If $f$ is nice what is $E[f(B(t)) \mid \mathcal F_s]$ for $s\leq t$? 
If $f$ is nice what is $E[f(B(t))\mid \mathcal F_s]$ for $s\leq t$ where $B$ is a Brownian motion?

Is there a general formula for this? I could Taylor expand $f$ at $B(s)$ to get
$$E[f(B(s))+f'(B(s))(B(t)-B(s))+\frac12f''(B(s))(B(t)-B(s))^2+\cdots\mid \mathcal F_s]$$
What is this?
 A: Assuming $F_s$ is filtration generated by process $B$, you can always write $B_t = B_s + (B_t - B_s)$. Then due to property of conditional expectation ($\mathbb E[F(X,Y)\mid\mathcal G] = \mathbb E[F(x,Y)]|_{x=X}$, when $X$ is $\mathcal G$ measurable, and $Y$ is independent of $\mathcal G$, while $F$ is borel such that $F(X,Y)$ is integrable), we get:
$$ \mathbb E[f(B_t)\mid\mathcal F_s] = \mathbb E[f(B_s + (B_t - B_s))\mid\mathcal F_s] = \mathbb E[f(x + B_t - B_s)]|_{x=B_s}$$
We used that fact with $F(X,Y) = f(X + Y)$ , $X = B_s$ and $Y = B_t-B_s$ (note $B_s$ is $\mathcal F_s$ measurable, while $B_t-B_s$ is indendent of $\mathcal F_s$). The latter can be rewritten as (let $G \sim \mathcal N(0,1)$ and use stationary increments $B_t - B_s \sim \sqrt{t-s}\cdot G$)
$$ \mathbb E[f(B_t)\mid\mathcal F_s] = \mathbb E[f(x + \sqrt{t-s} G)]|_{x = B_s}$$
So in fact it is enough for $f$ to be borel such that $f(B_t)$ is integrable.
A: If $s\le t$ then
$$
B_t \mid \mathcal F_s \sim \operatorname N(B(s), t-s).
$$
So the question is reduced to this: If $B\sim\operatorname N(b, \sigma^2),$ then what is $\operatorname E(f(B))\text{?}$
\begin{align}
\operatorname E(f(B)) = {} & \frac 1 {\sqrt{2\pi}} \int_{-\infty}^{+\infty} f(b +{}z\sqrt{t-s} ) e^{-z^2/2} \, dz \\[8pt]
= {} & \frac 1 {\sqrt{2\pi}} \int_{-\infty}^{+\infty} f(u) e^{-((u-b)/\sqrt{t-s})^2/2} \,\frac{du}{\sqrt{t-s}}.
\end{align}
