Why the Black–Scholes Equivalent Martingale Measure Is Unique

# Uniqueness of equivalent martingale measure in Black Scholes-Model


# 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 $d(LRS)_t \stackrel{mart}{=} (LRS)_t(b − r + \phi_t\sigma)dt$.

## Answer by emcor (score 1, accepted)

https://quant.stackexchange.com/a/15167

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.

## Answer by lkjldfkjhljk (score 1)

https://quant.stackexchange.com/a/15166

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.