examsoln2 - EECS 501 SOLUTIONS TO EXAM #2 Fall 2001 2 1a. E...

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EECS 501 SOLUTIONS TO EXAM #2 Fall 2001 1a. E [ k ] = n 2 . σ 2 k = n 1 2 (1 - 1 2 ) = n 4 . σ k = n/ 2 . (b): 1600( 1 2 ) = 800. 1b. Pr [790 k 810] = Φ[ 810 - 800 1600 / 2 ] - Φ[ 790 - 800 1600 / 2 ] = 2Φ[0 . 5] - 1 = 0 . 383. 1c. 0 . 9 = Pr [ k b ] = 1 - Φ[ b - 800 20 ] 800 - b 20 = Φ - 1 [0 . 9] = 1 . 28 b = 774. 1d. 0 . 95 = Pr [ k 1000] = 1 - Φ[ 1000 - n 2 n/ 2 ] b = 800 - 20(1 . 28) = 774. n 2 - 1000 n/ 2 = Φ - 1 [0 . 95] = 1 . 645 n - 1 . 645 n - 2000 = 0. 1e. Solving ( n ) 2 - 1 . 645 n - 2000 = 0 n = 2075 is the positive root. 1f. e = k n - 1 2 σ 2 e = 1 n 2 σ 2 k = 1 4 n . LIM n →∞ 1 4 n = 0. Note 1 2 has no effect. 2a. [1 , 1 , 1] 0 associated with 0 eigenvalue x 1 + x 2 + x 3 = 0 x 3 = - 5. 2b. f x 1 ,x 3 ( X 1 ,X 3 ) ∼ N ±• 0 0 , 2 - 1 - 1 2 ‚¶ . 2c. σ 2 y = [1 , 2 , 3] K [1 , 2 , 3] 0 = [1 , 2 , 3][ - 3 , 0 , 3] 0 = 6. 2d. λ 2 x 1 x 2 σ 2 x 1 σ 2 x 2 1 = ( - 1) 2 (2)(2) = 4 checks. 2e. z = e 2 · x and w = e 3 · x z,w independent z ∼ N (0 , 6), where σ 2 z = σ 2 x 1 + σ 2 x 2 - 2 λ x 1 ,x 2 = 2 + 2 - 2( - 1) = 6 (or do as in (c)). 2f. ( ˆ x 1 ) LSE ( X 3 ) = λ x 1 x 3 σ 2 x 3 X 3 = - 1 2 X 3 since all 0-mean. 3a. p r | a ( R | A ) = N R · A R (1 - A ) N - R ˆ a MLE ( R ) = R N from lecture. 3b. 0 = ∂A [log p r | a ( R | A ) + log f a ( A )] = R A - N - R 1 - A - 10 10 A 2 - (10 + N ) A + R = 0. Solve for A . 3c. Solving 10 A 2 - (92 + 10) A + 20 = 0 A = 10 , 1 5 ˆ a MAP (20) = 1 / 5. 3d. ˆ a LSE ( R ) = R Af r | a ( R | A ) f a ( A ) dA R f r | a ( R | A ) f a ( A ) dA
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This note was uploaded on 07/22/2011 for the course EECS 370 taught by Professor Lehman during the Spring '10 term at University of Florida.

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