please refer CMP ST6 2017, chapter 1, page 10. The first line says that combining the payoff graphs of long call and short put gives a diagonal line, representing the payoff from the underlying asset itself. In my view this line would give the payoff from a long forward, please explain.

Technically, you're correct. However, the shape of the payoff profile between the underlying share and its forward price are identical. Entering into a forward would leave you on risk from the moment you entered the transaction. Purchasing a forward merely defers when you would need to come up with the cash to pay for it. For completness (in light of the pedantic nature of the question ), the question assumes that the long call and short put are struck at the same strike and tenor. You'd have a risk reversal / diagonal type position if this wasn't the case.

Certainly both options are struck at the same price. The payoff from overall position is as follows If S(T) < K, 0 + [-{K - S(T)}] = S(T) - K If S(T) >. K, {S(T) - K} + 0 = S(T) - K This is payoff from a long forward ( a line with slope 1 passing through (K,0) ) The line representing payoff from the underlying would pass through the origin.

Is your question that you think there's a possibility for arbitrage? Using the ideas from put-call parity (European option with no dividends) the relationship between options and the forward is: c - p = exp(-r(T-t)) * [F - K]. The options are priced off the forward... where c = call, p = put, r = continously compounded risk-free rate, T-t time to expiration, F = forward price, K = strike price. There are a number of strike prices along an option chain and the payoff graph would intersect the x axis at the strike. The shape of the payoff however is unchanged and will be identical in shape to the underlying forward / share price. The key point is that you gain / lose 1-1 with changes in the underlying from the moment you enter a synthetic or actual long position in the underlying. K is a constant whilst S(T) will depend on where the underlying settles on expiration.

My original question was very simple in the sense that combining the payoffs from a long call and a short put would give the payoff f(S(T)) = S(T) - K, whereas the payoff from the underlying is f(S(T)) = S(T)

The profit / loss of the underlying from entering into a forward contract is NOT S(T). It's the difference between the terminal payoff and the amount of the underlying at the time of your original transaction. The profit / loss from entering a synthical long is the difference between the terminal payoff (S(T) in your notation) and the price it cost you to get long (i.e the strike) and the cost of putting on the original position. In the case of the option example, your payoff is positive if S(T) > K. If S(T) < K you make a loss.

That is exactly what I am saying. The statement made in the study material is not correct. S(T) - K Is the payoff from long forward and not from the underlying asset.

Prasanna, I don't have the notes and so am not able to comment on the wording. But is the point that the notes is (trying) getting at about the shape of the payoff... you make a profit 1-1 with the stock above your transaction price and lose 1-1 below your transaction price. Purchasing a stock at 100, 200, 300, 400 etc doesn't change this fact or the SHAPE of your payoff.

I'd still say it represents the payoff of the underlying asset, where you buy it at time 0 for K. It's pedantry, but in an exam I'm sure you'd get the marks either way.

Prasanna We’re on here taking time out of our day to help you. I thought your last msg was a little rude. Your training to be an Actuary... if ‘K’ was defined as the time 0 stock price would it be correct... show some flexibility*. Working out the profit / loss from stock trading is elementary. Raise your query about the error you say you’ve found with Acted and move on. *only the founder of a company would have a payoff that passed through 0 on the x axis. ALL investors would have paid a price K>0 to gain exposure to the stock. Who’s therefore wrong; you or the notes??

I am sorry if that hurts you, I just meant to say that I am not being pedantic. The notes talk of payoffs and not profit/loss. When we plot the payoff graph of any position, we plot the payoff at expiry against the underlying asset price at expiry{S(T)}. So if you combine the payoffs from a long call and short put (with same strike and tenor), you get the payoff from a long forward and we must plot f{S(T)} = S(T) - K against S(T), a line with slope 1 and through point (K,0). Notes say that the graph for combined position will represent the payoff from the underlying aset itself. The graph for the payoff from the underlying asset should plot f{S(T)} = S(T), a line with slope 1, through (0,0).

Prashanna I think you need to take a step back: In the case of a synthetic option position: c - p = max(0, S(T) - K) - max(0, K - S(T)). This is equivalent to being long the forward on inception, where the forward transaction is based on the underlining s(t) (not 0) at time t. As previously explained the options are priced off the forward - there would be an arbitrage if it wasn’t and you’d be able to make a riskless profit by buying / selling the synthetic and selling / buying the forward.* The ‘K’ notation in option parlance is typically reserved for describing option strikes. There is no ‘K’ in ‘forward parlance’. I suspect you’re over thinking this... *this is what you need to understand. I think about it all the time when I actually trade options (among many other things!)

Yeah Prasanna, your tone is a bit rude. so I'm going to leave it there. Good luck with the remaining entire course minus the first 10 pages!

Hi Patron I am a novice and I am trying to figure out that how can we deduce put-call parity from this observation as the notes claim.