HW5Solutions (dragged) 1

# HW5Solutions (dragged) 1 - 1 A Assuming the cost charged to...

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where Y is the random variable representing the cost per ounce of the commodity. We want to maximize E ( V ) where E ( V ) = 1000 - 2 x + x (1( . 5) + 4( . 5)) = 1000 + x 2 Since this is an increasing linear function of x , to maximize our expected amount of money, we should buy as much as possible. Thus let x = 500, i.e. buy all that one can. Problem 4.23 b. Now we want to maximize the expected amount of the commodity. By purchasing x at the beginning of the week, one is left with 1000 - 2 x cash to buy more at the end of the week. The amount of the commodity we have at the end of the week is given by A = x + 1000 - 2 x Y where Y is the random variable denoting the cost per ounce of the commodity at the end of the week. The expected value of A is E ( A ) = x + 1000 - 2 x 1 ( . 5) + 1000 - 2 x 4 ( . 5) = 625 - x 4 Which is a linear and decreasing function of x . Thus, the optimal x is x = 0, i.e. buy none of the commodity now and buy it all at the end of the week. Problem 4.27
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Unformatted text preview: . 1 A . Assuming the cost charged to each customer is C , the expected proFt of the company is â‡§ = C + p (-A ) + (1-p )(0) = C-pA Â±or this to be 0 . 1 A we should have C-pA = 0 . 1 A or C = âœ“ p + 1 10 â—† A Problem 4.29 If you Frstly check 1 then 2: E[cost]= C 1 + (1-p ) C 2 + pR 1 + (1-p ) R 2 If you Frstly check 2 then 1: E[cost]= C 2 + pC 1 + (1-p ) R 2 + pR 1 Thus, we should check 1 Frst if C 1 + (1-p ) C 2 C 2 + pC 1 Problem 4.31 DeFne X as the score. If he predicts p , then X = 1-(1-p ) 2 with probability p â‡¤ , and X = 1-p 2 with probability 1-p â‡¤ E [ X ] = p â‡¤ [1-(1-p ) 2 ] + (1-p â‡¤ )(1-p 2 ) = 2 pp â‡¤ + (1-p â‡¤ )-p 2 This is concave in p , so to maximize we can simply Fnd the Frst order condition, d dp = 2 p â‡¤-2 p = 0, which gives us p = p â‡¤ 2...
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