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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


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I'm confused. Doesn't the martingale property give $\mathbb E[W_{t+h}|\mathcal F_t]=W_t$? –  user31373 Jun 20 '12 at 23:44
@ Leonid Kovalev: Yes, of course, but writing the above conditional probability as an expectation, leads to: $P[W_{t+h}=W_t+(W_{t+h}-W_t)\in A|\mathcal{F}_t]=E[\mathbf1\{W_{t+h}=W_t+(W_{t+h}-W_t)\in A\}|\mathcal{F}_t]$. I do not see how to proceed from here –  math Jun 21 '12 at 6:50

1 Answer 1

up vote 3 down vote accepted

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.

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Thanks for your suggestion. I will have a look at the book. –  math Jun 21 '12 at 7:45
@Did could you give more details about the last step? Why can can you conclude it carring only about the distribution of $Z$? –  sky90 Jun 22 at 13:06
@sky90 Because $W_{t+h}$ is distributed as $W_t+Z_h$ where $W_t$ is $\mathcal F_t$-measurable and $Z_h$ is independent of $\mathcal F_t$, and because, for every $w$, $w+Z_h$ is normal with mean $w$ and variance $h$. –  Did Jun 22 at 15:31
I see thanks :) –  sky90 Jun 22 at 15:38

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