the formulation of the problem regardless of any conditions so I guess my inference is wrong. Why? $\endgroup$ – BCLC Nov 3 '15 at 20:20
  • 1
    $\begingroup$ It is not "another" scenario, because the forward does not need to be traded in a sense. This is only to compare a strategy to a product we know (the forward contract) that helps us understand what's going on with this strategy. You can always do that because since the price (of anything btw) is the discounted payoff, two products/strategy that have the exact same payoff (in any states of nature) should have the exact same price. I think I understand your concern, but the formal proof of that is the proof of the Put-Call parity, I guess Hull just wanted to give you some insight there... $\endgroup$ – Louis. B Nov 3 '15 at 23:18
  • 1
    $\begingroup$ Yes, I guess you can think about this that way. $\endgroup$ – Louis. B Nov 4 '15 at 0:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.