Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Probability and Statistics Grinshpan Chebyshev’s inequality Let X be a random variable taking nonnegative values and having a finite expectation. For simplicity, we will assume that X is discrete (the general case is treated similarly). Let xi ≥ 0 be the values of X, and consider the mean value ∑ E[X] = xi P (X = xi ). i Every term in the above sum is nonnegative. If we omit the terms with xi below a fixed bound t > 0, the net value of the sum would not be increased: ∑ ∑ xi P (X = xi ) ≥ xi P (X = xi ). i: xi ≥t i Observe that ∑ xi P (X = xi ) ≥ i: xi ≥t ∑ tP (X = xi ) = tP (X ≥ t). i: xi ≥t Therefore, putting the chain of inequalities together, E[X] , t > 0. P (X ≥ t) ≤ t Actually, this estimate is only meaningful for t > E[X]. We may also write it in the form E[X] . t Let now Y be a (real) random variable without value restrictions, and let µ be its mean. P (X < t) ≥ 1 − Then X = (Y − µ)2 is nonnegative. So, according to the derived estimate, E[(Y − µ)2 ] , t > 0. t √ Letting s = t, and noting that (Y − µ)2 ≥ t is equivalent to |Y − µ| ≥ s, we obtain P ((Y − µ)2 ≥ t) ≤ P (|Y − µ| ≥ s) ≤ Var(Y ) . s2 This is Chebyshev’s inequality. It is meaningful for any Y with finite mean and variance and s > σ, where σ is the standard deviation of Y. Var(Y ) For the complementary event |Y − µ| < s, we have P (|Y − µ| < s) ≥ 1 − . s2 A restatement of Chebyshev’s inequality is often in use. If s = kσ, the inequality becomes 1 P (|Y − µ| ≥ kσ) ≤ 2 . k For instance, the probability that Y falls two standard deviations away from its mean is under 1/4.