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?

  • 2
    $\begingroup$ 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. $\endgroup$ Jun 14, 2019 at 16:29
  • $\begingroup$ Please look at page 4, first line of text. According to this article it is martingale?? janroman.dhis.org/finance/SABR/ZABR%20Andreasen.pdf $\endgroup$
    – econmajorr
    Jun 14, 2019 at 17:42
  • 1
    $\begingroup$ 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. $\endgroup$ Jun 14, 2019 at 17:48

1 Answer 1


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 $s<t\leq T$: $$\begin{align} E^Q_s\left(c_t\right)&=E^Q_s\left(E^Q_t\left(D_{t,T}(S_T-K)^+\right)\right) \\[3pt] &=E^Q_s\left(D_{t,T}(S_T-K)^+\right) \\[3pt] &=E^Q_s\left(\frac{D_{s,t}}{D_{s,t}}D_{t,T}(S_T-K)^+\right) \\ &=\frac{c_s}{D_{s,t}}\end{align}$$ 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 $s<t$, 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.


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