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Unformatted text preview: David Ruppert Statistics and Finance: An Introduction Solutions Manual July 9, 2004 Springer Berlin Heidelberg NewYork HongKong London Milan Paris Tokyo 2 Probability and Statistical Models 1. (a) E (0 . 1 X + 0 . 9 Y ) = 1 . Var(0 . 1 X + 0 . 9 Y ) = (0 . 1 2 )(2) + 2(0 . 1)(0 . 9)(1) + (0 . 9 2 )(3) = 2 . 63 . (b) Var { wX + (1 w ) Y } = 3 w 2 4 w + 3 . The derivative of this expression is 6 w 4. Setting this derivative equal to 0 gives us w = 2 / 3. The second derivative is positive so the solution must be a minimum. In this problem, assets X and Y have same expected return. This means that regardless of the choice of w , that is, the asset allocation, the expected return of the portfolio doesnt change. So by minimizing the variance, we can reduce the risk without reducing the return. Thus the ratio w = 2 / 3 corresponds to the optimal portfolio 2. (a) Use (2.54) with w 1 = (1 1) T and w 2 = (1 1) T . (b) Use part (a) and the facts that Cov( 1 X, 2 X ) = 1 2 2 X , Cov( 1 X,Z ) = 0, Cov( Y, 2 X ) = 0, and Cov( Y,Z ) = 0. (c) Using (2.54) with w 1 and w 2 the appropriatelysized vectors of ones, it can be shown that Cov parenleftBigg n 1 summationdisplay i =1 X i , n 2 summationdisplay i prime =1 Y i prime parenrightBigg = n 1 summationdisplay i =1 n 2 summationdisplay i prime =1 Cov( X i ,Y i prime ) . 3. The likelihood is L ( 2 ) = n productdisplay i =1 1 2 2 e 1 2 2 ( Y i ) 2 . Therefore the loglikelihood is 2 2 Probability and Statistical Models log L ( 2 ) = 1 2 2 n summationdisplay i =1 ( Y i ) 2 n log( 2 ) / 2 + H where H consists of all terms that do not depend on . Differentiating the loglikelihood with respect to 2 and setting the derivative 1 with respect to 2 equal to zero we get 1 2( 2 ) 2 n summationdisplay i =1 ( Y i ) 2 n 2 2 = 0 whose solution is 2 = 1 n n summationdisplay i =1 ( Y i ) 2 . 4. Rearranging the first equation, we get = E ( Y ) 1 E ( X ) . (2.1) Substituting this into the second equation and rearranging, we get E ( XY ) E ( X ) E ( Y ) = 1 { E ( X 2 ) E ( X ) 2 } . Then using XY = E ( XY ) E ( X ) E ( Y ) and 2 X = E ( X 2 ) E ( X ) 2 we get 1 = XY 2 X , and substituting this into (2.1) we get = E ( Y ) XY 2 X E ( X ) . 5. E ( w T X ) = E parenleftBigg N summationdisplay i =1 w i X i parenrightBigg = N summationdisplay i =1 w i E ( X i ) = w T { E ( X ) } . Next Var( w T X ) = E braceleftbig w T X E ( w T X ) bracerightbig 2 = E bracketleftBigg N summationdisplay i =1 w i { X i E ( X i ) } bracketrightBigg 2 1 The solution to this problem is algebraically simpler if we treat 2 rather than as the parameter. 2 Probability and Statistical Models 3 = N summationdisplay i =1 N summationdisplay j =1 E [ w i w j { X i E ( X i ) }{ X j E ( X j ) } ] = N summationdisplay i =1 N summationdisplay j =1 w i w j Cov( X i ,X j ) ....
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