Can we have a Brownian motion under two different probability space?

Question

Is it possible to construct a stochastic process $B_t$ such that $B_t$ is a Brownian under $(\Omega, F, P)$ and $B_t$ is a Brownian under $(\Omega, F, \hat{P})$? If not, how to argue that $P=\hat{P}$?

Answer 1

Let $(B_t)_{t \geq 0}$ be a Brownian motion on a probability space $(\Omega_1,F_1,P_1)$ and let $(\Omega_2,F_2,P_2)$ be an arbitrary probability space. Define a new probability space $(\Omega,F,P)$ by $$\Omega := \Omega_1 \times \Omega_2 \qquad F := F_1 \otimes F_2 \qquad P := P_1 \otimes P_2.$$ If we set $$\tilde{B}_t(\omega_1,\omega_2) := B_t(\omega_1) \quad \text{for} \, (\omega_1,\omega_2) \in \Omega_1 \times \Omega_2,$$ then it is not difficult to see that $(\tilde{B}_t)_{t \geq 0}$ is still a Brownian motion on $(\Omega,F,P)$. Since this holds true for any probability space $(\Omega_2,F_2,P_2)$, we can easily construct a Brownian motion on $(\Omega,F,P)$ and $(\Omega,F,\hat{P})$ with $P \neq \hat{P}$.

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Note, however, that a Brownian motion $(B_t)_{t \geq 0}$ on $(\Omega,F,P)$ does determine the restricted probability measure $P|_{F_\infty}$ uniquely; here

$$F_{\infty} := \sigma(B_t; t \geq 0)$$

denotes the $\sigma$-algebra generated by $(B_t)_{t \geq 0}$. That is, for any probability space $(\Omega,F,\hat{P})$ such that $(B_t)_{t \geq 0}$ is a Brownian motion on $(\Omega,F,\hat{P})$, we have $\hat{P}(A) = P(A)$ for any $A \in F_{\infty}$.

Proof: Let $P,\hat{P}$ be two probability measures on $(\Omega,F)$ such that $(B_t)_{t \geq 0}$ is a Brownian motion on $(\Omega,F,P)$ and $(\Omega,F,\hat{P})$. If $A$ is a set of the form $$A = \bigcap_{j=1}^n \{B_{t_j} \in G_j\} \tag{1}$$ where $t_j \geq 0$ and $G_j \in \mathcal{B}(\mathbb{R})$ are Borel sets, $n \in \mathbb{N}$, then

$$\begin{align*} P(A) &= \mathbb{P}(B_{t_1} \in G_1,\ldots,B_{t_n} \in G_n) \\ &= \mu_{(B_{t_1},\ldots,B_{t_n})}(G_1 \times \ldots \times G_n) \\ &= \hat{P}(B_{t_1} \in G_1,\ldots,B_{t_n} \in G_n) \\ &= \hat{P}(A) \end{align*}$$

where $\mu_{(B_{t_1},\ldots,B_{t_n})}$ denotes the distribution of $(B_{t_1},\ldots,B_{t_n})$ (for a Brownian motion it is known that this distribution is centered Gaussian with a certain covariance matrix which can be calculated explicitly). Since sets of the form $(1)$ are a $\cap$-stable generator of $F_{\infty}$, the theorem on uniqueness of measures proves $P|_{F_{\infty}} = \hat{P}|_{F_{\infty}}$.