Brownian Motion Distributions and the Meaning of a Probability Measure

Summary

The note clarifies what it means for Brownian motion to be normally distributed under a probability measure. For a standard Brownian motion starting at zero, its value at time t has a normal distribution with mean zero and variance t. Dividing by the square root of time standardizes that value, so its probability of falling within an interval equals the corresponding standard normal cumulative probability difference.

The key distinction is that the probability measure is not itself a random variable or a normal distribution. Rather, it assigns probabilities to events involving the process, and the process has specified distributions under that measure. The answer also points out that changing the measure can change the process’s distributional properties. This is a conceptual clarification using the defining distribution of standard Brownian motion; it does not provide a trading strategy or address the additional conditions required to define a Brownian motion in full.

Key ideas

• At time t, standard Brownian motion has a normal distribution with mean zero and variance t.
• Dividing Brownian motion at time t by the square root of t gives a standard normal variable.
• A probability measure assigns probabilities to events; it is not itself a normally distributed variable.
• A process’s distributional properties depend on the probability measure under which it is considered.

Tags

# Measure of a Brownian motion = normal distribution?


# Measure of a Brownian motion = normal distribution?












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?

## Answer by phantagarow (score 6)

https://quant.stackexchange.com/a/44342

- 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.

- $\mathbf{P}$ 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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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.