Equivalent Martingale Measures and Risk-Neutral Asset Drifts
Summary
The document explains why, under an equivalent martingale measure, traded asset prices have drift equal to the short rate when expressed relative to a risk-free money-market account. The key definition is that each discounted asset price must be a martingale under the measure. Applying Itô’s product rule to divide an asset price by the bank account gives the relation between discounted and undiscounted dynamics.
The answer then sets the discounted process’s drift to zero and rearranges the result to obtain the asset’s risk-neutral drift. It also notes the integrability conditions needed for the transformed stochastic integral to remain a martingale. A second explanation reaches the same result by separating drift and martingale components. The discussion assumes the stated money-market account and suitable regularity; it is a proof sketch rather than a full treatment of market completeness, measure construction, or all technical conditions.
Key ideas
- An equivalent martingale measure is defined through martingale behavior of discounted traded asset prices.
- Dividing an asset price by the risk-free account connects its dynamics to the asset’s original drift.
- The discounted price has zero drift under the martingale measure.
- The resulting risk-neutral asset drift equals the short rate, subject to suitable integrability conditions.
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Full text
# Equivalent martingale measure price dynamics
# 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?
## Answer by Quantuple (score 2, accepted)
https://quant.stackexchange.com/a/34723
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)
## Answer by fni (score 1)
https://quant.stackexchange.com/a/34724
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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
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