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**Unformatted text preview: **P (-10 1 2 √ n ≤ S n-n 2 1 2 √ n ≤ 10 1 2 √ n ) = 2Φ( 10 1 2 √ n )-1 → (b) as n → ∞ P ( n 2-n 10 ≤ S n ≤ n 2 + n 10 ) = P (-1 10 ≤ S n n-1 2 ≤ 1 10 ) → 1 by weak law of large number. (c) as n → ∞ P ( n 2-√ n 2 ≤ S n ≤ n 2 + √ n 2 ) = P (-1 ≤ S n-n 2 1 2 √ n ≤ 1) → 2Φ(1)-1 = 0 . 6827 Prob. 4. Solution: (20pts) (a) X i are independent (b) X i are iid with mean 0 and variance 1 12 , let Y n = ∑ n i =1 X i , then EY n = 0 , V ar ( Y n ) = n 12 p = P ( | Y n | ≥ c ) = P ( | Y n √ n 12 | ≥ c √ n 12 ) ≈ 2(1-Φ( c √ n 12 )) Here c = 10 , n = 100,then p = 2(1-Φ( 10 √ 100 12 )) = 0 . 054% (c) No, according to CLT. Prob. 5. Solution. (20pts) var ( X ) = var ( ˆ X + Δ) = var ( ˆ X ) + var (Δ) + 2 cov ( ˆ X, Δ) = var ( ˆ X ) + var (Δ) 2...

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- Spring '18
- MAYANK