ma316f03hw4

ma316f03hw4 - MA 316 MATHEMATICAL PROBABILITY ASSIGNMENT#4...

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MA 316: MATHEMATICAL PROBABILITY ASSIGNMENT #4 SOLUTIONS: SECTIONS 1.8 AND 1.9 FALL 2003 1.80: If f ( x ) = ( x + 2) / 18 then E [ X ] = R 4 - 2 x ( x + 2) / 18 = 2 and E [( X + 2) 3 ] = Z 4 - 2 ( x + 2) 3 ( x + 2) / 18 = 86 . 4 . For the third part, since expectation is linear we have E [6 X - 2( X + 2) 3 ] = 6 E [ X ] - 2 E [( X + 2) 3 ] = 6 * 2 - 2 * 86 . 4 = - 160 . 8 . 1.81: If f ( x ) = 1 / 5 for x = 1 , 2 , 3 , 4 , 5 then E [ X ] = 4 x =1 x (1 / 5) = 3 and E [ X 2 ] = 4 x =1 x 2 (1 / 5) = 11. Using these results and the fact that expectation is linear we have E [( X + 2) 2 ] = E [( X 2 + 4 X + 4)] = E [ X 2 ] + 4 E [ X ] + E [4] = 11 + 4 * 3 + 4 = 27 . 1.83: For part a) if f (0) = 1 / 4 then f ( - 1)+ f (1) = 3 / 4 and so E [ X 2 ] = 1 x = - 1 x 2 f ( x ) = 1 f ( - 1)+0 f (0)+ 1 f (1) = 3 / 4 . For part b, if f (0) = 1 / 4 and E [ X ] = 1 / 4 then E [ X ] = ( - 1) f ( - 1) + 0 f (0) + 1 f (1) = f (1) - f ( - 1) = 1 / 4 . Since from part a) we have
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This note was uploaded on 09/10/2009 for the course STATS 517 taught by Professor Song during the Fall '07 term at Purdue.

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ma316f03hw4 - MA 316 MATHEMATICAL PROBABILITY ASSIGNMENT#4...

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