Conditional Brownian Expectations from the Normal Density

# Write expectation of brownian motion conditional on filtration as an integral?


# Write expectation of brownian motion conditional on filtration as an integral?












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.

## Answer by Quantuple (score 1, accepted)

https://quant.stackexchange.com/a/36622

Let $(W_t)_{t \geq 0}$ denote a standard Brownian motion and $\Bbb{F}=\{\mathcal{F}_t\}_{t \geq 0}$ the natural filtration it generates over some probability space $(\Omega, \Bbb{P})$.

By definition, we know that $$\forall 0 < s < t, W_t - W_s \sim \sqrt{t-s}\, N(0, 1)$$ under $\Bbb{P}$.

Noting that $W_t = (W_t - W_s) + W_s$ and conditioning on the knowledge of $\mathcal{F}_s$ we could further write that $$ W_t \vert \mathcal{F}_s \sim N(W_s, t-s)$$

Such that \begin{align} \Bbb{E}[W_t \vert \mathcal{F}_s] &= \int_{-\infty}^{+\infty} W_t p(W_t\vert\mathcal{F}_s) dW_t \\ &= \int_{-\infty}^{+\infty} x \frac{1}{\sqrt{2\pi(t-s)}}\exp\left(-\frac{1}{2}\left(\frac{x - W_s}{\sqrt{t-s}}\right)^2\right) dx \tag{A}\\ &= \underbrace{\int_{-\infty}^{+\infty} (x - W_s) \frac{1}{\sqrt{2\pi(t-s)}}\exp\left(-\frac{1}{2}\left(\frac{x - W_s}{\sqrt{t-s}}\right)^2\right) dx}_{=0} \dots \\ & \dots + W_s \int_{-\infty}^{+\infty}\underbrace{\frac{1}{\sqrt{2\pi(t-s)}}\exp\left(-\frac{1}{2}\left(\frac{x - W_s}{\sqrt{t-s}}\right)^2\right) dx}_{=1} \\ &= W_s \end{align} where $(A)$ is the result you're looking for.

Note that this is basically the hand-wavy way of writing: \begin{align} \Bbb{E}[W_t \vert \mathcal{F}_s] &= \Bbb{E}[W_t -W_s + W_s \vert \mathcal{F}_s] \\ &= \Bbb{E}[W_t -W_s \vert \mathcal{F}_s] + \Bbb{E}[ W_s \vert \mathcal{F}_s] \\ &= 0 + W_s = W_s \end{align}