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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?

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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.

Similarly, martingales are always defined with respect to a certain measure (expectation) and filtration. If you change the probability measure or the filtration, the considered process is not necessarily a martingale anymore.

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  • $\begingroup$ So just the usual Brownian motion with the measure changed? The numeraire does not come into play in the definition of $\mathbb{W}_t^Q$? $\endgroup$
    – Bravo
    Mar 5 '20 at 15:26
  • $\begingroup$ 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 $\mathbb{Q}$. $\endgroup$
    – Kevin
    Mar 5 '20 at 15:30

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