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Consider some model where the process increments are normally distributed, e.g. Vasicek: $$dr(t) = \left(\theta - ar(t)\right)dt + \sigma dW(t).$$

We usually say that $W(t)$ is a Brownian motion under a measure $\mathbb P$. $W(t)$ is a Brownian motion if, among other conditions, $W(t) \sim N(0, t)$ given $W(0)=0$. Does it mean that the measure $\mathbb P$ is actually a normal distribution, i.e. $$\mathbb P\left(\frac{W(t)}{\sqrt t} \in [a ,b]\right) = \Phi(b) - \Phi(a)$$ where $\Phi(\cdot)$ denotes the CDF of a standard normal random variable?

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  • $\begingroup$ The equation you wrote is correct. But it is only one example of the properties of the measure P. The measure P is more general than this and cannot be said "to be a normal distribution", but the normal distribution does come up in describing the measure. The measure applies to a stochastic procees, the normal distribution applies to a random variable, so they cannot be identical. $\endgroup$
    – nbbo2
    Commented Feb 28, 2019 at 19:33

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  1. It is correct that $$ \mathbf{P}(t^{-1/2}W(t) \in[a,b])=Φ(b)−Φ(a), \forall t\in(0,\infty) $$ due to the stationary increments property of the Wiener process and the fact that you normalized the random variable by dividing by its standard deviation.
  2. $\mathbf{P}$ is a probability measure on an abstract space, not a random variable. Hence, you probably mean that $W(t)$ is normally distributed under $\mathbf{P}$, NOT $\mathbf{P}$ is normally distributed. People tend to mention the probability measure, for if you change it the process will no longer be Gaussian.
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    $\begingroup$ Welcome to Quant SE. No need for comments :) This is (along with the rest of your answers) acceptable as an answer to the question. $\endgroup$
    – Sanjay
    Commented Feb 28, 2019 at 20:31
  • $\begingroup$ Could you elaborate more on “...measure on an abstract space”. Normal distribution is also a measure (of a rv). Also I could change to martingale measure $\mathbb Q$ and my process still would be Gaussian, i.e. $dW^{\mathbb Q}= dW^{\mathbb P} +\lambda dt$ by Girsanov. $\endgroup$
    – tosik
    Commented Mar 1, 2019 at 7:07
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    $\begingroup$ @tosik. A probability space is a triple $(\Omega, \mathcal{F}, \mathbf{P})$ where $\Omega$ is an abstract space. A normal distribution is a very specific type (with a specific density) on $\mathbb{R}^n$ for some $n \geq 1$. A random variable or vector $X: \Omega \rightarrow \mathbb{R}^n$ is said to be normal if it induces a normal distribution on the range space. $\endgroup$
    – phantagarow
    Commented Mar 1, 2019 at 14:01
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    $\begingroup$ @tosik. You might want to be precise with Girsanov's Theorem so as not conflate related but different concepts. If $W_{t}$ is a Brownian Motion under $\mathbf{P}$, the newly defined process $Z_{t} = W_{t} - \int_{0}^{t} \theta_{s} ds$ is no longer a BM under $\mathbf{P}$. Girsanov's Thm tells you how to construct a new measure so as that $Z_{t}$ becomes a BM. $\endgroup$
    – phantagarow
    Commented Mar 1, 2019 at 14:40
  • $\begingroup$ I agree that $Z_t$ is no longer BM under $\mathbb P$ but I think you agree that it is Gaussian both under $\mathbb P$ and $\mathbb Q$. I indeed probably confuse measures on $\Omega$ with distributions on $\mathbf R$. Your answer (and comments) do shed some light but I still don't have a clear picture. So, $\Omega$ is an abstract space where for each $\omega \in \Omega$ there is an associated function $f$ that we think of as a stochastic process? Can you elaborate more on the link between $\mathbb P$ and normal distribution for the case of BM? $\endgroup$
    – tosik
    Commented Mar 1, 2019 at 15:01

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