The mean square error given x u 2 is the variance of

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Unformatted text preview: e errors is greater than 5. 4.9. LAW OF LARGE NUMBERS AND CENTRAL LIMIT THEOREM Solution: The mean of each roundoff error is zero and the variance is E [S ] = 0 and V ar(S ) = 100 = 8.333. Thus, P [|S | ≥ 5] = P [| √8S333 | ≥ 12 . 2Q(1.73) = 2(1 − Φ(1.732)) = 0.083. 161 0.5 1 2 −0.5 u du = 12 . Thus, 5 √ ≈ 2Q( √85333 ) = 8.333 . Example 4.9.7 Suppose each day of the year, the value of a particular stock: increases by one percent with probability 0.5, remains the same with probability 0.4, and decreases by one percent with probability 0.1. Changes on different days are independent. Consider the value after one year (365 days), beginning with one unit of stock. Find (a) the probability the stock at least triples in value, (b) the probability the stock at least quadruples in value, (c) the median value after one year. Solution: The value, Y , after one year, is given by Y = D1 D2 · · · D365 , where Dk for each k is the growth factor for day k , with pmf pD (1.01) = 0.5, p...
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This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Tech.

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