HW5 Solution - MIE375H1F Financial Engineering HW5 Solutions Nick Yeung November 7 2010 Question 6.3 6.3a 1 = min 2 2 wi wj ij i,j =1 1 1 1 1 =

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MIE375H1F - Financial Engineering - HW5 Solutions Nick Yeung November 7, 2010 Question 6.3 6.3a σ = min 1 2 2 X i,j =1 w i w j σ ij = 1 2 α 2 (0 . 15) 2 + 1 2 ( α )(1 - α )(0 . 1)(0 . 15)(0 . 3) + 1 2 (1 - α )( α )(0 . 1)(0 . 15)(0 . 3) + 1 2 (1 - α ) 2 (0 . 3) 2 = 0 . 05175 α 2 - 0 . 0855 α + 0 . 045 Taking d and setting it = 0: d (0 . 05175 α 2 - 0 . 0855 α + 0 . 045) = 0 0 . 1035 α - 0 . 0855 = 0 α = 19 23 6.3b The minimum standard deviation is: σ = q α 2 σ 2 1 + 2 α (1 - α ) σ 12 + (1 - α ) 2 σ 2 2 = r ( 19 23 ) 2 (0 . 15) 2 + 2( 19 23 )(1 - 19 23 )(0 . 1 · 0 . 15 · 0 . 3) + (1 - 19 23 ) 2 (0 . 3) 2 = 0 . 137 = 13 . 7% 6.3c The expected return is: ¯ r = α ¯ r 1 + (1 - α ) ¯ r 2 = 19 23 (0 . 1) + (1 - 19 23 )(0 . 18) = 0 . 114 = 11 . 4% 1
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Question 6.7 6.7a The Conditions for an Efficient Set are: 2 w 1 + w 2 - 0 . 4 λ - μ = 0 (1) w 1 + 2 w 2 + w 3 - 0 . 8 λ - μ = 0 (2) w 2 + 2 w 3 - 0 . 8 λ - μ = 0 (3) 0 . 4 w 1 + 0 . 8 w 2 + 0 . 8 w 3 = ¯ r (4) w 1 + w 2 + w 3 = 1 (5) Using the hint that
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This note was uploaded on 04/20/2011 for the course MIE 375 taught by Professor R.kwon during the Spring '11 term at University of Toronto- Toronto.

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HW5 Solution - MIE375H1F Financial Engineering HW5 Solutions Nick Yeung November 7 2010 Question 6.3 6.3a 1 = min 2 2 wi wj ij i,j =1 1 1 1 1 =

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