Why Undiscounted Option Values Are Martingales Under a Forward Measure
Summary
The document explains why the undiscounted value of a European option can be a martingale, using the maturity bond as numeraire. Under the associated forward measure, the option value divided by the bond price equals the conditional expectation of the payoff at maturity. The tower property of conditional expectations then gives the martingale relation over time.
This argument is general and does not depend on the Bachelier option pricing formula; it applies to other models when the relevant payoff is integrable and the pricing setup is valid. With deterministic interest rates, bond prices and the bank account are deterministic, so the forward measure agrees with the usual risk-neutral measure. The explanation distinguishes the measure and numeraire under which a value process is a martingale; it does not assert that undiscounted prices are martingales under every measure or with stochastic rates.
Key ideas
- Under the maturity forward measure, an undiscounted payoff value is its conditional expected payoff.
- The tower property implies that this conditional expectation process is a martingale.
- The argument applies independently of whether option prices follow Bachelier, Black–Scholes, or another model.
- With deterministic rates, the forward measure coincides with the standard risk-neutral measure.
- The martingale statement depends on the chosen numeraire and measure, as well as integrability.
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# Is the Non-discounted Bachelier call option price a Martingale?
# Is the Non-discounted Bachelier call option price a Martingale?
My math finance professor once said someting that I can't make sense of. Hope you can answer:
> For a foward process the non-discounted price for a European call option under Bachelier is $$C_t = \left(f_t-K^*\right)\Phi\left(\frac{f_t-K^*}{v(t)}\right) +v(t)\,\phi\left(\frac{f_t-K^*}{v(t)}\right)$$ $C_t$ is a martingale
How come $C_t$ is a Martingale? I have been through most of Bjork's book Arbitrage Theory and I know that the fair valur of a derivative $X$ is $$E^Q_t\left[\frac{X}{B_t}\right]$$which is a Martingale, right?
How come the non-discounted Bachelier call option price is martingale?
However this is from my personal notes, so I might have it wrong.
## Answer by Kevin (score 4, accepted)
https://quant.stackexchange.com/a/46640
Let $P(t,T)$ denote the time $t$ price of a zero-coupon bond maturing at time $T$ and $\mathbb{Q}_T$ be the associated equivalent martingale measure which uses $P(t,T)$ as numeraire. Then, for any $\mathcal{F}_T$-measurable payoff $\xi$, the time $t$ value of $\xi$ is given by $$V_t=P(t,T)\cdot\mathbb{E}^{\mathbb{Q}_T} [\xi\mid\mathcal{F}_t].$$ The undiscounted time $t$ price is given by $$\tilde{V}_t = \frac{V_t}{P(t,T)} = \mathbb{E}^{\mathbb{Q}_T} [\xi\mid\mathcal{F}_t].$$ And indeed, $(\tilde{V}_t)$ is a $\mathbb{Q}_T$-martingale. Assuming integrability and adaptness (trivial), we need to show the martingale property. To this end, let $0\leq s<t\leq T$. Then, by the tower law, \begin{align*} \mathbb{E}^{\mathbb{Q}_T}[\tilde{V}_t\mid\mathcal{F}_s] &= \mathbb{E}^{\mathbb{Q}_T}\left[\mathbb{E}^{\mathbb{Q}_T} [\xi\mid\mathcal{F}_t]\bigg|\mathcal{F}_s\right] \\ &= \mathbb{E}^{\mathbb{Q}_T}[\xi\mid\mathcal{F}_s] \\ &= \tilde{V}_s. \end{align*}
Please note the following:
- This result is completely independent of the Bachelier model and equally applies to the Black-Scholes model, the Heston model and others.
- If interest rates are deterministic, so are bond prices and back accounts. Thus, the forward measure $\mathbb{Q}_T$ coincides with the ``standard'' risk-neutral measure $\mathbb{Q}$ which uses a risk-free bank account $B_t=e^{\int_0^t r(s)\mathrm{d}s}$ as numeraire.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.