I have the following proposition in my textbook:
Let B be a standard BM. Then, whenever the integral exists, $$E[h(B_{t+s} | \mathcal{F}_s ] = \int_{\mathbb{R}} h(y)p(t;B_s,y)dy =E^{B_s}[h(B_t)].$$
My question is, what is the transition probability density function $ p(t;x,y)$?
I know that for a standard Brownian motion,
$$ p(t;x,y) = \frac{1}{\sqrt{2\pi t}} \text{exp}\big(-\frac{(y-x)^2}{2t} \big)$$
But I'm not sure what the arguments t, x and y is.
I suppose $(\mathscr{F}_t)_{t \geq 0}$ is the natural filtration of the Brownian motion. In such case, we know that $B_t-B_s$ is independent of $\mathscr{F}_s$ for all $s<t$; and of course $B_s$ is $\mathscr{F}_s$-measurable. We have (standard assumption: $h$ is measurable and bounded) $$\begin{aligned}E[h(B_t)|\mathscr{F}_s]&=E[h(B_t-B_s+B_s)|\mathscr{F}_s]=\\&=E[h(B_t-B_s+b)]|_{b=B_s}=\\ &=E[h(B_{t-s}+b)]|_{b=B_s} \end{aligned}$$ where the last equality follows from $B_t-B_s\sim B_{t-s}$. Also $E[h(B_t)|\mathscr{F}_s]=E[h(B_t)|B_s]$ by the same reasoning (recall: $B_t-B_s$ is independent of $B_s-B_0=B_s$ by properties of Brownian motion) so $$\begin{aligned}E[h(B_t)|B_s=b]&=E[h(B_{t-s}+b)]=\\ &=\int_\mathbb{R}h(x+b)\frac{1}{\sqrt{2\pi (t-s)}}\exp\bigg(-\frac{x^2}{2(t-s)}\bigg)dx=\\ &=\int_\mathbb{R}h(y)\frac{1}{\sqrt{2\pi (t-s)}}\exp\bigg(-\frac{(y-b)^2}{2(t-s)}\bigg)dy \end{aligned}$$