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11ProbHw4sol - IEOR 3658 Probability Prof Mariana...

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IEOR 3658 Assignment #4 Solutions Probability October 7, 2011 Prof. Mariana Olvera-Cravioto Page 1 of 2 Assignment #4 Solutions 1. Let S =$1000, r =0.1, p =0.05. Then the PMF of the total premium paid up to and including the first claim year is: P X ( x ) = (1 - p ) k p if x = S × 1 - (1 - r ) k r k = 0 , 1 , . . . 2. P X ( x ) = 0 . 18 x = 0 0 . 27 x = 1 0 . 34 x = 2 0 . 14 x = 3 0 . 07 x = 4 3. (a) K = 1 2(1 2 +2 2 +3 2 ) = 1 28 (b) p Y ( y ) = y 2 14 , if x = 0 , 1 , 2 , 3 , 0 , otherwise , 4. P X ( x ) = Binomial(5 , 1 2 , x - 2) = 5 x - 2 ( 1 2 ) 5 , if x = 2 , 3 , ..., 7 5. The number of candy bars that you need to eat to find a ticket follows Geometric Distribution with parameter p . So the mean and variance are 1 p and 1 - p p 2 . 6. Let σ Y = var( Y ), then (a) For Y = 4 X : E [ Y ] = E [4 X ] = 4 E [ X ] = 20 , var( Y ) = var(4 X ) = 16var( X ) = 144 , σ Y = 12 . And for Y = - 2 X : E [ Y ] = E [ - 2 X ] = - 2 E [ X ] = - 10 , var( Y ) = var( - 2 X ) = 4var( X ) = 36 , σ Y = 6 . (b) E [ Y ] = E [ X - E [ X ]] = E [ X ] - E [ X ] = 0, var( Y ) = var( X - E [ X ]) = var( X ) = 9, σ Y = var( Y ) = 3. (c) The mean is given by: E [ Y ] = E X - E [ X ] var( X ) = 1 var( X ) E [ X - E [ X ]] = 0 , the variance by: var( Y ) = var X - E [ X ] var( X ) = var( X - E [ X ]) var( X ) = var( X ) var( X ) = 1 and σ Y = 1.
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2 IEOR 3658, Assignment #4 Solutions 7. (a) On a street cleaning day, you will get a ticket if the parking inspector checks your street
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