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 integration
Search options not deleted
user 501568
Questions related to various forms of integration including the Riemann integral, Lebesgue integral, Riemann–Stieltjes integral, double integrals, line integrals, contour integrals, surface integrals, integrals of differential forms, ...
0
votes
Accepted
Random variable as an integral of an indicator function
The layer cake representation of a non-negative measurable function, $X$, is applied in the proof of proposition 2.1 here.
- 115
-1
votes
1
answer
933
views
Random variable as an integral of an indicator function
My thinking is that for a probability space $(\Omega, F, \mu)$, $\{X > t\} \in F \implies \int_\Omega I_{\{X > t\}}\mathrm{d}\mu = \mu(\{X > t\})$ but it's the wrong variable of integration. …
- 115