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Jensen's inequality generalizes the statement that a secant line of a convex function lies above its graph. Visualizing convexity and Jensen's inequality. In mathematics, Jensen's inequality, named after the Danish mathematician Johan Jensen, relates the value of a convex function of an integral to the integral of the convex function.
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.
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 ...
Download as PDF; Printable version; ... for example. Löwner–Heinz theorem ... satisfies Jensen's Operator Inequality if the following holds ...
Download as PDF; Printable version; In other projects ... In mathematics, Jensen's theorem may refer to: Johan Jensen's inequality for convex functions; Johan Jensen ...
Firstly, while the sample variance (using Bessel's correction) is an unbiased estimator of the population variance, its square root, the sample standard deviation, is a biased estimate of the population standard deviation; because the square root is a concave function, the bias is downward, by Jensen's inequality.
Instead, he was a successful engineer for the Copenhagen Telephone Company between 1881 and 1924, and became head of the technical department in 1890. All his mathematics research was carried out in his spare time. Jensen is mostly renowned for his famous inequality, Jensen's inequality. In 1915, Jensen also proved Jensen's formula in complex ...
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}.