Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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...?

share|cite|improve this question
@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
up vote 2 down vote accepted

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..

share|cite|improve this answer
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
I saw your explanations about conditionings by events of probability zero but I think this is not the direction to go to. The reason why (1) and (2) are equivalent, when (1) and (2) have the meaning people give them to, is the one I explain in my post. – Did Jan 10 '13 at 12:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.