I want to verify that the discounted stock price process $\mathrm{e}^{-r(T-t)}V(S_t,t)$ is a martingale in the BS-model. Using Ito's formula and the BS-PDE I get that

$$ \mathrm{d}\mathrm{e}^{-r(T-t)}V(S_t,t)= \mathrm{e}^{-r(T-t)}\sigma S_t\frac{\partial V}{\partial S}(S_t,t)\mathrm{d}W_t $$

The Ito integral is a martingale if

$$ \mathbb{E}\left[\int_0^T\left(S_t\frac{\partial V}{\partial S}(S_t,t)\right)^2\right]<\infty $$

Unfortunately, I am not able to show this as I cannot apply Jensen, Hölder or Cauchy-Schwartz to eliminate the square. How do I get around this issue. A related question is whether the delta for an arbitrary option is bounded in the BS-model.

  • 1
    $\begingroup$ If you are interested in the discounted stock price, then $V \left( S_t, t \right) = S_t$..? As for your related question, see quant.stackexchange.com/questions/30177 - the delta is bounded by the slope of the payoff function in the B/S model. $\endgroup$ Commented Jan 1, 2019 at 22:46
  • $\begingroup$ Are you interested in the discounted option price, $V(S_t, t)$ or the discounted stock price, $S_t$? Note the difference... $\endgroup$ Commented Jan 2, 2019 at 0:00
  • $\begingroup$ You can check the technical point on my answer, where I use the Black-Scholes formula and the fact that delta is a probability. $\endgroup$ Commented Jan 2, 2019 at 10:38

1 Answer 1


First of all, it is part of the Ito formula theorem that if $S_t$ is an Ito process then $V(t,S_t)$ is also an Ito process. This includes the regularity property you mention for the integral you mention and only assumes $V$ is continuous in time and $C^2$ in space. See Oksendal theorem 4.1.2


In order to give a bit more color notice that since $S_t$ is continuous a.s. then $\frac{\partial V}{\partial S}$ is bounded over the interval $[0,T]$ which is enough to ensure the integral is finite following assumption made on $S_t$ itself.


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