Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with brownian-motion
Search options not deleted
user 29212
In mathematics, Brownian motion is described by the Wiener process; a continuous-time stochastic process named in honor of Norbert Wiener.
0
votes
0
answers
1k
views
Proof that integral of Brownian motion wrt time is not a martingale
Let $X_t=\int_0^t W_s ds$ where $W_s$ is Brownian motion, so $E[W_s]=0$.
Then $E[X_t]=\int_0^t E[W_s] ds=\int_0^t 0 ds=0$.
So $E[X_t|{\cal F}_s]=0\neq X_s$, almost everywhere. So by previous senten …
- 187
1
vote
1
answer
827
views
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
=\i …
- 187