HW1 - Mathematics Department UCLA T Richthammer spring 09...

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Mathematics Department, UCLA T. Richthammer spring 09, sheet 1 Mar 30, 2009 Homework assignments: Math 170B Probability, Sec. 1 01. Let X be a normal RV with parameters μ, σ 2 , and a, b R . (a) Show that Y = aX + b again is a normal RV (with which parameters?) if a ± = 0. (Hint: consider the cases a > 0 and a < 0 separately.) (b) What happens if a = 0? Answer: (a) For a > 0 we have F Y ( c ) = P ( Y c ) = P ( aX + b c ) = P ( X c - b a ) = F X ( c - b a ), so f Y ( c ) = d dc F X ( c - b a ) = f X ( c - b a ) 1 a = 1 2 πσ e - ( c - b a - μ ) 2 / 2 σ 2 1 a = 1 2 πaσ e - ( c - b - ) 2 / 2 a 2 σ 2 , so Y is a normal RV with parameters μ ± = + b and σ ± 2 = a 2 σ 2 . Similarly for a < 0 we have F Y ( c ) = P ( aX + b c ) = P ( X c - b a ) = 1 - F X ( c - b a ), so f Y ( c ) = - d dc F X ( c - b a ) = - f X ( c - b a ) 1 a = 1 2 π | a | σ e - ( c - b - ) 2 / 2 a 2 σ 2 , so again Y is a normal RV with parameters μ ± = + b and σ ± 2 = a 2 σ 2 . (b) For a = 0 we have Y = b , so Y is a discrete RV with all the mass on the value b . 02. Let X 1 , X 2 be independent exponential RVs with parameter 1. Determine the PDF of (a) Y = max { X 1 , X 2 } (b) Z = min { X 1 , X 2 } . Answer: (a) R ( Y ) = (0 , ) For c > 0 we have F Y ( c ) = P ( Y c ) = P (max { X 1 , X 2 } ≤ c ) = P ( X 1 c, X 2 c ) = P ( X 1 c ) P ( X 2 c ) = (1 - e - c ) 2 . So f Y ( c ) = d dc F Y ( c ) = 2(1 - e - c ) e - c , so f Y ( y ) = 2( e - y - e - 2 y )1 { y> 0 } . (b) R ( Z ) = (0 , ) For c > 0 we have 1 - F Z ( c ) = P ( Z > c ) = P (min { X 1 , X 2 } > c ) = P ( X 1 > c, X 2 > c ) = P ( X 1 > c ) P ( X 2 > c ) = e - 2 c . So f Z ( c ) = d dc F Z ( c ) = 2 e - 2 c , so f Z ( z ) = 2 e - 2 z 1 { z> 0 } . 03. The joint PDF of X and Y is given by f ( x, y ) = ( y - x ) e - y 1 { 0 <x<y } . (a) Check that f ( x, y ) is a density function. (b) Find the PDFs of X and Y . Are X and Y independent? (c) Calculate E ( Y ), P ( 1 3 Y X 2 3 Y ) and P ( X + Y 1). Answer: (a) We have f ( x, y ) 0 and f is normalized: R dy R dxf ( x, y ) = R dy R dx ( y - x ) e - y 1 { 0 <x<y } = R 0 dye - y R y 0 dx ( y - x ) = R 0 dye - y y 2 2 = 1. (b) f Y ( y ) = R dxf ( x, y ) = e - y y 2 2 1 { y> 0 } by the same calculation as in (a). f X ( x ) = R dyf ( x, y ) = R x dy ( y - x ) e - y 1 { x> 0 } = R 0 dy ± y ± e - y ± - x 1 { x> 0 } = e - x 1 { x> 0 } , so X is an exponential RV with parameter 1. f ( x, y ) ± = f X ( x ) f Y ( y ), so X and Y are not independent. (c) E ( Y ) = R dyyf Y ( y ) = R 0 dye - y y 3 2 = 3. P ( 1 3 Y X 2 3 Y ) = R 0 dye - y R 2 y/ 3 y/ 3 dx ( y - x ) = R 0 dye - y 1 6 y 2 = 1 3 .
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P ( X + Y 1) = R dy R dx ( y - x ) e - y 1 { 0 <x<y,x + y 1 } = R 1 0 dye - y R min { y, 1 - y } 0 dx ( y - x ) = R 1 / 2 0 dye - y R y 0 dx ( y - x ) + R 1 1 / 2 dye - y R 1 - y 0 dx ( y - x ) = R 1 / 2 0 dye - y y 2 2 + R 1 1 / 2 dye - y
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HW1 - Mathematics Department UCLA T Richthammer spring 09...

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