STAT 400 Lecture AL1 Examples for 5.8 Spring 2015 Dalpiaz Markov’s Inequality: Let u ( X ) be a non-negative function of the random variable X. If E [ u ( X ) ] exists, then, for every positive constant c, P ( u ( X ) ≥ c ) ≤ ( )[ ]cu XE. Chebyshev’s Inequality: Let X be any random variable with mean µ and variance σ 2. For any ε > 0, P ( | X – µ | ≥ ε ) ≤ 22 εσ or, equivalently, P ( | X – µ | < ε ) ≥ 22 εσ 1 − Setting ε = k σ, k > 1, we obtain P ( | X – µ | ≥ k σ ) ≤ 2 1k or, equivalently, P ( | X – µ | < k σ ) ≥ 2 1 1k− That is, for any k > 1, the probability that the value of any random variable will be within k standard deviations of its mean is at least 2 1 1k−. Example 1: Suppose µ = E ( X ) = 17, σ = SD ( X ) = 5. Consider interval ( 9, 25 ) = ( 17 – 8, 17 + 8 ). ⇒ k = 58 = 1.6. ⇒ P ( 9 < X < 25 ) = P ( | X – µ | < 1.6 σ ) ≥ 2 6.11 1 − = 6439 = 0.609375.Example 2: Suppose µ = E ( X ) = 17, σ = SD ( X ) = 5. Suppose also that the distribution of X is symmetric about the mean. Consider interval ( 10, 30 ) = ( 17 – 7, 17 + 13 ) = ( µ – 1.4 σ, µ + 2.6 σ ). P ( 10 < X < 24 ) = P ( | X – µ | < 1.4 σ ) ≥ 2 4.11 1− ≈ 0.490. P ( 4 < X < 30 ) = P ( | X – µ | < 2.6 σ ) ≥ 2 6.21 1 − ≈ 0.852. Since the distribution of X is symmetric about the mean, P ( 10 < X < 17 ) ≥ 2490.0 = 0.245, P ( 17 < X < 30 ) ≥ 2852.0 = 0.426. ⇒ P ( 10 < X < 30 ) ≥ 0.245 + 0.426 = 0.671. Example 3: Consider a discrete random variable X with p.m.f. P ( X = – 1 ) = ½, P ( X = 1 ) = ½. Then µ = E ( X ) = 0, σ 2 = Var ( X ) = E ( X 2 ) = 1. ⇒ P ( | X – µ | ≥ σ ) = P ( | X | ≥ 1 ) = 1. ( k = 1 ) P ( | X – µ | < σ ) = P ( | X | < 1 ) = 0. Example 4: ( Chebyshev’s Inequality cannot be improved ) Let a > 0, 0 < p < ½. Consider a discrete random variable X with p.m.f. P ( X = – a ) = p, P ( X = 0 ) = 1 – 2 p, P ( X = a ) = p. Then µ = E ( X ) = 0, σ 2 = Var ( X ) = E ( X 2 ) = 2 p a 2. Let k = p 21 > 1. Then k σ = a. ⇒ P ( | X – µ | ≥ k σ ) = P ( | X | ≥ a ) = 2 p = 2 1k. P ( | X – µ | < k σ ) = P ( | X | < a ) = 1 – 2 p = 2 1
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