Static Replication of Linear Derivative Payoffs
Summary
The document clarifies the relationship between a derivative’s payoff shape, its delta, and the need for a pricing model. A claim with payoff linear in the underlying price, expressed as a times the stock price plus b, can be replicated by a position in the stock and a zero-coupon bond. This static hedge provides the same maturity payoff, allowing arbitrage-based pricing without a model such as Black–Scholes.
The discussion corrects the proposed delta criterion: a linear payoff does not necessarily have delta equal to one, since its slope may be another value. The relevant feature is replicability, rather than whether the payoff is nonnegative. Payoffs that lack a static hedge generally require a model for valuation. The explanation is brief and framed around European claims; it does not detail market assumptions, hedging costs, or how to value claims when replication is unavailable.
Key ideas
- A linear payoff of the form a times the underlying price plus b can be replicated with stock and a zero-coupon bond.
- A linear payoff does not imply a delta of one because its slope can differ from one.
- Static replication supports arbitrage-based pricing without a stochastic pricing model.
- Payoff replicability, rather than nonnegativity, is the central distinction in the explanation.
- Claims without a static hedge generally require a pricing model.
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Full text
# Delta of a derivative with a linear payoff # 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? ## Answer by dm63 (score 4) https://quant.stackexchange.com/a/37464 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.
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