How Changing Probability Measures Changes Brownian Motion

# How can there be Brownian motions under different measures?


# How can there be Brownian motions under different measures?












According to the definition, for a Brownian motion it holds that

$W_0 = 0$, and $W_t - W_s \in N(0, t-s), \quad t > s$.

This implies that $W_t \in N(0, t)$, for all $t \geq 0$. Hence, the definition gives us the distribution of every single value in the process, if I'm not misunderstanding something. Wouldn't that mean that the definition uniquely defines the probability space (including the measure) for the process? Then, how can there be different Brownian motions under different measures?

I have read the answer to What is a Brownian motion "under the risk-neutral measure"?, but I still don't understand this. I am new to the subject, and do not know a lot about measure theory, so if it's possible that someone could give a somewhat simple explanation, perhaps in plain English, it would be very helpful.

## Answer by algebruh (score 4, accepted)

https://quant.stackexchange.com/a/68373

Maybe this example helps:

$$\Omega=\{\omega_1,\omega_2\} $$

Consider the following two probability measures

$$ \mathbb{P_1}(\{ \omega_1\}) = 0.4\ \ \ ; \ \ \mathbb{P_2}(\{ \omega_1\}) = 0.6 \ $$

and $\mathbb{P_i}(\{ \omega_2\})=1-\mathbb{P_i}(\{ \omega_1\})$ Consider the two random variables: $$ X_1: \Omega \to \{0,1 \}, X_1(\omega_1) \mapsto 0 , X_1(\omega_2) \mapsto 1 $$ $$ X_2: \Omega \to \{0,1 \}, X_2(\omega_1) \mapsto 1 , X_2(\omega_2) \mapsto 0 $$

Note that $X_1$ is under $\mathbb{P_1}$ bernoulli distributed with $p=0.6$ while $X_2$ is under $\mathbb{P_1}$ bernoulli distributed with $p=0.4$

But if we change the measure from $\mathbb{P_1}$ to $\mathbb{P_2}$ , $X_1$ and $X_2$ swap their distributions. Therefore by changing the measure we change the distribution of random variables.We thus have $$ X_1^{\mathbb{P_1}} \stackrel{\mathcal{D}}{=} X_2^{\mathbb{P_2}} $$

The same principle applies for the brownian motion. Though in this case $\Omega$ is the set of continuous functions. Being a brownian motion is a matter of distribution and not of the strucutre of the probability space.

## Answer by joshdalton (score 1)

https://quant.stackexchange.com/a/76468

A simple answer that our lecturer once mentioned is that a room can be both measured in inches and meters. Both would describe the exact same process (the room), but in different measures $P$ or $Q$. However, when considering stochastic process under different measures, the room itself may show different properties. That is why a Brownian Motion under $P$ may not be under $Q$.