# Unique risk neutral measure for Brownian Motion

For a standard geometric Brownian motion model of stock prices: $$dS = a S dt + \sigma S dZ$$ we can transform the process to be under risk neutral measure: $$dS = r S dt + \sigma S d \tilde{Z}$$ and from the references I found, this risk neutral measure is "unique".

If we make a transform, say $$dS = r S dt + \tau S d \hat{Z}$$ where $\tau$ is different from $\sigma$, this equation gives the correct price of stock. but Black-Scholes equation will fail as we have changed volatility.

However, for a discrete model, e.g. a tree model, if there are $n$ states of world, then we need $n-1$ assets plus cash to uniquely pin down risk neutral measure.

Question:The Brownian motion model in effect has infinite number of states and only one asset, then where does uniqueness of risk neutral measure come from?

The uniqueness of the risk-neutral measure comes from the abundance of tradable assets. Let $B_t$ be the money-market account at time $t$. Let $Q_1$ and $Q_2$ be two risk-neutral measures. Then, for any tradable asset $X$ with maturity $T$, \begin{align*} E^{Q_1}\left(\frac{X_T}{B_T}\right) &= E^{Q_2}\left(\frac{X_T}{B_T}\right)\\ &=\frac{X_0}{B_0}. \end{align*} For any $A\in \mathcal{F}_T$, we define an asset with payoff $$\mathbb{I}_{A} B_T.$$ Note that, this deal may not be exchange traded, however, it can be made over-the-counter. Then \begin{align*} Q_1(A) &= E^{Q_1}\left(\frac{\mathbb{I}_{A} B_T}{B_T}\right)\\ &= E^{Q_2}\left(\frac{\mathbb{I}_{A} B_T}{B_T}\right)\\ &= Q_2(A). \end{align*} That is, $Q_1=Q_2$.
• @user3785097, $E(\mathbb{I}_{A} \mid \mathcal{F}_t)$ is a martingale, on which the differential is taken. – Gordon Jun 26 '15 at 12:35
essentially it comes down the fact that the dyadic quadratic variation of $W_t$ is $t$ with probability 1 and any measure change has to preserve this fact. Changing volatility would violate this invariance.