Let $W(t)$ be a standard brownian motion in $E \triangleq \mathbb{R}^d$. The transition probability from a state $x \in E$ at time $t$ to a state $y \in E$ at time $T$ is $$ p(x,t;y,T) = \frac{1}{\sqrt{2 \pi (T-t)}^d} \exp \left ( -\frac{1}{2} \frac{\|x-y \|^2}{T-t}\right ). $$ Using the Chapman-Kolmogorov equation, it is clear that for all $n \geq 0$ and for all finite sequence of states $x_0 = x, x_1, ... , x_n, x_{n+1} = y$ and $t_0 = t < t_1 < ... < t_n < t_{n+1} = T$, we have $$ p(x,t;y,T) = \int_{E^n} \frac{1}{\sqrt{2 \pi}^{nd}} \exp \left ( -\frac{1}{2} \sum_{i=0}^{n} \frac{\|x_{i+1}-x_i \|^2}{t_{i+1}-t_i}\right ) \prod_{i=1}^n \frac{d x_i}{\sqrt{t_{i+1}-t_i}^d}. $$ I am not so sure that the limit as $n \to \infty$ makes sense in the right hand side above but I believe that it can be interpreted in terms of the Wiener measure on $\mathcal{C}([0,\infty);E)$ and $$ C(x,t;y,T) \triangleq \{ \omega \in \mathcal{C}([0,\infty);E) \: | \: w(t) = x, \: w(T) = y \}. $$ (if no mistake) how to interpret $p(x,t;y,T)$ in terms of $\mu$ and $C(x,t;y,T))$?
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1$\begingroup$ This isn't really correct. Your set $C(x,t;y,T)$ has Wiener measure zero. The "transition probability" is really a probability density and needs to be understood as such. $\endgroup$– Nate EldredgeJul 2, 2017 at 15:03
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1$\begingroup$ Thank you Nate, could you help me to show that $C(x,t;y,T)$ has Wiener measure zero? $\endgroup$– megaprobaJul 2, 2017 at 15:06
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2$\begingroup$ That would be a better question for Math.SE. But you are asking about $P(\{B_t = x\} \cap \{B_T = y\})$. The random variable $B_t$ has a nondegenerate Gaussian distribution, which is a continuous distribution, so $P(B_t = x)=0$ for any $x$. $\endgroup$– Nate EldredgeJul 2, 2017 at 15:08
1 Answer
I have find in [equation 1.1.51, page 24,Chaichian M, Demichev A. Path integrals in physics, vol I : stochastic processes and quantum mechanics. Bristol (Philadelphia): Institute of Physics Publishing; 2001] that when $N$ goes to $\infty$, then $$ \lim_{\Delta t \to 0,\\ N \to \infty} \exp \left ( -\frac{1}{2} \sum_{i=1}^N \frac{(x_i-x_{i-1})^2}{t_i-t_{i-1}}\right ) \prod_{i=1}^N \frac{d x_i}{\sqrt{2 \pi (t_i -t_{i-1})}} =\\ ``\exp \left ( -\frac{1}{2} \int_0^t |\dot{x}(s)|^2 d s \right ) \prod_{\tau=0}^t \frac{d x(\tau)}{\sqrt{2 \pi d \tau}}" $$ thus, if $A_y$ is a non-empty open set containing $y$, $$ \mathbb{P} (x(T) \in A_y | x(t) = x) = \int_{A_y} p(x,t;y,T) d y\\ = "\int_{C(x,t;A_y,T)} \int_{E^\infty}\exp \left ( -\frac{1}{2} \int_0^t |\dot{x}(s)|^2 d s \right ) \prod_{\tau=0}^t \frac{d x(\tau)}{\sqrt{2 \pi d \tau}}" $$ Here we don't really know what's inside the integral but, if no mistake, what should be correct is to write $$ \mathbb{P} (x(T) \in A_y | x(t) = x) = \mu (C(x,t;A_y,T)). $$