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Jensen's inequality can be proved in several ways, and three different proofs corresponding to the different statements above will be offered. Before embarking on these mathematical derivations, however, it is worth analyzing an intuitive graphical argument based on the probabilistic case where X is a real number (see figure).
Since the square root is a strictly concave function, it follows from Jensen's inequality that the square root of the sample variance is an underestimate. The use of n − 1 instead of n in the formula for the sample variance is known as Bessel's correction , which corrects the bias in the estimation of the population variance, and some, but ...
There are three inequalities between means to prove. There are various methods to prove the inequalities, including mathematical induction, the Cauchy–Schwarz inequality, Lagrange multipliers, and Jensen's inequality. For several proofs that GM ≤ AM, see Inequality of arithmetic and geometric means.
Hölder's inequality; Jackson's inequality; Jensen's inequality; Khabibullin's conjecture on integral inequalities; Kantorovich inequality; Karamata's inequality; Korn's inequality; Ladyzhenskaya's inequality; Landau–Kolmogorov inequality; Lebedev–Milin inequality; Lieb–Thirring inequality; Littlewood's 4/3 inequality; Markov brothers ...
The exponential function is convex, so by Jensen's inequality (). It follows that the bound on the right tail is greater or equal to one when a ≤ E ( X ) {\displaystyle a\leq \operatorname {E} (X)} , and therefore trivial; similarly, the left bound is trivial for a ≥ E ( X ) {\displaystyle a\geq \operatorname {E} (X)} .
In mathematics, Jensen's theorem may refer to: Johan Jensen's inequality for convex functions; Johan Jensen's formula in complex analysis;
Toggle Jensen's operator and trace inequalities subsection. 12.1 Jensen's trace inequality. 12.2 Jensen's operator inequality. 13 Araki–Lieb–Thirring inequality.
The finite form of Jensen's inequality is a special case of this result. Consider the real numbers x 1, …, x n ∈ I and let := + + + denote their arithmetic mean.Then (x 1, …, x n) majorizes the n-tuple (a, a, …, a), since the arithmetic mean of the i largest numbers of (x 1, …, x n) is at least as large as the arithmetic mean a of all the n numbers, for every i ∈ {1, …, n − 1}.