# What is the martingale measure requirement when $\mu(t,S(t)) = \mu(t)$?

It is known that if the numeraire is the bank account, then the martingale measure is determined by the fact that every asset has $r$ as its local rate of return.

However, the local rate of return is the "multiplier" of the stock price in the dt-term of the SDE. That is, if $$dS_t = S_t\alpha dt + S_t\sigma dW$$ then $\alpha$ is the local rate of return, which must be equal to $r$ under the martingale measure.

However, imagine if under the $P$-measure, $S_t$ satisfied the SDE$$dS_t = \alpha dt + \sigma dW,$$ in this case, what is the equivalent martingale measure requirement?

Let $B_t$ denote the $t$-value of a riskless money market account in which 1 unit of currency has been invested at time $t=0$. We have $B_t = e^{rt}$, where $B_t$ solves $$dB_t = B_trdt,\quad B_0=1 \tag{0}$$
If $S_t$ represents the $t$-value of a tradable asset, then in the absence of arbitrage opportunity we must have that (fundamental theorem of asset pricing): $$\frac{S_t}{B_t} \text{ is a } \Bbb{Q} \text{-martingale}$$ in other words \begin{align} d\left(\frac{S_t}{B_t}\right) &= \frac{dS_t}{B_t} - \frac{S_t}{B_t^2} dB_t + 0 \tag{Itô} \\ &= \frac{1}{B_t} \left( dS_t - S_t r dt \right) \tag{using $(0)$} \\ &= \dots dW_t^\Bbb{Q} \tag{1} \end{align} by the martingale representation theorem.
Note that for $(1)$ to hold it is necessary to have: $$dS_t = \color{blue}{rS_t} dt + \dots dW_t^\Bbb{Q}$$ This is the equivalent martingale measure requirement you're looking for.
Now, to reach that $rS_t$ drift under $\Bbb{Q}$, you shall not use the same Girsanov kernel depending on whether $S_t$ follows a GBM or an ABM under $\Bbb{P}$ but that's another question.
In the first case you'll have: $$dS_t = \alpha S_t dt + \sigma S_t dW_t^\Bbb{P} \to dS_t = \color{blue}{rS_t} dt + \sigma S_t dW_t^\Bbb{Q}$$ with $$\left. \frac{d\Bbb{Q}}{d\Bbb{P}} \right\vert_{\mathcal{F}_t} = \mathcal{E}\left[ -\lambda W_t^\Bbb{P}\right], \quad \lambda=\frac{\alpha-r}{\sigma}$$ In the second: $$dS_t = \alpha dt + \sigma dW_t^\Bbb{P} \to dS_t = \color{blue}{rS_t} dt + \sigma dW_t^\Bbb{Q}$$ with $$\left. \frac{d\Bbb{Q}}{d\Bbb{P}} \right\vert_{\mathcal{F}_t} = \mathcal{E}\left[ -\lambda W_t^\Bbb{P}\right], \quad \lambda=\frac{\alpha-rS_t}{\sigma}$$ See this paper if you want more mathematical details.
• +1 Nice answer. Indeed, to be a martingale, under the risk-neutral measure, the drift term has to be of the form $rS_t$. – Gordon Feb 24 '17 at 15:01