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Bernoulli's inequality can be proved for case 2, in which is a non-negative integer and , using mathematical induction in the following form: we prove the inequality for r ∈ { 0 , 1 } {\displaystyle r\in \{0,1\}} ,
In mathematics, the Lebedev–Milin inequality is any of several inequalities for the coefficients of the exponential of a power series, found by Lebedev and Milin and Isaak Moiseevich Milin . It was used in the proof of the Bieberbach conjecture , as it shows that the Milin conjecture implies the Robertson conjecture .
The bounds these inequalities give on a finite sample are less tight than those the Chebyshev inequality gives for a distribution. To illustrate this let the sample size N = 100 and let k = 3. Chebyshev's inequality states that at most approximately 11.11% of the distribution will lie at least three standard deviations away from the mean.
Bennett's inequality, an upper bound on the probability that the sum of independent random variables deviates from its expected value by more than any specified amount Bhatia–Davis inequality , an upper bound on the variance of any bounded probability distribution
A more frequently encountered inequality measure is the Gini coefficient which is based on the summation, over all income-ordered population-percentiles, of the cumulative income up to each percentile. That sum is divided by the maximum value that it could have (its value with complete equality), to express it as a percentage of its maximum ...
The inequality was first proven by Grönwall in 1919 (the integral form below with α and β being constants). [1] Richard Bellman proved a slightly more general integral form in 1943. [2] A nonlinear generalization of the Grönwall–Bellman inequality is known as Bihari–LaSalle inequality. Other variants and generalizations can be found in ...
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Since all the inequalities are in the same form (all less-than or all greater-than), we can examine the coefficient signs for each variable. Eliminating x would yield 2*2 = 4 inequalities on the remaining variables, and so would eliminating y. Eliminating z would yield only 3*1 = 3 inequalities so we use that instead.