Assume $S_0(t)=\exp(\int_0^t r(s) ds)$. Then $\mathbb{Q}\sim \mathbb P$ is a martingale measure $\iff$ every asset price process $S_i$ has price dynamics under $\mathbb Q$ of the form


where $M_i$ is a $\mathbb Q$ - martingale.

I read the following proof for this theorem:

Let $\tilde{S}_i(t)=\dfrac{S_i(t)}{S_0(t)}$.

$\dfrac{1}{S_0(t)}=\exp(-\int_0^t r(s) ds)$



By Itó's product rule

$d\left(\dfrac{S_i(t)}{S_0(t)}\right)=-r(t)S_i(t)\dfrac{1}{S_0(t)}dt+\dfrac{1}{S_0(t)}dS_i(t)+d\langle S_i,\dfrac{1}{S_0}\rangle_t= -r(t)\tilde{S}_i(t)dt+\dfrac{1}{S_0(t)}dS_i(t).$

I understand every mathematical step of the proof but why does this proof the theorem? Can anyone explain?

  • $\begingroup$ Do you know what book this is from? $\endgroup$
    – Alex C
    Commented Jun 16, 2017 at 2:14

2 Answers 2


As it stands, the assertion "$\Bbb{Q} \sim \Bbb{P}$ is a martingale measure" is not complete. It omits to tell you what process(es) should emerge as martingale(s) under $\Bbb{Q}$. These processes are $\tilde{S}_i(t) = S_i(t)/S_0(t)$ for any traded asset $S_i$.

That being said, starting from the last equation: $$d\left(\dfrac{S_i(t)}{S_0(t)}\right)= -r(t)\tilde{S}_i(t)dt+\dfrac{1}{S_0(t)}dS_i(t).$$ For $\tilde{S}_i(t)=S_i(t)/S_0(t)$ to emerge as a $\Bbb{Q}$ martingale, you should have, $$ d\tilde{S}_i(t) = -r(t)\tilde{S}_i(t)dt+\dfrac{1}{S_0(t)}dS_i(t) = d M_i(t) $$ with $M_i(t)$ a $\Bbb{Q}$-martingale.

Isolating $dS_i(t)$ in the second equality gives \begin{align} dS_i(t) &= r(t) \tilde{S}_i (t) S_0(t) dt + S_0(t) d M_i(t) \\ &= r(t) S_i(t) dt + d M^*_i(t) \\ \end{align} assuming usual integrability conditions hold such that $$ M^*_i(t) = \int_0^t S_0(u) dM_i(u) $$ is a well-defined Itô-integral and hence also a martingale (see hints here)


What is the definition of Equivalent Martingale Measure? It is a measure $\mathbb{Q} \sim \mathbb{P}$ s.t. $\frac{S_i}{S_0}$ is martingale under $\mathbb{Q}$. In the last step of your prove assume $S_i$ has some drift $a$ and volatility $b$, i.e. $dS_i=adt+bdZ^\mathbb{Q}$ and substitute to obtain: $$d\left(\frac{S_i}{S_0}\right)=-r\left(\frac{S_i}{S_0}\right)dt +\left(\frac{1}{S_0}\right)dS_i=-r\left(\frac{S_i}{S_0}\right)dt +\left(\frac{1}{S_0}\right)(adt+bdZ^\mathbb{Q})$$ To guarante that the process is indeed a martingale notice that: $$d\left(\frac{S_i}{S_0}\right)=\left(-r\frac{S_i}{S_0}+\frac{a}{S_0}\right)dt+ \text{martingale part} $$ Therefore set the drift equal to zero and obtain $a=rS_i$ as requested.

  • $\begingroup$ I agree with the idea but there is no reason to explicitly introduce a driving Brownian motion (i.e. continuous paths). Though indeed for diffusion processes any martingale can be represented as the stochastic integral of some predictable process with respect to the Brownian motion. $\endgroup$
    – Quantuple
    Commented Jun 16, 2017 at 13:42
  • $\begingroup$ You can add a jump component $dJ$ and nothing would change. If $dS_i=adt+bdZ^\mathbb{Q}+dJ$ still leads to $d\left(\frac{S_i}{S_0}\right) = \left(-r\frac{S_i}{S_0}+\frac{a}{S_0}\right)dt+ \text{ martingale part }$. So it seems totally unconsequential to me. $\endgroup$
    – fni
    Commented Jun 16, 2017 at 16:04
  • $\begingroup$ A "naive" jump process is not a martingale unless it is compensated, so it may in fact have an impact on the drift. What you say does not work for any $J_t$. $\endgroup$
    – Quantuple
    Commented Jun 16, 2017 at 16:11
  • $\begingroup$ I don't see your point given that I did not say anything about $adt$, hence it seems totally fine to me the fact that the jump part may affect the drift. For every point of the derivation if you don't like $bdZ^\mathbb{Q}$ just write $dM$ and everything goes through. As you can see I did not even bother writing the non-drift part in the second equation but I just left a generic "martingale part" $\endgroup$
    – fni
    Commented Jun 16, 2017 at 16:20
  • $\begingroup$ Right that's exactly my point you don't need to introduce the Brownian. The rest I completely agree with. It's just what you said in the comments that I don't agree with ie "you can add $dJ_t$ ..." which is usually not a martingale so no you cannot do that, at least without changing the drift. But then indeed you can introduce any $dM_t$. Or if you changed the drift adequately you'd still end up with what you wrote as well. $\endgroup$
    – Quantuple
    Commented Jun 16, 2017 at 16:31

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