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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.

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    $\begingroup$ You are right. If we analyze a stochastic process $S_t$ and come to the conclusion that "It is a BM under the measure $M$" then we can rule out that it is a BM under any other measure $P$, $Q$, etc. If we change the measure to $P$ then $S_t$ will no longer be a BM. $\endgroup$
    – noob2
    Oct 13 at 16:38
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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.

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