# 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?

• No it isn't. Only discounted asset prices can be martingales. A discounted European call is a martingale under any model. To prove it, note that the price of an European call is a conditional expectation, then use the Law of Iterated Expectations. Jun 14, 2019 at 16:29
• Please look at page 4, first line of text. According to this article it is martingale?? janroman.dhis.org/finance/SABR/ZABR%20Andreasen.pdf Jun 14, 2019 at 17:42
• In the article you link rates are assumed to be null, so in that case yes because the undiscounted call price is equal to the discounted call price. Jun 14, 2019 at 17:48

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.