assign05soln

Assign05soln - (5.10a Consider the problem without nonnegativity constraints min{− tμ ′ x 1 2 x ′ Σ x | l ′ x = 1 whose solution is given

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Unformatted text preview: (5.10a) Consider the problem without nonnegativity constraints, min {− tμ ′ x + 1 2 x ′ Σ x | l ′ x = 1 } , whose solution is given by x ( t ) = h + th 1 , where h = Σ − 1 l l ′ Σ − 1 l and h 1 = Σ − 1 μ − l ′ Σ − 1 μ l ′ Σ − 1 l Σ − 1 l. Since Σ is positive definite and diagonal, we know that x (0) = h > 0. Therefore, for all nonnegative t sufficiently small, x ( t ) ≥ 0, and thus must be optimal for the problem with the constraints x ≥ 0. (b) Recall from question 2.6 from Text (appeared in Assignment 4) that when Σ is diagonal, x ( t ) is given by x i ( t ) = [ θ 1 + t ( μ i − θ 2 )] /σ i , i = 1 , . . . , n, where θ 1 = 1 σ − 1 1 + ··· + σ − 1 n and θ 2 = θ 1 parenleftbigg μ 1 σ 1 + ··· + μ n σ n parenrightbigg . For x ( t ) to be optimal, we need x ( t ) ≥ 0. That is, [ θ 1 + t ( μ i − θ 2 )] /σ i ≥ , i = 1 , . . . , n, which is satisfied when t ( θ 2 − μ i ) ≤ θ 1 , for all i . Since we have μ 1 < μ 2 < ··· < μ n , then θ 2 − μ 1 > θ 2 − μ i , for all i . So, if t ( θ 2 − μ 1 ) ≤ θ 1 , then t ( θ 2 − μ i ) ≤ t ( θ 2 − μ 1 ) ≤ θ 1 , for all i . Note that θ 2 − μ 1 = θ 1 parenleftbigg μ 1 σ 1 + ··· + μ n σ n parenrightbigg − θ 1 μ 1 θ 1 = θ 1 bracketleftbigg μ 1 σ 1 + ··· + μ n σ n − μ 1 parenleftbigg 1 σ 1 + ··· + 1 σ n parenrightbiggbracketrightbigg = θ 1 bracketleftbigg μ 2 − μ 1 σ 2 + ··· + μ n − μ 1 σ n bracketrightbigg > . 1 Therefore, we just need t ≤ θ 1 θ 2 − μ 1 in order to satisfy x ( t ) ≥ 0. Thus, t 1 = θ 1 θ 2 − μ 1 . (c) When t is slightly larger than t 1 , the optimal solution must satisfy x 1 = 0. Notice that if x 1 = 0, then we are just solving the optimization problem min {− t ¯ μ ′ ¯ x + 1 2 ¯ x ′ ¯ Σ¯ x | l ′ ¯ x = 1 , ¯ x ≥ } , where ¯ x = ( x 2 , . . . , x n ) ′ , ¯ μ = ( μ 2 , . . . , μ n ) ′ , and ¯ Σ = diag( σ 2 , . . . , σ n ). As in part (b), for t sufficiently small, the efficient portfolios are given by x i ( t ) = bracketleftbig ¯ θ 1 − t ( ¯ θ 2 − μ i ) bracketrightbig /σ i , i = 2 , . . . , n, where ¯ θ 1 = 1 σ − 1 2 + ··· + σ − 1 n and ¯ θ 2 = ¯ θ 1 parenleftbigg μ 2 σ 2 + ··· + μ n σ n parenrightbigg . Moreover, this solution is valid for t ≤ t 2 , where t 2 = ¯ θ 1 ¯ θ 2 − μ 2 . Therefore, for t 1 < t ≤ t 2 , the optimal solution of the original problem is given by x 1 ( t ) = 0 , x i ( t ) = bracketleftbig ¯ θ 1 − t ( ¯ θ 2 − μ i ) bracketrightbig /σ i , i = 2 , . . . , n. In order to verify that this is indeed the optimal solution, we should check the optimality conditions, which, since Σ is diagonal, are given by l ′ x = 1 , x ≥ , tμ i − σ i x i = c − u i , for i = 1 , . . . , n, u ≥ , u ′ x = 0 ....
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This note was uploaded on 03/29/2010 for the course CO CO372 taught by Professor Michael during the Winter '09 term at Waterloo.

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Assign05soln - (5.10a Consider the problem without nonnegativity constraints min{− tμ ′ x 1 2 x ′ Σ x | l ′ x = 1 whose solution is given

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