solution09

# solution09 - Solutions to Homework 09 1 a Since X is...

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Solutions to Homework 09 1. a. Since X is uniform in the range [ - b, b ), we have F X ( x ) = 0 x < - b x + b 2 b - b x < b 1 x 1 (1) and m = E [ X ] = 0 and σ 2 = V AR [ X ] = b 2 / 3. Now, P [ | X - m | > c ] = P [ c - m > X > m + c ]. Thus we get P [ | X - 0 | > c ] = ( F X ( c ) - F X ( - c ) 0 c b 0 x > b = ( 1 - c b 0 c b 0 x > b (2) The Chebychev bound gives P [ | X - m | > c ] σ 2 c 2 = b 2 3 c 2 (3) b. For the Laplacian random variable, m = E [ X ] = 0 and σ 2 = V AR [ X ] = 2 2 and its pdf is given by f X ( x ) = α 2 e - α | x | - ∞ < x < (4) Then, the exact probability is given by P [ | X - m | > c ] = Z c + m c - m α 2 e - α | x | dx = e - αc (5) The Chebychev bound gives P [ | X - m | > c ] σ 2 c 2 = 2 α 2 c 2 (6) c. For our Gaussian random variable, m = E [ X ] = 0 and σ 2 = V AR [ X ] and its pdf is given by f X ( x ) = 1 2 πσ 2 e - x 2 2 σ 2 - ∞ < x < (7) Then, the exact probability is given by P [ | X - 0 | > c ] = F X ( c ) - F X ( - c ) = Φ c σ - Φ c σ = 2 Q c σ (8) 1

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The Chebychev bound gives P [ | X - m | > c ] σ 2 c 2 (9) 2. a. k observed Expected ( O - E ) 2 /E 0 0 10.5 10.5 1 0 10.5 10.5 2 24 10.5 17.36 3 2 10.5 6.88 4 25 10.5 20.02 5 3 10.5 5.36 6 32 10.5 44.02 7 15 10.5 1.93 8 2 10.5 6.88 9 2 10.5 6.88 105 D 2 = 130 . 33 Degrees of freedom = 10 - 1 = 9, chi-square value at 1% signifance level = 21.7. D 2 > 21 . 7 and so we reject the hupothesis that the numbers are drawn from a Uniform distribution in { 0 , 1 , · · · , 9 } . b. k observed Expected ( O - E ) 2 /E 2 24 13.125 9.01 3 2 13.125 9.43 4 25 13.125 10.74 5 3 13.125 7.81 6 32 13.125 77.41 7 15 13.125 0.27 8 2 13.125 9.43 9 2 13.125
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