hw4vqn - Vinh Nguyen Stat 120A(Fall 2007): HW4 Solutions...

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Vinh Nguyen Stat 120A(Fall 2007): HW4 Solutions November 7, 2007 Text: Probabilty and Statistics , by Degroot and Schervish (3 rd ed) Problems: 4.5.6, 4.5.11, 4.8.5, 5.6.3, 5.6.13, 5.7.2, and 5.7.11 4.5.6 Suppose X has the following pdf: f ( x ) = ± 2 x for 0 < x < 1, 0 otherwise. (a) We know from section 4.5 that E [( X - d ) 2 ] is minimized at d = E ( X ) = μ . E ( X ) = Z 1 0 x 2 xdx = ² 2 x 3 3 ³ 1 0 = 2 3 (b) We know from section 4.5 that E [ | X - d | 2 ] is minimized at d = F - 1 X (1 / 2) = median ( X ). F X ( x ) = Z x 0 2 tdt = t 2 | x 0 = x 2 which implies F - 1 X ( q ) = q so that F - 1 X (1 / 2) = r 1 2 . 4.5.11 Suppose X Bin ( n = 7 ,p = 1 / 4) and Y Bin ( n = 5 ,p = 1 / 2). Recall that the M.S.E. for a r.v. X is defined as E [( X - μ ) 2 ] = σ 2 . For a Bin ( n,p ) distribution, we know that μ = p and σ 2 = np (1 - p ). Applying this, we have σ 2 X = 7(1 / 4)(3 / 4) = 21 / 16 and σ 2 Y = 5(1 / 2)(1 / 2) = 5 / 4. Since σ X > σ Y , Y can be predicted with a smaller M.S.E.
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hw4vqn - Vinh Nguyen Stat 120A(Fall 2007): HW4 Solutions...

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