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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.

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  • $\begingroup$ Got something from an answer? $\endgroup$ – Did Jan 4 '15 at 15:40
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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.

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  • $\begingroup$ Well, lets say I want the Brownian motion to be normal only, then are my propositions correct? $\endgroup$ – user3001408 Nov 29 '14 at 13:45
  • $\begingroup$ No-- and actually I am afraid that it is less and less possible to understand the question you really want to ask. $\endgroup$ – Did Nov 29 '14 at 13:48
  • $\begingroup$ Edited my question. I just want to check my understanding of the concept. $\endgroup$ – user3001408 Nov 29 '14 at 13:50
  • $\begingroup$ But if the new proability measures differ on probabilities on going up or down, then the brownian motion can not be considered random, that is we inject a trend in it. Besides, it is a martingale and W($0$)=$0$, the probabilites are not symmetric, then the notion may go down. $\endgroup$ – user3001408 Nov 29 '14 at 13:53
  • $\begingroup$ What a mishmash (and obviously you did not even read my answer)... If $W$ is a Brownian motion under $P$ and if $Q$ is another measure such that $W$ is still a Brownian motion under $Q$, then of course $E_Q(W_t)=0$ for every $t$ (this is the question I can most clearly see in your text, and I am sorry but it is just absurd). Your last comment is really offtopic since one of the commonest changes of measures on the Wiener space is precisely used to introduce a given drift... $\endgroup$ – Did Nov 29 '14 at 14:00
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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.

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