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 |
A stochastic process is a collection of random variables usually indexed by a totally ordered set.
6
votes
Accepted
The conditions in the definition of Poisson process (and a Lévy process generalization)
You cannot define a Lévy process by the individual distributions of its increments, except in the trivial case of a deterministic process Xt − X0 = bt with constant b. In fact, you can't identify it b …
11
votes
Correlated Brownian motion and Poisson process
To further elaborate on my comment, it is a theorem that if $X^1,X^2,\ldots,X^n$ are Lévy processes with respect to a common filtration, all starting from zero, then they are independent if and only i …
11
votes
Accepted
The conditions in the definition of Brownian motion
No, it is not true that a process W satisfying the properties (1), (3) and (4) has to be a Brownian motion. We can construct a counter-example as follows.
This construction is rather contrived, and I …
11
votes
Accepted
Filtrations generated by cadlag martingales.
No, that is not true. Consider the following, defined on a filtered probability space $(\Omega,\mathcal{F},\{\mathcal{F}_t\}\_{t\in[0,T]},\mathbb{P})$.
$W$ is a standard Brownian motion.
$U$ is an $ …
18
votes
Accepted
A Markov process which is not a strong markov process?
Consider the following continuous Markov process X, starting from position x
if x = 0 then Xt = 0 for all times.
if x ≠ 0 then X is a standard Brownian motion starting from x.
This is not strong M …
8
votes
Accepted
Does infinite-dimensional Brownian motion live in hyperplanes?
As suggested in my comment, here's a simple fact which applies to any probability measure $\mu$ on (the Borel σ-algebra of) a second countable topological space $X$. There is a unique minimal closed s …
8
votes
Accepted
$L^\infty$ properties of an infinite-dimensional Gaussian semigroup
As suggested in the question, $P_t$ need not be a well defined operator on $L^\infty(W,\mu)$. That is, $F$ can be zero $\mu$-almost everywhere, but $P_tF$ is nonzero on a set of positive $\mu$-measure …
23
votes
Gaussian processes, sample paths and associated Hilbert space.
No, it is not true for simple examples such as standard Brownian motion or a sequence of independent random variables.
Suppose $W$ is a standard Brownian motion on the interval $[0,T]$. The covarian …
10
votes
Accepted
Fictitious density of paths of diffusion processes outside the Cameron--Martin space
Here's a proof of the statement for $f=0$, so that $X=W$ is a Wiener process. (The proof with general $f$ is a bit more involved, and I give this further below). I'll base the proof on the following s …
6
votes
Accepted
Stochastic integrals as honest martingales -- comparison criterion
No, $\rho$ need not be a proper martingale. To guarantee that $p_t=\int_0^ta_sd\rho_s$ is a martingale for all predictable $0\le a_t\le 1$ you need the additional property that $\sup_{s\le t}\rho_s$ i …
4
votes
Accepted
Stochastic integrals as honest martingales — exponential damping
Yes, in this case it is true that $p$ is a proper martingale! Note that your integrand $\exp\left(-\int_0^tr_u du\right)$ is an adapted, continuous, and decreasing process bounded by 1. So, the follow …
17
votes
Accepted
Bochner integral of stochastic process = path by path Lebesgue integral?
Yes, the Bochner integral does agree with the Lebesgue integral of the sample paths of the process. We can prove this in a slightly more general situation than that asked for in the question.
For a p …
7
votes
Accepted
Compactness of the set of densities of equivalent martingale measures
The set $Z_{\mathcal{P^\ast}}$ is never compact except in the case where it is a singleton (or empty). This is for the general case with $S=(S^1,S^2,\ldots,S^d)$ being an $\mathbb{R}^d$-valued semimar …
11
votes
Accepted
Extending state space to make a process Feller
Yes, it is possible to extend the state space with respect to which $Y$ is a Feller process. Then, $X$ will be a dense open subset of the extension $\hat X$. Furthermore, for any initial distribution …