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 levy-processes
Search options answers only
not deleted
user 85303
Theory and applications of Lévy processes (stochastic processes with stationary and independent increments): e.g. path properties, stochastic differential equations driven by jump-type processes, fluctuation theory of Lévy processes, queuing theory.
1
vote
Accepted
Lévy measure and Lévy process
No, for a counterexample, just take
$$
\nu(dx) = \left(C_1 x^{-2} I(0<x<1) +C_2 |x|^{-2} I(-1<x<0)\right)dx,
$$
where $C_1\neq C_2$. Then,
$$
\int_{r<|x|\le 1} x\nu(dx) = (C_2-C_1) \int_r^1 x^{-1} dx …
- 718