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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.10 : Let D be the random outcome of rolling a dice once. A new dice has values of D* = D - 3.5. There is a total of n rolls of a dice. (a) Find the variance for D* by using the formula 6 V = [ D* (D = 1) ] [ D* (D = 1) ] + [ D* (D = 2) ] [ D* (D = 2) ] + [ D* (D = 3) ] [ D* (D = 3) ] + [ D* (D = 4) ] [ D* (D = 4) ] + [ D* (D = 5) ] [ D* (D = 5) ] + [ D* (D = 6) ] [ D* (D = 6) ]. (b) Calculate the standard deviation of D* as a square root of V. (c) Another new dice has values of D** = kD*. (i) Find the value of k so that D** has a standard deviation of 1. (ii) Find the values of D** for each outcome of D = 1, 2, 3, 4, 5 and 6, when the standard deviation is 1. (iii) Given that the average score of a dice is 3.5, find the equivalent, new and improved model of a dice, Sn in term of n and D**. (iv) Find the expected value of D** as the average of D**.
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Answer Posted / kangchuentat
ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.10 : (a) 6 V = (1 - 3.5) (1 - 3.5) + (2 - 3.5) (2 - 3.5) + (3 - 3.5) (3 - 3.5) + (4 - 3.5) (4 - 3.5) + (5 - 3.5) (5 - 3.5) + (6 - 3.5) (6 - 3.5) = 17.5. V = 17.5 / 6 = 2.92. (b) Standard deviation = square root of V = (2.92) ^ (0.5) = 1.71. (c) (i) k = 1 / (standard deviation of D*) = 1 / 1.71 = 0.585. (ii) D** (D = 1) = -2.5 / 1.71 = -1.462, D** (D = 2) = -1.5 / 1.71 = -0.877, D** (D = 3) = -0.5 / 1.71 = -0.292, D** (D = 4) = 0.5 / 1.71 = 0.292, D** (D = 5) = 1.5 / 1.71 = 0.877, D** (D = 6) = 2.5 / 1.71 = 1.462. (iii) Sn = 3.5 n + 1.71 nD**. (iv) Expected value of D** = [ D** (D = 1) + D** (D = 2) + D** (D = 3) + D** (D = 4) + D** (D = 5) + D** (D = 6) ] / 6 = (- 1.462 - 0.877 - 0.292 + 0.292 + 0.877 + 1.462) / 6 = 0. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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