Hi,

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!