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SolutionsManual

# SolutionsManual - David Ruppert Statistics and Finance An...

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David Ruppert Statistics and Finance: An Introduction Solutions Manual July 9, 2004 Springer Berlin Heidelberg NewYork Hong Kong London Milan Paris Tokyo

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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 doesn’t 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( α 1 X, α 2 X ) = α 1 α 2 σ 2 X , Cov( α 1 X, Z ) = 0, Cov( Y, α 2 X ) = 0, and Cov( Y, Z ) = 0. (c) Using (2.54) with w 1 and w 2 the appropriately-sized 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 ( σ 2 ) = n productdisplay i =1 1 2 πσ 2 e - 1 2 σ 2 ( Y i - μ ) 2 . Therefore the log-likelihood is

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2 2 Probability and Statistical Models log L ( σ 2 ) = - 1 2 σ 2 n summationdisplay i =1 ( Y i - μ ) 2 - n log( σ 2 ) / 2 + H where H consists of all terms that do not depend on σ . Differentiating the log-likelihood with respect to σ 2 and setting the derivative 1 with respect to σ 2 equal to zero we get 1 2( σ 2 ) 2 n summationdisplay i =1 ( Y i - μ ) 2 - n 2 σ 2 = 0 whose solution is σ 2 = 1 n n summationdisplay i =1 ( Y i - μ ) 2 . 4. Rearranging the first equation, we get β 0 = E ( Y ) - β 1 E ( X ) . (2.1) Substituting this into the second equation and rearranging, we get E ( XY ) - E ( X ) E ( Y ) = β 1 { E ( X 2 ) - E ( X ) 2 } . Then using σ XY = E ( XY ) - E ( X ) E ( Y ) and σ 2 X = E ( X 2 ) - E ( X ) 2 we get β 1 = σ XY σ 2 X , and substituting this into (2.1) we get β 0 = E ( Y ) - σ XY σ 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 σ 2 rather than σ 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 ) . One can easily check that for any N × 1 vector X and N × N matrix A X T AX = N summationdisplay i =1 N summationdisplay j =1 X i X j A ij , whence w T COV( X ) w = N summationdisplay i =1 N summationdisplay j =1 w i w j Cov( X i , X j ) .

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