Let $W_t$ be a Brownian motion, so $W_t=z_t \sqrt{t}$ where $z_t \in N(0,1)$ and the pdf of $z$ is $f(z)=\frac{e^{-\frac{z^2}{2}}}{\sqrt{2\pi}}$. So

$$E(W_t)=\int_{-\infty}^{\infty} W_t f(z) dz =\int_{-\infty}^{\infty} z \sqrt{t} \frac{e^{-\frac{z^2}{2}}}{\sqrt{2\pi}} dz =\int_{0}^{\infty} (z+(-z)) \sqrt{t} \frac{e^{-\frac{z^2}{2}}}{\sqrt{2\pi}} dz=0$$

Now suppose ${\cal F}_t$ is the natural filtration for $W_t$. By construction of Brownian motion, we are given that $E(W_t|{\cal F}_s)=W_s, 0\leq s\leq t$.

Question: How do I write $E(W_t|{\cal F}_s)$ as a Riemann integral expression similar to the Riemann integral expression of $E(W_t)$ given above?

Note: I have done extensive Google search on this, without finding any responsive exposition. If this question is beside the point, please explain why. If it's on point, please answer with the Riemann integral expression.

Note: Cross posted.


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