Search results
Results from the WOW.Com Content Network
Fefferman's inequality; Fréchet inequalities; Gauss's inequality; Gauss–Markov theorem, the statement that the least-squares estimators in certain linear models are the best linear unbiased estimators; Gaussian correlation inequality; Gaussian isoperimetric inequality; Gibbs's inequality; Hoeffding's inequality; Hoeffding's lemma; Jensen's ...
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\}} ,
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 ...
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.
Proof [2]. Since + =, =. A graph = on the -plane is thus also a graph =. From sketching a visual representation of the integrals of the area between this curve and the axes, and the area in the rectangle bounded by the lines =, =, =, =, and the fact that is always increasing for increasing and vice versa, we can see that upper bounds the area of the rectangle below the curve (with equality ...
For instance, to solve the inequality 4x < 2x + 1 ≤ 3x + 2, it is not possible to isolate x in any one part of the inequality through addition or subtraction. Instead, the inequalities must be solved independently, yielding x < 1 / 2 and x ≥ −1 respectively, which can be combined into the final solution −1 ≤ x < 1 / 2 .
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.
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 .