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midterm2_solution

# midterm2_solution - W and Z are uncorrelated which since...

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IE 300 / GE 331 Spring 2009 Midterm Exam Solution Problem 1. (a) T. X and Y are independent. (b) T. W and Z are uncorrelated. (c) F. cov(2 W, Z - W ) = - 2var( W ) (d) F. X and Y might not have the same variance. (e) T. W and Z are uncorrelated, then the cov( W, Z ) = E ( WZ ) - E ( W ) E ( Z ) = 0. (f) F. W and Z might not be independent. (g) T. X and Y are independent. Problem 2. P ( X + Y > 1) = Z 1 v =0 Z 1 u =1 - v u + v d u d v = Z 1 v =0 u 2 2 + uv 1 u =1 - v d v = Z 1 v =0 1 - (1 - v ) 2 2 + v 2 d v = v 2 + (1 - v ) 3 6 + v 3 3 1 0 = 2 3 . Problem 3. (a) Since distribution of X is Poisson, with mean 3 and variance 3, then according to Chebyshev’s inequality, P ( | X - 3 | ≥ 9) 3 9 2 = 1 27 . 1

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IE 300 / GE 331: Midterm Exam Solution 2 (b) Let S n be total number of watches repaired in 100 days, according to Central Limit Theorem, P ( S n - 680 - ) Φ( 680 - ) = Φ( 680 - 100 · 7 100 · 2 ) = Φ( - 1) = 0 . 1587 . Problem 4. (a) cov( W, Z ) = 2cov( X, X ) + (3 + 2 a )cov( X, Y ) + 3 a cov( Y, Y ) = 2var( Y ) + (3 + 2 a ) σ X σ Y ρ X,Y + 3 a var( Y ) = 2 · 4 + (3 + 2 a ) · 2 · 3 · 0 . 5 + 3 a · 9 = 17 + 33 a. (b) For all a , because linear transformations preserve the property of being jointly Gaussian. (c) For a = - 17 33 , because for that value of a , cov( W, Z
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Unformatted text preview: W and Z are uncorrelated) which, since they are jointly Gaussian, makes them independent. Problem 5. (a) Assuming hypothesis H 1 is twice more likely than hypothesis H , then P ( H ) = 1 3 , P ( H 1 ) = 2 3 . τ = P ( H ) P ( H 1 ) = 1 2 . Therefore, L ( x ) = P ( x | H 1 ) P ( x | H ) H 1 > < H 1 2 . Plugging in the value, we have L ( x ) = | x | 4 1 4 H 1 > < H 1 2 = ⇒ | X | H 1 > < H 1 2 . (b) P (error) = P (error | H was true) P ( H ) + P (error | H 1 was true) P ( H 1 ) IE 300 / GE 331: Midterm Exam Solution 3 P (error | H was true) = Z | x | > 1 2 1 4 d x = 2 Z 2 1 2 1 4 d x = 2 · 1 4 · (2-1 2 ) = 3 4 . P (error | H 1 was true) = Z | x | < 1 2 | x | 4 d x = 2 Z 1 2 x 4 d x = 2 · x 2 8 ± ± ± ± ± 1 2 = 1 4 · ( 1 4-0) = 1 16 . = ⇒ P (error) = 3 4 · 1 3 + 1 16 · 2 3 = 1 4 + 1 24 = 7 24 ....
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