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Let $(\Omega,\mathcal{A},\mathbb{P})$ a probability space and $(B_t)_{t \geq 0}$ a Brownian motion (started in $x=0$). Then one can define a probability measure $\mathbb{P}^x$, $x \in \mathbb{R}$, on $(\Omega,\sigma(B_t; t \geq 0))$ by

$$\mathbb{P}^x(B_{t_1} \in A_1, \ldots, B_{t_n} \in A_n) := \mathbb{P}(x+B_{t_1} \in A_1, \ldots, x+B_{t_n} \in A_n)$$ where $A_j \in \mathcal{B}(\mathbb{R})$, $t_j \geq 0$.

This implies in particular $\mathbb{P}^x(B_0=x)=\mathbb{P}(B_0=0)=1$, i.e. $B_0 \sim \delta_x$ with respect to $\mathbb{P}^x$.

But what's wrong about the following argumentation? Assume that $B_0(w)=0$ for all $w \in \Omega$. Then $B_0(\cdot)^{-1}(\{x\})=\emptyset$ for all $x \not= 0$, hence in particular $\mu(B_0=x)=0$ for any measure $\mu$. This clearly contradicts $\mathbb{P}^x(B_0=x)=1$.

Thanks!

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This remark is a good reason to avoid assuming that $B_0(\omega)=0$ for all $\omega$ in $\Omega$. Instead, one usually defines different measures $\mathbb P^x$ on a common probability space $\Omega$, and each $\mathbb P^x$ sees a different part of $\Omega$.

For example, one can choose for $\Omega$ the space of continuous functions $[0,+\infty)\to\mathbb R$ and define $(B_t)_{t\geqslant0}$ by $B_t:\omega\mapsto\omega(t)$, for every $t\geqslant0$. Then each $\mathbb P^x$ is a probability measure on the entire $\Omega$ but $\mathbb P^x(\Omega_x)=1$ for every $x$, where $\Omega_x=[B_0=x]$, that is, $\Omega_x\subset\Omega$ is the space of all the continuous functions $\omega:[0,+\infty)\to\mathbb R$ such that $\omega(0)=x$ (note that each $\Omega_x$ is measurable). In particular, $\mathbb P^x$ and $\mathbb P^y$ are mutually singular for every $x\ne y$ since $\Omega_x\cap\Omega_y=\varnothing$.

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  • $\begingroup$ I know that it works fine if I consider the canonical Wiener process, i.e. for your given example. My problem is how to prove (rigorously) that any Brownian motion is a Markov process, i.e. that there exists a measure $\mathbb{P}^x$ such that $$\mathbb{P}(B_t \in B \mid \mathcal{F}_s)= \mathbb{P}^{B_s}(B_{t-s} \in B) \, \, (s \leq t)$$ Clearly $\mathbb{P}(B_t \in B \mid \mathcal{F}_s)=p_{t-s}(B_s,B)$ for some transition function $p_r(x,\cdot)$, so by Kolmogorov's theorem I have $\mathbb{P}(B_t \in B \mid \mathcal{F}_s) = \mathbb{Q}^{W_s}(W_{t-s} \in B)$ [...] $\endgroup$ – saz Apr 28 '13 at 8:16
  • $\begingroup$ where $\mathbb{Q}^x$ is a probability measure on the space of continuous functions and $W_t(w) := w(t)$ the canonical process. But I still don't grasp how to combine this with the definiton of a Markov process (where -as I already wrote- one usually requires the existence of a measure $\mathbb{P}^x$ on the probability space of the given process.) $\endgroup$ – saz Apr 28 '13 at 8:22
  • $\begingroup$ As I wrote, the way out is to consider that the probability space of the given process is the full space $\Omega$ in my answer, not some $\Omega_x$ depending on $B_0=x$. $\endgroup$ – Did Apr 28 '13 at 8:31
  • $\begingroup$ So in the end, a process on an arbritrary probability space is a Markov process iff the corresponding canonical process is a Markov process? $\endgroup$ – saz Apr 28 '13 at 9:56
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    $\begingroup$ Sure, try $\Omega=\Omega_0\times\mathbb R$ with the obvious product sigma-algebra and $X_t(\omega)=x+B_t(\omega_0)$ for every $\omega=(\omega_0,x)$ in $\Omega$. Then, if $(B_t)$ is a Brownian motion starting from $0$ with respect to some probability $P$ on $\Omega_0$ then $(X_t)$ is a Brownian motion starting from $x$ with respect to the distribution $P\otimes\delta_x$ on $\Omega$. $\endgroup$ – Did Apr 28 '13 at 12:35

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