How to show $F(x_1,\ldots,x_n) = \int_0^{x_1}\cdots\int_0^{x_n}\prod p(t_k,t_{k-1},y_{k-1},y_k)dy_1\cdots dy_n$? Suppose $B_t$ a Brownian motion, and I fix $0=t_0<t_1<\cdots<t_n$. I want to show that $$P(B_{t_1} \leq x_1, \ldots, B_{t_n} \leq x_n) = F(x_1,\ldots,x_n) = \int_0^{x_1}\hspace{-.1in}\cdots\int_0^{x_n}\prod_{k=1}^n p(t_k-t_{k-1},y_{k-1},y_k)dy_1\cdots dy_n$$
where $p(t,x,y)=\frac{1}{\sqrt{2\pi t}}\exp(\frac{1}{2t}(x-y)^2)$ is the Gaussian kernel.
I have tried setting $Y_{t_i} := B_{t_i}-B_{t_{i-1}}$ and writing $P(B_{t_1} \leq x_1, \ldots, B_{t_n} \leq x_n) = P(Y_{t_1} \leq x_1, Y_{t_2} + X_{t_2} \leq x_2, \ldots, Y_{t_n} + X_{t_n} \leq x_n)$ and some other change of variable tricks, but I can't seem to fit it into the form above.

I figured this out myself. Simply write $B=(B_{t_1},\ldots,B_{t_n})$, $U=(U_1,\ldots,U_n)$ with $U_i = B_{t_i}-B_{t_{i-1}}$, so $B_{t_i}=\sum_1^i U_k$, i.e. $B=AU$ for $A_{ij}=1$ if $i\leq j$. The distribution of $U$ is known since the $U_i$ are independent, and the change of variables formula then tells us the distribution of $B$.
