# Delta of a derivative with a linear payoff

The Black-Scholes PDE can be used to price any European contingent claim with a payoff that is only dependent on the underlying's price at maturity, for instance forwards and vanilla options. In the case of forwards, the PDE is not required, c.f. the cash-and-carry technique. Is it possible to show that a delta of a European derivative is equal to one if and only if the payoff is linear in stock price, i.e. in the form $f(S) = S - K$?

I might be wrong about this, but I think any derivative with linear payoff is just a forward (even in the case where the coefficient on $S$ is not one, the derivative is a portfolio of forwards plus/minus cash). I'm basically trying to justify why one needs a model for the stock in order to price an option but not a forward, i.e. is it the non-negativity or the non-linearity of the payoff that makes pricing hard?

• Did you mean to say "...that the delta of a European derivative is constant if and only if the payoff is linear..."? Dec 22 '17 at 7:45
• Even when changing from one to constant, this is still not entirely correct as it misses the discount factor
– Bram
Dec 22 '17 at 14:00
• @LocalVolatility I clarified the question with $f(S) = S - K$, which should have delta 1. Dec 23 '17 at 4:05
• @Bram the delta of a forward contract is 1 and independent of the discount factor. I am computing the delta with respect to the forward's value not price. Dec 23 '17 at 4:06

The statement "the delta of a European derivative is equal to one if and only if the payoff is linear in stock price" is false (eg $f(S) = 2S)$. The statement "every European derivative with payoff of the form $f(S) = aS+b$ is replicable by a combination of forwards on the stock and zero coupon bonds and is therefore priceable by arbitrage without resorting to a model such as Black Scholes " is true. The latter also has a static hedge- a portfolio of stocks and bonds, purchased today, which replicate the derivative at maturity. Any payoff that does not have a static hedge requires a model to price.
• Yes, if $f(S)$ is non linear then $f'(S)$ is non constant, meaning that the hedge needs to change as a function of stock price.