Pricing European Puts from a Risk-Neutral Return Distribution

Summary

The document asks whether a European put’s price can be calculated as the expected payoff when the underlying’s return over the option’s life follows a specified density. Assuming zero interest rates, it writes the put payoff as the strike minus the terminal underlying price when that price is below the strike, and zero otherwise. It then expresses the price as an integral over the return distribution, restricted to outcomes where the put finishes in the money.

The answer says this method gives the correct price when the density is risk-neutral, and that a risk-neutral terminal distribution can also be used to derive the Black–Scholes formula. The key qualification is the measure: an ordinary real-world return distribution does not generally give the arbitrage-free option price directly. The document assumes a finite expectation and zero rates, and it does not develop how to obtain the risk-neutral density or handle discounting when rates are nonzero.

Key ideas

• A European put payoff depends on whether the terminal underlying price is below its strike.
• With zero interest rates, the risk-neutral price is the expected payoff under the risk-neutral distribution.
• The return density must be risk-neutral; a real-world distribution does not generally determine the arbitrage-free price.
• The integral representation assumes the payoff expectation is finite and omits nonzero-rate discounting.

Tags

# Option pricing under distribution assumption


# Option pricing under distribution assumption












For simplicity assume zero interest rates in the following.

Given the price of a (European) put option with strike K and maturity T at time point t. $P_t(K, T)$ for a given underlying S with values $S_t$ at time point t.

Assuming you are currently at timepoint $t_0$. The return of S over the lifespan of the option is given by $r_T=\frac{S_T-S_0}{S_0}$.

Assuming that $r_t \sim f$. For some density $f$. Is the risk-neutral price of the option equal to the expected value of the option (if it exists, i.e. is finite)? So is this true? If not, why not?

$$P_{t_0}(K, T) = \mathbb{E}(P_T(K,T)=\int_{0}^{K} K-S_T \: dP(S_T) = \int_{-1}^{\frac{K-S_0}{S_0}} K-(1+r_T)S_0\: dP(r_T) \\ = \int_{-1}^{\frac{K-S_0}{S_0}} (K-(1+x)S_0) \: f(x)\: dx$$

## Answer by MrLCh (score 1)

https://quant.stackexchange.com/a/77504

After reviewing further literature, I have come to the conclusion that indeed this method gives the correct answer. This thought process can be used to derive BS-formula, given the (risk-neutral) density of the "results" of the underlying stochastic process.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.