How Brownian Motion Changes Under a Risk-Neutral Measure
Summary
The document clarifies that Brownian motion is defined relative to a probability measure and filtration. A process that is Brownian under the physical measure has continuous paths and normally distributed, independent increments with the specified time-based variance under that measure. If the probability measure changes to an equivalent risk-neutral measure, continuity is preserved almost surely, but the increment distributions and expectations need not remain the same.
Thus, a Brownian motion under the risk-neutral measure means a process satisfying the Brownian-motion definition when evaluated using that measure. The original Brownian process does not automatically retain that property after a measure change; it may acquire a drift, and a suitable transformed process may be Brownian under the new measure. The same measure and filtration dependence applies to martingales, including the discounted asset-price condition used in risk-neutral pricing. The explanation is conceptual and does not derive a change of measure or specify a particular asset model.
Key ideas
- Brownian motion is defined relative to a probability measure and its associated expectation.
- Equivalent measures preserve almost-sure path continuity but may alter increment distributions.
- A process Brownian under one measure need not remain Brownian under another.
- Martingale status also depends on the probability measure and filtration.
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# What is a Brownian motion "under the risk-neutral measure"?
# 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?
## Answer by Kevin (score 10, accepted)
https://quant.stackexchange.com/a/51500
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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.