Change of measure on Brownian motion

Question

I have a small doubt as I am currently self-studying stochastic calculus.

In Brownian motion part, the author talked about change of probability measure over Brownian motions. Now we we know that Brownian motion is normally distributed. As well as each step in the Brownian motion has to be symmetric, otherwise the mean $0$ criteria (of the increments) would be violated.

The following is my understanding.

1) $P(H)=0.5, \overline P(H)=0.5$ where $P$ and $\overline P$ are two probability measures, and $H$ is getting Head. Here I am assuming that Brownian motion is simulated by coin tosses. Then these 2 probability meausures would always agree on the above probability values.

2) In addition, I think due to normality of Brownian motions, the change of measure can only change the mean of the Brownian distribution.

Am I right in point # 1 and # 2. Would be grateful if someone can help in this regard.

Answer 1

Brownian motion is often defined as a probability measure on the space $C([0,\infty),\mathbb R)$ of real valued continuous functions defined on $[0,\infty)$. As such, a change of probability measure on this space can yield processes pretty different from the ones you have in mind: the means can change, the variances can change, the gaussianity can fail, etc.

Answer 2

As @Did wrote, let's talk about the space $\Omega = C([0,\infty),\Bbb R)$. This is the space of continuous trajectories, so if we define a measure on $\Omega$ we are saying which of the trajectories are more likely than others. For example, we can define a Wiener measure $P$ on $\Omega$ - that is what you call a Brownian motion. If we define another measure $Q$ on $\Omega$, the very same process may have different distribution, and may not be a Brownian motion at all.