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 $$ p(x,t;y,T) = \frac{1}{\sqrt{2 \pi}^d} \exp \left ( -\frac{1}{2} \frac{\|x-y \|^2}{T-t}\right ). $$ 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 $$ p(x,t;y,T) = \int_{E^n} \frac{1}{\sqrt{2 \pi}^{nd}} \exp \left ( -\frac{1}{2} \sum_{i=0}^{n} \frac{\|x_{i+1}-x_i \|^2}{t_{i+1}-t_i}\right ) d x_1 ... d x_n. $$ 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 $C(x,t;y,T))$?

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 $C(x,t;y,T))$?

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 $$ p(x,t;y,T) = \frac{1}{\sqrt{2 \pi}^d} \exp \left ( -\frac{1}{2} \frac{\|x-y \|^2}{T-t}\right ). $$ 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 $$ p(x,t;y,T) = \int_{E^n} \frac{1}{\sqrt{2 \pi}^{nd}} \exp \left ( -\frac{1}{2} \sum_{i=0}^{n} \frac{\|x_{i+1}-x_i \|^2}{t_{i+1}-t_i}\right ) d x_1 ... d x_n. $$ 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 should be interpreted as $$ \mu (C(x,t;y,T)) $$ where $\mu$ is 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 establish rigorously that $p(x,t;y,T) = \mu (C(x,t;y,T))$?

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 $$ p(x,t;y,T) = \frac{1}{\sqrt{2 \pi}^d} \exp \left ( -\frac{1}{2} \frac{\|x-y \|^2}{T-t}\right ). $$ 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 $$ p(x,t;y,T) = \int_{E^n} \frac{1}{\sqrt{2 \pi}^{nd}} \exp \left ( -\frac{1}{2} \sum_{i=0}^{n} \frac{\|x_{i+1}-x_i \|^2}{t_{i+1}-t_i}\right ) d x_1 ... d x_n. $$ 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 $C(x,t;y,T))$?