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### SaraswatVishal.FM5011.HW06

Course: MATH 5011, Fall 2008
School: Minnesota
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5011 Mathematical FM Background for Finance Fall 2008 For # 1,2,3 Dene a PCRV (piecewise continuous random variable) 2, if 4, if Y () = 3, if 15, if (10/16)-1-1a Find the center of mass of the distribution of Y . Y : [0, 1] R by 0.00 0.25 0.50 &lt; 0.70 &lt; 0.25 0.50 0.70 &lt; 1.00 Solution: The center of mass of the distribution of Y is E[Y ] = 3.60 . (10/16)-1-1b Compute (Y...

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5011 Mathematical FM Background for Finance Fall 2008 For # 1,2,3 Dene a PCRV (piecewise continuous random variable) 2, if 4, if Y () = 3, if 15, if (10/16)-1-1a Find the center of mass of the distribution of Y . Y : [0, 1] R by 0.00 0.25 0.50 < 0.70 < 0.25 0.50 0.70 < 1.00 Solution: The center of mass of the distribution of Y is E[Y ] = 3.60 . (10/16)-1-1b Compute (Y ())(3, 5), where is the Lebesgue measure on [0, 1]. Solution: (Y ())(3, 5) = (Y 1 (3, 5)) = (Y 1 ({2, 4})) = ([0.00, 0.25)[0.25, 0.50]) = 0.25+0.25 = 0.50 . (10/16)-1-2a Compute P r[2 Y 2 9]. Solution: P r[2 Y 2 9] = P r[(3 Y 2) OR ( 2 Y 3)] = P r[(Y = 3) OR (Y = 2)] = 0.45 . (10/16)-1-2b Let be the distribution of Y 2 . Compute ([2, 9]). Solution: ([2, 9]) = ({4}) + ({9}) = 0.25 + 0.20 = 0.45 . (10/16)-1-3a Compute E[Y ] Solution: E[Y ] = xImY x[(Y 1 (x))] = x{2,4,3,15} x[(Y 1 (x))] = 2(0.25 0.00) + 4(0.50 0.25) 3(0.70 0.50) 15(1.00 0.70) = 3.60 . (10/16)-1-3b Compute E[Y 2 ] Solution: E[Y 2 ] = xImY 2 x[((Y 2 )1(x))] = xImY x2 [(Y 1 (x))] = x{2,4,3,15} x2 [(Y 1 (x))] = 4(0.25 0.00) + 16(0.50 0.25) + 9(0.70 0.50) + 225(1.00 0.70) = 74.30 . (10/16)-1-3c Let be the distribution of Y 2 . Compute the center of mass of . Solution: The center of mass of is E[Y 2 ] = 74.30. Vishal Saraswat HW 6 Solutions Page 1 of 2 FM 5011 Mathematical Background for Finance Fall 2008 (10/16)-1-4 Find the Lebesgue measure of R \ Q. Solution: Let denote the Lebesgue measure on R. The set of rational numbers Q is countable, so we can write Q = {a1 , a2 , a3 , ...}. Let > 0 be an arbitrary real number. For each n = 1, 2, 3, . . . , dene In to be the open interval In = an n+1 , an + n+1 . 2 2 Then ) (In = 2n and for each n = 1, 2, 3, . . . , an In . Thus, Q In so that n=1 (Q) < n=1 In n=1 (In ) = n=1 2n = . Since we can take arbitrarily small, (Q) = 0. Thus, (R \ Q) = (R) = . (10/16)-1-5 Find the Lebesgue measure (in R) of [0, 1/3] \ (1/9, 2/9). Solution: Let denote the Lebesgue measure. Then, ([0, 1/3] \ (1/9, 2/9)) = ([0, 1/9] [2/9, 1/3]) = ([0, 1/9]) + ([2/9, 1/3]) = 1/9 + 1/9 = 2/9 . (10/16)-1-6 Find the Lebesgue measure (in R2 ) of ([1, 3] [5, 10]) ([2, 4] [6, 11]). Solution: Let denote the Lebesgue measure. Then (([1, 3] [5, 10]) ([2, 4] [6, 11])) = (([1, 3] [5, 10]) + (([2, 4] [6, 11])) (([2, 3] [6, 10])) = 2 5 + 2 5 1 4 = 10 + 10 4 = 16 . (10/16)-1-7 Find the Lebesgue measure (in R2 ) of n=1 [n, n + 2n ] [n, n + 3n ]. Solution: Let denote the Lebesgue measure and for each n = 1, 2, 3, . . . , denote by In = [n, n + 2n ] [n, n + 3n ]. For each n = 1, 2, 3, . . . , 2n , 3n < 1, so that for any two distinct positive integers i and j the intervals Ii and Ij are disj...

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