# Equivalent martingale measure price dynamics

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

$dS_i(t)=r(t)S_i(t)dt+dM_i(t)$,

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

I read the following proof for this theorem:

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

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

Hence

$d\left(\dfrac{1}{S_0(t)}\right)=-r(t)\dfrac{1}{S_0(t)}dt.$

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?

• Do you know what book this is from? Jun 16 '17 at 2:14

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.

• 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. Jun 16 '17 at 13:42
• 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.
– fni
Jun 16 '17 at 16:04
• 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 $J_t$. Jun 16 '17 at 16:11
• 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"
– fni
Jun 16 '17 at 16:20
• 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. Jun 16 '17 at 16:31