In probability theory, Hoeffding's inequality provides an upper bound on the probability that the sum of bounded independent random variables deviates from its expected value by more than a certain amount. Hoeffding's inequality was proven by Wassily Hoeffding in 1963. Hoeffding's inequality is a … Se mer Let X1, ..., Xn be independent random variables such that $${\displaystyle a_{i}\leq X_{i}\leq b_{i}}$$ almost surely. Consider the sum of these random variables, $${\displaystyle S_{n}=X_{1}+\cdots +X_{n}.}$$ Se mer The proof of Hoeffding's inequality follows similarly to concentration inequalities like Chernoff bounds. The main difference is the use of Se mer • Concentration inequality – a summary of tail-bounds on random variables. • Hoeffding's lemma Se mer The proof of Hoeffding's inequality can be generalized to any sub-Gaussian distribution. In fact, the main lemma used in the proof, Hoeffding's lemma, implies that bounded random variables are sub-Gaussian. A random variable X is called sub-Gaussian, if Se mer Confidence intervals Hoeffding's inequality can be used to derive confidence intervals. We consider a coin that shows heads with probability p and tails with probability 1 − p. We toss the coin n times, generating n samples Se mer Nettet15. jan. 2002 · Hoeffding's inequality is a key tool in the analysis of many problems arising in both probability and statistics. Given a sequence Y ≡ (Y i: i⩾0) of independent and bounded random variables, Hoeffding's inequality provides an exponential bound …
Hoeffding
http://chihaozhang.com/teaching/SP2024spring/notes/lec8.pdf NettetSimilar results for Bernstein and Bennet inequalities are available. 3 Bennet Inequality In Bennet inequality, we assume that the variable is upper bounded, and want to estimate its moment generating function using variance information. Lemma 3.1. If X EX 1, then 8 0: lnEe (X ) (e 1)Var(X): where = EX Proof. It suffices to prove the lemma when ... melanie olmstead yellowstone cast and crew
Understanding the Hoeffding Inequality - Open Data Science
NettetX0 is obtained from X by replacing the kth coordinate Xk with an independent copy X0 k and leaving all of the other coordinates alone. Then E(Y 0jFk) ˘E(Y jFk¡1) ˘Yk¡1 But by hypothesis, jY 0 ¡Y j•¾k.This implies that jYk ¡Yk¡1j˘jE(Y ¡Y 0jF k)j•¾k. Given this, the result follows immediately from the Azuma-Hoeffding inequality, because Y ˘ E(Y jFn) … Nettet23. jan. 2024 · The inequality I'm having trouble with is the following : The first line is clearly true by the law of total expectation, and I understand that the second line is a direct application of Hoeffding's inequality since, conditional on the data, is a sum of i.i.d … NettetHoeffding's lemma: Suppose x is an random variable, x∈ [a, b] , and E (x)=0 , then for any t>0 , the following inequality holds： E (e^ {tx})\leq exp\frac {t^2 (b-a)^2} {8} We prove the lemma first: Obviously, f (x)=e^ {tx} is a convex function, so for any α∈ [0,1] , we have: f (αx_1+ (1-α)x_2)\le αf (x_1)+ (1-α)f (x_2) melanie orlins where is she going