Uniqueness of equivalent martingale measure in Black Scholes-Model

Let's consider standard Black-Scholes model with price process $S_t$ satisfying SDE $$dS_t = S_t(bdt + \sigma dB_t)$$, where $B_t$ is standard Brownian Motion for probability $\mathbb{P}$. I understand the proof of existence of martingal measure $\mathbb{Q}$ equivalent to $\mathbb{P}$ based on Girsanov theorem, but I can't see how to derive uniqueness of $\mathbb{Q}$. Can you help?

Edit: In Jeanblanc, Yor, Chesney $\textit{Mathematical Methods for Financial Markets}$ I found the following proof:

If $\mathbb{Q}$ is equivalent to $\mathbb{P}$ then there exists strictly positive martingale $L_t$ such that $\mathbb{Q}|_{F_t} = L_t\mathbb{P}|_{F_t}$. From the predictable representation property under $\mathbb{P}$, there exists a predictable $\psi$ such that $$dL_t = \psi_tdB_t = L_t\phi_tdB_t,$$ where $\psi_t = \phi_tL_t$. It follows that $$d(LRS)_t \stackrel{mart}{=} (LRS)_t(b − r + \phi_t\sigma)dt$$ (where $dX_t \stackrel{mart}{=} dY_t$ for semimartingales $X$ and $Y$ means that $X-Y$ is a local martingale, $R_t=e^{-rt}$ is a discount process). Hence, in order for $\mathbb{Q}$ to be an e.m.m., or equivalently for $LRS$ to be a $\mathbb{P}$-local martingale, there is one and only one process $\phi$ such that the bounded variation part of LRS is null, that is $$\phi_t = \frac{r − b}{\sigma}=−\theta.$$

Now Girsanov theorem gives us the existence of such e.m.m. and fact that $\phi$ is unique gives us uniqueness of $\mathbb{Q}$. Unfortunately, I don't understand where $\stackrel{mart}{=}$ equality comes from and why for $LRS$ to be a $\mathbb{P}$-local martingale, there must be process $\phi$ such that the bounded variation part of LRS is null. Do you have any idea how to proceed with these steps?

I am especially interested in the proof of $d(LRS)_t \stackrel{mart}{=} (LRS)_t(b − r + \phi_t\sigma)dt$.

A martingale must have constant expectation, such that adding a deterministic finite variation process $(b-r)dt$ would break the martingale property (except for when its a constant, which it is not by multiplication with $dt$).

Hence the finite variation process must be eliminated under $Q$ for LRS to be an (equivalent) martingale measure, and as shown the only unique choice in this case is $$\phi_t=-\theta.$$

The assertion $\stackrel{mart}{=}$ does not represent an equality per se, it is the postulated martingale requirement under $Q$. $Q$ is then chosen by Girsanov theorem with $\phi_t=-\theta$ such that $\stackrel{mart}{=}$ holds.

• Thank you for your answer! If I understand well: if $dX_t \stackrel{mart}{=} dY_t,$ then $X_t$ is martingale if and only if when $Y_t$ is, so in our case it's enough to show that $\int_0^t LRS_t(b-r+\phi_t \sigma)dt$ is martingale, but why this happens only when $b-r+\phi_t \sigma$ is constantly zero? And again, I don't see where postulated $\stackrel{mart}{=}$ comes from. – Adam Oct 26 '14 at 13:58
• @Adam The section is about to find $Q$, such that LRS becomes martingale: "in order for ℚ to be an e.m.m.", therefore $\stackrel{mart}{=}$. And note that $(b-r)dt$ is a deterministic increasing process. Therefore, it must be eliminated to achieve constant expectation (martingale property). Think of $X=(b-r)t$. $X$ is not a martingale, because $E(X_t|F_s)=(b-r)t>(b-r)s=X_s$. So $X$ must be eliminated under $Q$. – emcor Oct 26 '14 at 14:46
• emcor, is it wrong to say that the required result immediately follows from uniqueness of the Radon-Nikodym derivative? – BCLC Dec 17 '15 at 2:45
• @BCLC Given that you have an equivalent measure $Q$, it follows that there exists only one Radon-Nikodym derivative. However, this does not mean that $Q$ itself is unique. There can be many $Q$ (if markets are incomplete) and each $Q$ then has a unique Radon-Nikodym derivative. – emcor Dec 17 '15 at 10:22
• @BCLC I did not find a source that Girsanov-Theorem delivers a $Q$ which is also unique. So you may use Girsanov-Theorem and then additionally proof that this $Q$ is unique. In your example, you have a martingale plus a deterministic time-changing process, which is not a martingale. So the only option to have a martingale is to eliminate the $\sin_t$ term and Girsanov gives you the correct term $+\cos t$. – emcor Dec 18 '15 at 11:53

I saw a quote from Brigo & Mercurio "IR models" (page 26, 2.1 No-Arbitrage in Continuous Time) . May be it will help you to find answer:

Harrison and Pliska (1983) proved the following fundamental result. A financial market is (arbitrage free and) complete if and only if there exists a unique equivalent martingale measure.

• Thanks a lot @Sasha, but I want to prove that market in B-S model is arbitrage free and complete by proving that unique e.m.m. exists. But maybe the other way is simpler? – Adam Oct 26 '14 at 10:36
• So how do you prove the market is arbitrage-free? – BCLC Dec 17 '15 at 0:42