Jump to content

Talk:Bernstein inequalities (probability theory)

Page contents not supported in other languages.
From Wikipedia, the free encyclopedia

Untitled

[edit]

Isn't it more accurate to say that the proofs are based on Markov's inequality rather than Chebyshev's? I haven't read seen the proofs for Benstein's inequalities but the proofs of the closely related Chernoff and Azuma-Hoeffding inequalities are all based on Markov's inequality. --129.34.20.23 (talk) 23:29, 28 September 2010 (UTC)[reply]

"Markov inequality" is a misnomer (although often used). Chebyshev was the supervisor of Markov, and he applied the inequality to X, X2, and other functions of X. Sasha (talk) 15:58, 23 February 2012 (UTC)[reply]

I don't see how the Hoeffding and Azuma inequalities are special cases of Bernstein's inequalities. They are proven in a similar manner, of course.--Olethros (talk) 15:21, 23 February 2012 (UTC)[reply]

when I wrote this, I think I checked Bernstein's paper vs. Hoeffding's, and Hoeffding's result was indeed contained in that of Bernstein (ie my claim is that H's result is contained in B's paper, not that it follows from the simplest case stated in the lead of our article). This was long ago, and now I do not have either paper at hand, but I will try to check again and provide better references (preferably, secondary ones). Sasha (talk) 15:58, 23 February 2012 (UTC)[reply]

Something is clearly inconsistent: the article on Hoeffding ineq. says Hoeffding is more general than Bernstein. But the article on Bernstains says that "special cases of Bernstein is also known as ... Hoeffding ineq." -- 194.126.99.204 (talk) 13:14, 14 June 2012 (UTC)[reply]

It could well depend on what is meant by Hoeffding's inequality. Hoeffding's paper contains several general results that are more complicated but stronger than what is typically called Hoeffding's inequality: see paper cited in Hoeffding's inequality. Melcombe (talk) 22:45, 14 June 2012 (UTC)[reply]

The second inequality is incorrect as stated. (For example, if one chooses all X_i to be constant at 5, I think the stated inequality is false for very small t). It's possible that the inequality is true when the X_i have 0 expected value. I tried searching the references for these inequalities but couldn't find the statement of the second one :-( — Preceding unsigned comment added by 198.102.153.2 (talk) 20:20, 10 July 2015 (UTC)[reply]

Assessment comment

[edit]

The comment(s) below were originally left at Talk:Bernstein inequalities (probability theory)/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

The article is a sea of formula with little supporting text, desperately needs some more explanation. --Salix alba (talk) 14:50, 20 May 2007 (UTC)[reply]

Last edited at 14:50, 20 May 2007 (UTC). Substituted at 19:49, 1 May 2016 (UTC)