elemprob-page26 - Proposition 13.2. If Y 0, then for any A,...

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Proposition 13.2. If Y 0 , then for any A , P ( Y > A ) E Y A . Proof. Let B = { Y > A } . Recall 1 B is the random variable that is 1 if ω B and 0 otherwise. Note 1 B Y/A . This is obvious if ω / B , while if ω B , then Y ( ω ) /A > 1 = 1 B ( ω ). We then have P ( Y > A ) = P ( B ) = E 1 B E ( Y/A ) = E Y A . We now prove the WLLN. Theorem 13.3. Suppose the X i are i.i.d. and E | X 1 | and Var X 1 are finite. Then for every a > 0 , P ±² ² ² S n n - E X 1 ² ² ² > a ³ 0 as n → ∞ . Proof. Recall E S n = n E X 1 and by the independence, Var S n = n Var X 1 , so Var ( S n /n ) = Var X 1 /n . We have P ±² ² ² S n n - E X 1 ² ² ² > a ³ = P ±² ² ² S n n - E ± S n n ³² ² ² > a ³ = P ±² ² ² S n n - E ± S n n ³² ² ² 2 > a 2 ³ E | S n n - E ( S n n ) | 2 a 2 = Var ( S n n ) a 2 = Var X 1 n a 2 0 . The inequality step follows from Proposition 13.2 with A = a 2 and Y = | S n n - E ( S n n ) | 2 . We now turn to the central limit theorem (CLT). Theorem 13.4. Suppose the X i are i.i.d. Suppose E X 2 i < . Let μ = E X i and σ 2 = Var X i . Then P ± a S n - σ n b ³
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