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 |
Theory and applications of probability and stochastic processes: e.g. central limit theorems, large deviations, stochastic differential equations, models from statistical mechanics, queuing theory.
1
vote
measurability of integrated functions
I know in computable analysis, which is closely related to the descriptive set theoretic questions you are asking, that the Lévy–Prokhorov metric on the weak topology is useful. (I think the narrow t …
11
votes
2
answers
2k
views
De Finetti's theorem, the pointwise ergodic theorem, and reverse martingales
De Finetti's theorem says that an exchangeable sequence of random variables $X_i$ is a mixture of i.i.d. random variables. In other words, if $\mu$ is a measure on $\mathbb{R}^\infty$ that is invaria …
13
votes
1
answer
3k
views
Does this metric have an official name? Lévy metric? Ky Fan metric?
Let $X$ and $Y$ be random variables taking values in a separable metric space $(S,d)$. The metric I have in mind is
$$\rho(X,Y) = \mathbb{E}[\min\{d(X,Y),1\}]$$
if $X$ and $Y$ take values in the a me …
5
votes
2
answers
639
views
Is the Hausdorff metric on sub-$\sigma$-fields separable?
Let $(X,\mu,\mathcal{F})$ be a probability space. The paper Equiconvergence of Martingales by Edward Boylan introduced a pseudometric on sub-$\sigma$-fields (sub-$\sigma$-algebras) of $\mathcal{F}$ a …
13
votes
Accepted
How do we express measurable spaces using type theory?
If you clarify your question, I can modify this answer to be better. (I am sure you won't like my answer. The best answer is this is not practical, except in the simplest of settings. Or you are in …
9
votes
2
answers
535
views
What mode of convergence is this?
I'm interested in a new (to me) mode of convergence which is stronger than convergence in measure/probability. I want to know if it has a name and if it is used much in the literature. I will write …
4
votes
0
answers
309
views
Conditional expectation with respect to random closed sets
Short question
If $X$ is a random closed set, and $Y$ is an integrable random variable, I would like a definition of the "conditional expectation" $$\mathbf{E}[Y \mid X\ni x].$$ Has this been worked …
4
votes
Continuity on a measure one set versus measure one set of points of continuity
If $X$ and $Y$ are Polish and $Y$ is compact, then YES. (I think my proof can be fixed to handle the noncompact setting, but I don't see how right now.)
My proof involves this lemma.
Lemma. Assume …
3
votes
Symmetries of the standard probability space
I can't explain the group theoretic structure of $\Gamma$, but I can explain the topological structure. (Warning, this post is mostly a continuous stream of thoughts. I hope it is well organized and …
2
votes
Is There An Algorithmic Complexity Of A Random Distribution
The short answer is it depends on what one means by extending Kolmogorov complexity. The details are below. First, when I say Kolmogorov complexity, I will mean prefix-free complexity since that is …
5
votes
De Finetti's theorem, the pointwise ergodic theorem, and reverse martingales
(My understanding of this material has significantly gone up in the months since I asked it, and I will attempt to answer my own question.)
In general, if $(\Omega,\mathcal{B},\mathbb{P},\{T_g\})$ is …
11
votes
0
answers
223
views
Savings property: A transformation which turns nonnegative martingales into uniformly integr...
Background
I work in a subfield of computability theory called algorithmic randomness. We have been using martingales as long as probability theory (going back to work of von Mises). However, since …
1
vote
0
answers
1k
views
What conditions on a filtration guarantee that a (sub)martingale has a continuous modification?
There is a theorem as follows:
Theorem. Let $\mathcal{F}_t$ be a filtration which is right-continuous and complete. Assume $M_t$ is a submartingale adapted to $\mathcal{F}_t$ such that $t \mapsto \m …
7
votes
1
answer
389
views
Reference request: Martingale decompositions (positive/negative and u.i./singular)
For a paper I am writing, I need these two facts. The proofs are fairly short, but I would rather just cite them. This is for martingales index by natural numbers. Also, I call a martingale which co …