I am just starting to study the theory of Brownian motion and I was wondering whether the following was true.

We consider a one-dimensional, one sided, Brownian motion process.

For $A$ an event on $\mathcal{C}(\mathbb{R^+})$ (space of continuous functions from $\mathbb{R^+}$ to $\mathbb{R}$) we denote, for all $x \in \mathbb{R}$, by $\mathbb{P}(x,A)$ the probability that a Brownian path starting at $x$ (at time $0$) belongs to $A$. For a given time $\tau$, we denote by $A_{\tau}$ the event $t \mapsto B(t -\tau) \in A$" ($t$ varying from $\tau$ to $\infty$).

Now, given an event $A$ which only depends on the values of $B(t)$ for $t > \varepsilon$ ($\varepsilon > 0$ being fixed), then for any $x$ and any $0 \leq \tau < \varepsilon$, the following relation holds: $$ \mathbb{P}(x,A) = \int_{y \in \mathbb{R}} \mathbb{P}(x+y,A_\tau) \cdot \frac{e^{-y^2/(2\tau)}}{\sqrt{2\pi \tau} } dy $$

My intuition (hopefully not too naive) is that if we take a path with value $x$ at time $0$ at random, and observe its value at time $\tau$, this value, as a random variable, obeys a normal distribution of mean $x$ and variance $\tau$. Now, knowing that the value of the path is $x+y$ at time $\tau$, the probability that the path is in $A$ is $\mathbb{P}(x+y,A_\tau)$.

So first of all: is this indeed correct? If so, the next question: can someone give me a short proof of this fact or, even better, a reference for this formula? (I cannot find it in the books I've been reading on the subject).

Many thanks in advance!

  • $\begingroup$ This does not seem to be right. Fix some $x$ and let $A = \{ \omega : \omega(0) = x\}$, i.e. $A$ is the event $\{B_0 = x\}$. Then $P(x,A) = 1$. Now $A_\tau$ is the event $\{B_\tau = x\}$; so for any $\tau \ne 0$ and any $y \in \mathbb{R}$ we have $P(x+y, A_\tau) = 0$, so the equation doesn't hold. $\endgroup$ – Nate Eldredge May 20 '12 at 0:43
  • $\begingroup$ You are right, thanks. I fixed my hypotheses to address your comment. $\endgroup$ – Laurent Bienvenu May 20 '12 at 15:46
  • $\begingroup$ It is still does not seem correct with the correction, you should request that $A$ depends on $(B_t)_{t\ge\tau}$ only. Anyway, I think what you are trying to formulate is the \emph{Markov property} of Brownian motion, I therefore recommend you to inform yourself about the basics of Markov processes (or, to start with, about Markov chains). $\endgroup$ – Pascal Maillard May 21 '12 at 7:43
  • $\begingroup$ I did request that. And no, it does not seem to just be a reformulation of the Markov property, even though it must be of crucial use to prove the equality (if correct). $\endgroup$ – Laurent Bienvenu May 23 '12 at 21:18

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