hw505-07-2-solution

# hw505-07-2-solution - IE 505 SKETCH OF SOLUTIONS TO...

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IE 505 SKETCH OF SOLUTIONS TO HOMEWORK # 2 1. The possible values are { 1 , 2 , . . . , 6 } with probability 1 / 6: P ( s ) = ( s + s 2 + s 3 + . . . + s 6 ) / 6, and P 0 ( s ) = (1 + 2 s + 3 s 2 + . . . + 6 s 5 ) / 6. Then, E [ X ] = P 0 (1) = 7 / 2. 2. (a) Pr { all items marked } = X k =0 Pr { X = k, all k items are marked } = X k =0 p k Pr { all k items are marked } = X k =0 p k s k = P ( s ) . For part (b), Pr { X T } = X k =0 Pr { T k } Pr { X = k } = X k =0 (1 - s k +1 ) p k = 1 - sP ( s ) . 3. Part (i): Pr { S n +1 = k } = Pr { S n +1 = k, X n +1 = 1 } + Pr { S n +1 = k, X n +1 = 0 } = Pr { S n = k - 1 , X n +1 = 1 } + Pr { S n = k, X n +1 = 0 } = p Pr { S n = k - 1 } + q Pr { S n = k } For part (ii), let P n ( s ) = E [ s S n ] and P n +1 ( s ) = n +1 X k =0 Pr { S n +1 = k } s k = q n +1 + n +1 X k =1 ( p Pr { S n = k - 1 } + q Pr { S n = k } ) s k = q n +1 + psP n ( s ) + q ( P n ( s ) - q n ) = P n ( s )( ps + q ) = P 1 ( ps + q ) n . Since P 1 ( s ) = ( ps + q ) (Bernoulli trial), P n ( s ) = ( ps + q ) n . We can invert by Binomial expansion: Pr { S n = k } = ( n k ) p k (1 - p ) n - k .

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4. Since P (1) = 1, a + b + c = 1. Also m = b + 2 a = a + 1 - c and m > 1 if and only if c < a . P ( s ) = s yields: (1 - b ± q (1 - b ) 2 - 4 ac ) / 2 a = (1 - b ± q ( c + a ) 2 - 4 ac ) / 2 a = (1 - b ± p c 2 - 2 ac + a 2 ) / 2 a = (1 - b ± | c - a | ) / 2 a = (1 - b ± ( a - c )) / 2 a = { 1 , c/a } . So, the smallest root is π = c/a . 5. Let Z 1 be the number of members in the first population. Let S i , i = 1 , 2 , . . . , Z 1 be the total population generated by the i th member of the first generation. Then S = 1 + Z 1 X i =1 S i
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