6
$\begingroup$

I understand that the risk-neutral measure is one under which the discounted price (acc. to the risk-free rate) of any asset is a martingale. But we also see notation like $\mathbb{W}^Q_t$ to denote a Brownian motion under the risk-neutral measure. 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?

$\endgroup$
6
$\begingroup$

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 expectation 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. 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 martingale anymore.

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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 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 $\mathbb{Q}$. $\endgroup$ – KeSchn Mar 5 at 15:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.