Why a Discounted Option Price Is a Martingale
Summary
The document explains why risk-neutral valuation does not imply that the discounted call payoff expectation equals the discounted intrinsic value calculated from today’s stock price. The payoff function of a call is nonlinear, so the positive-part payoff at a future date is not itself the price of a traded asset and does not generally follow the martingale property.
In the Black–Scholes setting described, the option can be replicated using the stock and money-market account. Its value is the conditional risk-neutral expectation of its discounted future payoff, grown to the current date. Once that replicable option is treated as an asset, its price divided by the money-market account is a martingale. The derivation uses conditional expectation and the law of iterated expectations to connect option values across dates. The explanation relies on the stated replication and risk-neutral framework; it does not discuss dividends, transaction costs, or deviations from Black–Scholes assumptions.
Key ideas
- The discounted price of a traded asset is a martingale under the risk-neutral measure in the framework described.
- A call payoff alone is not the price process of a traded asset.
- A replicable option’s value is the conditional risk-neutral expectation of its discounted payoff, multiplied by the money-market account value.
- Conditional expectation shows that the discounted option value satisfies the martingale property.
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Full text
# Discounted price of an option
# Discounted price of an option
If the discounted price of any asset is a martingale under risk neutral measure, why is $E^Q[e^{-rT} (S_T-K)_+ | F_t]$, not merely $e^{-rt} (S_t-K)_+$?
This is something I wanted to clarify, since that's the definition of a martingale. Instead we use the lognormal distribution of the stock price and solve the expectation completely to get the black Scholes call price.
## Answer by Daneel Olivaw (score 6)
https://quant.stackexchange.com/a/75454
The process $Y_t:=(S_t-K)^+$ cannot be the price of a traded asset because of Jensen's inequality. Instead, it is the price of the option which is a martingale.
In the Black-Scholes model, the primitive market model has only two assets: the stock with price $S_t$ and the money market account (MMA) with price $B_t:=e^{rt}$. Within this market, Black and Scholes prove that it is possible to replicate a European (call) option with payoff $(S_T-K)^+$ at some future expiry $T>t$. The value $V$ of this claim is given by its risk-neutral expectation: $$V_t=B_tE^{\mathbb{Q}}\left(\left.\frac{(S_T-K)^+}{B_T}\right|\mathscr{F}_t\right)\tag{1}$$ where $B_t/B_T=e^{-r(T-t)}$.
Given the option can be replicated, it can be viewed as an asset. One can then consider an "augmented" market model with the stock, the MMA and the option. Per risk-neutral theory, it is the discounted price of an asset which is a martingale, that is if $P$ is the price of an asset then the process $$\frac{P_t}{B_t}$$ is a martingale. The price of the option is $V$ therefore letting $s<t$: \begin{align} B_sE^\mathbb{Q}\left(\left.\frac{V_t}{B_t}\right|\mathscr{F}_s\right) &=B_sE^\mathbb{Q}\left(\left.\frac{B_tE^{\mathbb{Q}}\left(\left.\frac{(S_T-K)^+}{B_T}\right|\mathscr{F}_t\right)}{B_t}\right|\mathscr{F}_s\right) \\ &=B_sE^\mathbb{Q}\left(\left.E^{\mathbb{Q}}\left(\left.\frac{(S_T-K)^+}{B_T}\right|\mathscr{F}_t\right)\right|\mathscr{F}_s\right) \\[7pt] &\overbrace{=}^{\text{LIE}}B_sE^\mathbb{Q}\left(\left.\frac{(S_T-K)^+}{B_T}\right|\mathscr{F}_s\right) \\[3pt] &\overbrace{=}^{\text{(1)}}V_s \end{align} where we have used the Law of Iterated Expectations (LIE) and the definition $(1)$. Dividing by $B_s$: \begin{align} E^\mathbb{Q}\left(\left.\frac{V_t}{B_t}\right|\mathscr{F}_s\right) &=\frac{V_s}{B_s}, \end{align} hence the discounted price of the option is indeed a martingale.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.