16.12.2019 · Reverse Markov Inequality on the Unit Interval for Polynomials Whose Zeros Lie in the Upper Unit Half-Disk M. A. Komarov 1 Analysis Mathematica volume 45 , pages 817–821 ( 2019 ) Cite this article
Title: reverse Markov inequality: Canonical name: ReverseMarkovInequality: Date of creation: 2013-03-22 17:48:08: Last modified on: 2013-03-22 17:48:08: Owner
Reverse Markov inequality 175 In this note, we are interested in a general form of the reverse Markov in-equality for a polynomial phaving all of its zeros in K: (6) kp0k K‚C(K)(degp)bkpkK: Note that any polynomial p(z) = czn+1, where jcjis su–ciently small, provides a counterexample to any form of the reverse Markov inequality for polynomials
Full-text available Dec 2015 MATH INEQUAL APPL Polina Yu. Glazyrina Szilárd Révész ... In this estimate, which is reverse to the classical Markov inequality [8], the order √ n is sharp because of...
For example, Markov's inequality tells us that as long as X doesn't take negative values, the probability that X is twice as large as its expected value is ...
REVERSE MARKOV INEQUALITY. Norman Levenberg and Evgeny A. Poletsky. University of Auckland, Department of Mathematics. Private Bag 92019, Auckland, ...
Reverse Markov Inequality for non-negative unbounded random variables. Ask Question Asked 10 years, 1 month ago. Active 4 years, 9 months ago. Viewed 7k times 6 5 $\begingroup$ I need to lower bound the tail probability of a non-negative random variable. I have a lower bound on ...
07.03.2016 · The usual trick for this type of question is to use indicator function. Given the assumptions, We claim that the following inequality is true. a I ( Z > 1 − a) ≥ Z − ( 1 − a). Then we discuss in case that (i) Z > 1 − a and (ii) Z ≤ 1 − a. For (i) since Z > 1 − a, we have a > Z − 1 + a which is true; For (ii), the conclusion is ...
Title: reverse Markov inequality: Canonical name: ReverseMarkovInequality: Date of creation: 2013-03-22 17:48:08: Last modified on: 2013-03-22 17:48:08: Owner
Dec 16, 2019 · Reverse Markov Inequality on the Unit Interval for Polynomials Whose Zeros Lie in the Upper Unit Half-Disk M. A. Komarov Analysis Mathematica 45 , 817–821 ( 2019) Cite this article 50 Accesses 4 Citations Metrics Abstract We prove that there is an absolute constant A > 0 such that
Let K be a compact convex set in C. For each point z O ε ∂K and each holomorphic polynomial p = p(z) having all of its zeros in K, we prove that there exists a point z ε K with |z -z O | ≤ 20 diam K/ √deg p such that |p′(z)| ≥ (deg p) 1/2 /20(diam K) |p(z O)| i.e., we have a pointwise reverse Markov inequality.
In probability theory, Markov's inequality gives an upper bound for the probability that a ... taking expectation of both sides of an inequality cannot reverse it.
Request PDF | Reverse Markov inequality | Let K be a compact convex set in C. For each point zO ε ∂K and each holomorphic polynomial p = p(z) having all of its zeros in K, we prove that ...
Reverse Markov inequality 175 In this note, we are interested in a general form of the reverse Markov in-equality for a polynomial phaving all of its zeros in K: (6) kp0k K‚C(K)(degp)bkpkK: Note that any polynomial p(z) = czn+1, where jcjis su–ciently small, provides a counterexample to any form of the reverse Markov inequality for polynomials
Request PDF | Reverse Markov inequality | Let K be a compact convex set in C. For each point zO ε ∂K and each holomorphic polynomial p = p(z) having all of ...
as reverse (or inverse) Markov- and Bernstein-type inequalities for incomplete polynomials on the interval [0,1], but I have not been aware of any such inequalities in the literature. This short paper is a result of an effort to answer the questions raised by A. Eskenazis and