# Translational invariance of Brownian motion

Let $(\Omega,\mathcal{A},\mathbb{P})$ a probability space, $(X_t,\mathcal{F}_t)_{t \geq 0}$ a time-homogeneous Markov process. A paper I read defines a probability measure $\mathbb{P}^x$ by

$$\mathbb{P}^x(C) := \mathbb{P}(C \mid X_0=x) \qquad \qquad (C \in \mathcal{A})$$

Applying this in particular to a Brownian motion $(B_t)_{t \geq 0}$ we obtain

$$\mathbb{P}^x(C) = \mathbb{P}(C \mid B_0 = x)$$

so in particular for $C:=[B_t \in B]$ (where $B \in \mathcal{B}(\mathbb{R})$)

$$\mathbb{P}^x[B_t \in B] = \mathbb{P}[B_t \in B \mid B_0=x] \tag{1}$$

But I know the following definition of $\mathbb{P}^x$:

$$\mathbb{P}^x[B_t \in B] := \mathbb{P}[x+B_t \in B] \tag{2}$$

But I don't see why the measures defined in (1) and (2) are the same (for $x \not= 0$)...

Edit: Let $x \mapsto g(x):=\mathbb{P}[B_t \in B \mid B_0=x]$. Then $g$ is only $\mathbb{P}_{B_0}$-a.s. uniquely defined, i.e. $\delta_0$-a.s. uniquely defined. Therefore an arbritary measurable function $h: \mathbb{R} \to \mathbb{R}$ such that $h(0) = \mathbb{P}[B_t \in B]$ would fulfill $h(B_0)=\mathbb{P}[B_t \in B|B_0]$ - and this would imply that (1) is (for $x \not= 0$) not well-defined. Am I correct...?

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@DavideGiraudo: the left-hand side, you meant? – Ilya Jan 9 '13 at 17:29
Also, $C$ shall be a measurable subset of the path space $\Omega$, not of the state space $\Bbb R$. Anyway, the definition of measures $\Bbb P^x$ as conditional on the initial state is handy, but I am not sure it is perfectly precise. – Ilya Jan 9 '13 at 17:49
@Ilya I rewrote my question, maybe it's clearer now. I'm somehow confused about this... (@all I'm sorry that I totally rewrote it, but hopefully it's easier to answer now.) – saz Jan 9 '13 at 18:06
@DavideGiraudo Could you take a look at the question again? I rewrote it, because the former example was not a good one. I'm simply confused about this definition of $\mathbb{P}^x$... – saz Jan 9 '13 at 18:22
From where do you have $(2)$? – Stefan Hansen Jan 9 '13 at 18:38

The heart of the matter here is that, considering a standard Brownian motion $(W_t)_{t\geqslant0}$ (in particular, such that $W_0=0$ almost surely), for every $x$, one can realize a Brownian motion $B^x=(B^x_t)_{t\geqslant0}$ starting from $x$ by $B^x_t=x+W_t$ for every $t\geqslant0$.
The rest is a matter of notations. Let $\Omega=\mathcal C([0,+\infty),\mathbb R_+)$ and $W_t:\Omega\to\mathbb R$ defined by $W_t(\omega)=\omega(t)$. Then the Wiener measure $\mathbb P$ on $\Omega$ is concentrated on $\Omega_0=\{\omega\in\Omega\mid\omega(0)=0\}$ and, for every $x$ in $\mathbb R$, one can define a probability measure $\mathbb P^x$ on $\Omega$ as the image of $\mathbb P$ by the translation $\sigma_x:\Omega\to\Omega$ defined by $\sigma_x(\omega)(t)=x+\omega(t)$.
Thus $\mathbb P=\mathbb P^0$ and, for every $x$, $\mathbb P^x$ is concentrated on $\Omega_x=\sigma_x(\Omega_0)=\{\omega\in\Omega\mid\omega(0)=x\}$. Furthermore, for every measurable $C$ in $\Omega$, $\mathbb P^x(C)=\mathbb P(\sigma_x^{-1}(C))$. For example $\mathbb P^x(W_t\in D)=\mathbb P(x+W_t\in D)$ for every Borel set $D$ and, more generally, $W$ under $\mathbb P^x$ is distributed like $B^x$ under $\mathbb P$ and is such that $W_0=x$, $\mathbb P^x$-almost surely..
As far as I can see your definition coincides with the definition given in $(2)$. But my question was more about whether one can define the measure $\mathbb{P}^x$ also by $(1)$. In my oppinion $(1)$ and $(2)$ don't define the same measure (as I explained in the edit above). Am I correct? – saz Jan 10 '13 at 12:34