calculate the conditional distribution of brownian motion Suppose $W=(W_t)$ is a Brownian Motion with respect to a filtration $(\mathcal{F}_t)$. How can I compute the conditional distribution of $W_{t+h}$ given $\mathcal{F}_t$.
I started like this: $W_{t+h}-W_t$ is idependent of $\mathcal{F}_t$ and normald distributed with mean $0$ and variance $h$. Then I wrote $W_{t+h}=W_t+ (W_{t+h}-W_t)$, hence I have to compute:
$$P[W_{t+h}=W_t+ (W_{t+h}-W_t)\in A|\mathcal{F}_t]$$
For $A\in \mathcal{B}(\mathbb{R})$. I wrote the conditional probability as a expectation of an indicator function. The result should be a normal distribution with mean $W_t$ and variance $h$. Thanks for your help
math
 A: The conditional distribution of $W_{t+h}$ conditionally on $\mathcal F_t$ is a random distribution, that is, a map $M:\Omega\to\mathcal M_1^+(\mathbb R,\mathcal B(\mathbb R))$, measurable with respect to $\mathcal F_t$, and such that, for every bounded measurable function $u$,
$$
\mathrm E(u(W_{t+h})\mid\mathcal F_t)=\int_{\mathbb R} u\mathrm dM\quad\text{almost surely}.
$$
The OP explains why
$$
\mathrm E(u(W_{t+h})\mid\mathcal F_t)=\mathrm E(u(W_{t}+Z_h)\mid W_t)\quad\text{almost surely},
$$
where $Z_h$ is centered normal with variance $h$ and independent of $W_t$. Thus, for every bounded measurable function $u$,
$$
\int_{\mathbb R} u\mathrm dM=\int_{\mathbb R} u(W_t+z)\mathrm d\gamma_h(z)\quad\text{almost surely},
$$
where $\gamma_h$ is the centered normal distribution with variance $h$. This proves that, for $\mathrm P$-almost every $\omega$, the distribution $M(\omega)$ is normal with mean $W_t(\omega)$ and variance $h$.
Now, my suggestion would be to forget what is written above and, instead, to find and read (and meditate) an excellent (and most congenial) presentation of this stuff (and much more) given in the little (but excellent) blue book called Probability with martingales by David Williams.
