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 |
Stochastic calculus provides a consistent theory of integration for stochastic processes and is used to model random systems. Its applications range from statistical physics to quantitative finance.
3
votes
Accepted
Girsanov theorem and the density of a process
Short answer: it follows directly from the Radon-Nikodym theorem.
Longer answer: Let $\mu$ denote the law of $Y(t)$ under $\mathbb P_{y_0}$ and let $\nu$ denote the law of $Y(t)$ under $\mathbb Q_{y_ …
3
votes
1
answer
340
views
Example of Girsanov change of density with finite relative entropy, but with infinite integr...
Let $(\Omega, (\mathcal F_t), \mathbb P)$ denote the usual Wiener space where $\Omega = C[0,\infty)$, etc., and where $(W_t)_{t \geq 0}$ denotes the Wiener process.
Let $Z \in L^1(\mathbb P)$ with $Z …
2
votes
Accepted
Example of Girsanov change of density with finite relative entropy, but with infinite integr...
I can now answer my own question. There exists no such counterexample. I can show that finite relative entropy in this setting implies $\mathbb E^{\mathbb Q} \int_0^{\infty} \theta_t^2 \ d t < \infty$ …
0
votes
On the existence and uniqueness of solution to SPDE with nonlinear growth coefficients
flawed, see Martin Hairer's comment below.
Step I) Perform substitution $u_t(x) = \exp(\psi_t(x))$. The SPDE for $\psi$ becomes, after dividing by $u_t(x)$,
\begin{equation}
\frac{\partial}{\partial …
4
votes
A Stochastic Taylor Expansion/Asymptotics
Write $f(t) := \mathbb E \left[ \exp\left( - \int_0^t r_s \ d s \right) \right]$. Define stochastic processes $y_t = \exp \left( -\int_0^t r_s \ d s \right)$ and $z_t = r_t y_t$. Then
$y_t = 1 - \int_ …