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Unformatted text preview: P (40 < X < 60). Solution: 1. By Markov’s inequality, P ( X > 75) ≤ 50 75 = 2 3 2. Write P (40 < X < 60) = 1P (  X50  ≥ 10). By Chebyshev’s inequality, P (  X50  ≥ 10) ≤ 25 / 100 = 0 . 25. Hence, P (40 < X < 60) = 1P (  X50  ≥ 10) ≥ 1. 25 = 0 . 75 . Michael Akritas The Law of Large Numbers and The Central Limit Theorem Outline Markov’s Inequality; Chebyshev’s Inequality The WLLN and the CLT Theorem (The Weak Law of Large Numbers) Let X 1 , X 2 , . . . be iid with ﬁnite mean μ , and set X n = 1 n ∑ n i =1 X i . Then, for any ± > , P (  X nμ  > ± ) → , as n → ∞ Theorem (The Central Limit Theorem) Let X 1 , X 2 , . . . be iid with mean μ and ﬁnite variave σ 2 , and set X n = 1 n ∑ n i =1 X i . Then X n ˙ ∼ N ± μ, σ 2 n ² X 1 + ··· + X n ˙ ∼ N ( n μ, n σ 2 ) Michael Akritas The Law of Large Numbers and The Central Limit Theorem...
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 '08
 G.JOGESHBABU
 Central Limit Theorem, Law Of Large Numbers, Probability, Variance, Probability theory, Michael Akritas

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