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lecture19

# lecture19 - Stat 302 Introduction to Probability Jiahua...

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Unformatted text preview: Stat 302, Introduction to Probability Jiahua Chen January-April 2011 Jiahua Chen () Lecture 19 January-April 2011 1 / 24 What LLNs do not answer Let X 1 , X 2 , . . . be a sequence of iid random variables with mean μ and variance σ 2 . Denote ¯ X n = n- 1 ( X + 1 + X 2 + ··· + X n ) . According to WLLN, lim n → ∞ P ( | X n- μ | ≥ ǫ ) = 0 for any ǫ > 0. According to SLLN, P ( lim n → ∞ ¯ X n = μ ) = 1. Both Laws tell us that ¯ X n ≈ μ . Yet how close are they? Jiahua Chen () Lecture 19 January-April 2011 2 / 24 More detailed error bound By Chebyshev’s inequality, we have P ( | ¯ X n- μ | ≥ ǫ ) ≤ σ 2 n ǫ 2 . Letting ǫ = n- 1 / 3 , it becomes P ( | ¯ X n- μ | ≥ n- 1 / 3 ) ≤ σ 2 n 1 / 3 . Note the limit of the upper bound is still 0 as n → ∞ . Jiahua Chen () Lecture 19 January-April 2011 3 / 24 More detailed error bound By Chebyshev’s inequality, we have P ( | ¯ X n- μ | ≥ ǫ ) ≤ σ 2 n ǫ 2 . Letting ǫ = n- 1 / 3 , it becomes P ( | ¯ X n- μ | ≥ n- 1 / 3 ) ≤ σ 2 n 1 / 3 . Note the limit of the upper bound is still 0 as n → ∞ . That is, the precision for ¯ X n as an estimator of μ is within a range of n- 1 / 3 . Jiahua Chen () Lecture 19 January-April 2011 3 / 24 More detailed error bound Keep the inequality here again: P ( | ¯ X n- μ | ≥ ǫ ) ≤ σ 2 n ǫ 2 . Letting ǫ = n- 2 / 3 , it becomes P ( | ¯ X n- μ | ≥ n- 2 / 3 ) ≤ n 1 / 3 σ 2 . Note the limit is ∞ as n → ∞ . Jiahua Chen () Lecture 19 January-April 2011 4 / 24 More detailed error bound Keep the inequality here again: P ( | ¯ X n- μ | ≥ ǫ ) ≤ σ 2 n ǫ 2 . Letting ǫ = n- 2 / 3 , it becomes P ( | ¯ X n- μ | ≥ n- 2 / 3 ) ≤ n 1 / 3 σ 2 . Note the limit is ∞ as n → ∞ . The largest possible probability is 1, so the infinity upper bound is not useful. Yet it does tell us that the difference | ¯ X n- μ | will exceed n- 2 / 3 . Jiahua Chen () Lecture 19 January-April 2011 4 / 24 Tossing a die 6000 times. These analyses should reveal that n- 2 / 3 < | ¯ X n- μ | < n- 1 / 3 . When n = 6000, it becomes: 0.003 < | ¯ X n- μ | < 0.055. In general, it is easy to see that n- 1 / 2 gives more tight bounds: P ( √ n | ¯ X n- μ | σ ≥ 3 ) ≤ 1 9 . That is, with 89% probability, for this die tossing example, | ¯ X n- μ | < 0.01443376. Jiahua Chen () Lecture 19 January-April 2011 5 / 24 Tossing a die 6000 times: exact value We may directly compute P ( | ¯ X n- μ | > 0.01443376 ) = 0.0030543 = 0.3% which is much lower than the Chebyshev upper bound: 11%....
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