Skip to main content
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
Search options not deleted user 41082

Theory and applications of probability and stochastic processes: e.g. central limit theorems, large deviations, stochastic differential equations, models from statistical mechanics, queuing theory.

7 votes
1 answer
2k views

Exponential tail bounds without the moment generating function

As an exercise, I thought I would try to prove some classical Chernoff bounds without ever using the moment generating function, but then found myself getting stuck in certain places. Before I state m …
Jelani Nelson's user avatar
3 votes

Exponential tail bounds without the moment generating function

Ok, I figured out the answer to my own question, though without the $\ln(ep K^2/\sigma^2)$ in the denominator. I'll see whether I can get that later (or maybe someone else sees it?). We can assume wit …
Jelani Nelson's user avatar
2 votes

bilinear form tail bound

The Hanson-Wright inequality says $$ \mathop{\mathbb{P}}_{u,v}(|u^T A v| > t) \le C\cdot \max\left\{e^{-c t^2/\|A\|_F^2}, e^{-c t/\|A\|}\right\} $$ where $\|\cdot\|$ is the largest singular value and …
Jelani Nelson's user avatar
1 vote

bound the tail distribution

Exponential tail bounds automatically imply moment bounds and vice versa. That is to say, $(a)$ is equivalent to $(A)$ for $a\in \{j,k,l\}$ below where $X$ is a nonnegative random variable and $\|X\|_ …
Jelani Nelson's user avatar