Dr. Hackney STA Solutions pg 42

# Dr. Hackney STA Solutions pg 42 - Chebychev exact...

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Second Edition 3-15 3.45 a. M X ( t ) = Z -∞ e tx f X ( x ) dx Z a e tx f X ( x ) dx e ta Z a f X ( x ) dx = e ta P ( X a ) , where we use the fact that e tx is increasing in x for t > 0. b. M X ( t ) = Z -∞ e tx f X ( x ) dx Z a -∞ e tx f X ( x ) dx e ta Z a -∞ f X ( x ) dx = e ta P ( X a ) , where we use the fact that e tx is decreasing in x for t < 0. c. h ( t, x ) must be nonnegative. 3.46 For X uniform(0 , 1), μ = 1 2 and σ 2 = 1 12 , thus P ( | X - μ | > kσ ) = 1 - P ± 1 2 - k 12 X 1 2 + k 12 ² = ³ 1 - 2 k 12 k < 3, 0 k 3, For X exponential( λ ), μ = λ and σ 2 = λ 2 , thus P ( | X - μ | > kσ ) = 1 - P ( λ - X λ + ) = ³ 1 + e - ( k +1) - e k - 1 k 1 e - ( k +1) k > 1. From Example 3.6.2, Chebychev’s Inequality gives the bound P ( | X - μ | > kσ ) 1 /k 2 . Comparison of probabilities k u(0 , 1) exp( λ
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Unformatted text preview: ) Chebychev exact exact .1 .942 .926 100 .5 .711 .617 4 1 .423 .135 1 1.5 .134 .0821 .44 √ 3 0.0651 .33 2 0.0498 .25 4 0.00674 .0625 10 0.0000167 .01 So we see that Chebychev’s Inequality is quite conservative. 3.47 P ( | Z | > t ) = 2 P ( Z > t ) = 2 1 √ 2 π Z ∞ t e-x 2 / 2 dx = r 2 π Z ∞ t 1+ x 2 1+ x 2 e-x 2 / 2 dx = r 2 π ´Z ∞ t 1 1+ x 2 e-x 2 / 2 dx + Z ∞ t x 2 1+ x 2 e-x 2 / 2 dx µ ....
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