# Is the undiscounted value process of a Euro call option under Bachelier model a Martingale? [duplicate]

Assume that $$c_t$$ is the UNDISCOUNTED price process for a European call option in Bachelier model. In Bachelier model call option pricing formula the formulas is discussed. The undiscounted value process is $$c_t = (S_t-K)\Phi( \frac{S_t-K}{\sigma\sqrt{T-t}})+\sigma\sqrt{T-t}\phi( \frac{S_t-K}{\sigma\sqrt{T-t}})$$.

Is $$c_t$$ a martingale process?

My personal guess is YES, because of the first fundamental theorem of asset pricing. Am I correct?

## marked as duplicate by Daneel Olivaw, skoestlmeier, byouness, Helin, Attack68♦Jun 19 at 5:51

Let $$c_t$$ be the price of an European call with maturity $$T$$ and $$D_{t,T}$$ the discount factor from $$T$$ to $$t$$. We assume deterministic rates. Then note that for $$s: \begin{align} E^Q_s\left(c_t\right)&=E^Q_s\left(E^Q_t\left(D_{t,T}(S_T-K)^+\right)\right) \\[3pt] &=E^Q_s\left(D_{t,T}(S_T-K)^+\right) \\[3pt] &=E^Q_s\left(\frac{D_{s,t}}{D_{s,t}}D_{t,T}(S_T-K)^+\right) \\ &=\frac{c_s}{D_{s,t}}\end{align} because $$D_{s,t}D_{t,T}=D_{s,T}$$. The second inequality stems from the fact that: $$E^Q_s(E^Q_t(\cdot))=E^Q_s(\cdot)$$ if $$s, this is the Law of Iterated Expectations. From the last equation you see $$c_t$$ is not a martingale, however rearranging: $$E^Q_s\left(D_{s,t}c_t\right)=D_{s,s}c_s=c_s$$ Thus the discounted call price is a martingale. As you can see this is a model-free result.