# What is a Brownian motion "under the risk-neutral measure"?

I understand that the risk-neutral measure associated with the money-market Numeraire is one under which the discounted price (acc. to the risk-free rate) of any asset is a martingale.

Brownian motion under the risk-neutral measure is often denoted $$\mathbb{W}^Q_t$$. What exactly is the definition of $$\mathbb{W}^Q_t$$? How does $$Q$$ alter the stationary and independent increments properties of the standard Wiener process?

A Brownian motion is always defined with repect to a given probability space. Let $$(\Omega,\mathcal{F},\mathbb{P})$$ be a probability space and $$X_t=W_t^\mathbb{P}$$ a Brownian motion, i.e. a stochastic process with i.i.d. increments $$X_t-X_s\sim N(0,t-s)$$ and continuous sample paths $$\mathbb{P}$$-a.s. and with $$X_0=0$$.
Now, let $$\mathbb{Q}\sim\mathbb{P}$$ be a new probability measure defined on the measurable space $$(\Omega,\mathcal{F})$$. Due to the equivalence, the sample paths of $$X_t$$ are continuous $$\mathbb{Q}$$-almost surely but what about the distribution of the increments? $$\mathbb{E}^\mathbb{P}[X_t-X_s]=0$$ does not imply $$\mathbb{E}^\mathbb{Q}[X_t-X_s]=0$$. Thus, in general, $$W_t^\mathbb{P}$$ is not a Brownian motion anymore if you alter the probability measure and hence the associated expectation operator etc.
When you say that $$W_t^\mathbb{Q}$$ is a $$\mathbb{Q}$$-Brownian motion, you mean that it satisfies the definition with respect to the given probability space $$(\Omega,\mathcal{F},\mathbb{Q})$$. If you alter any component of the probability space, the process may not satisfy the original definition anymore.
• So just the usual Brownian motion with the measure changed? The numeraire does not come into play in the definition of $\mathbb{W}_t^Q$? Mar 5 '20 at 15:26
• The numeraire may come into play when defining the new measure $\mathbb{Q}$. Recall that $\mathbb{E}^\mathbb{P}[X]=\mathbb{E}^\mathbb{Q}[\frac{\mathrm{d}\mathbb{P}}{\mathrm{d}\mathbb{Q}}X]$ and the Radon Nikodym derivative can be related to a change of numeraire. But when you say that $W_t^\mathbb{Q}$ is a $\mathbb{Q}$-Brownian motion, you literally only mean that $W_0=0$ $\mathbb{Q}$-a.s., that $t\mapsto W_t(\omega)$ is continuous for $\mathbb{Q}$-almost all $\omega\in\Omega$ and that the increments $W_t-W_s$ are i.i.d. normal with mean $0$ and variance $t-s$, with respect to $\mathbb{Q}$. Mar 5 '20 at 15:30