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assign03soln - (2.11 The optimality conditions for this...

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(2.11) The optimality conditions for this problem are - Σ x = ul and l 0 x = 0 , which implies that x = - u Σ - 1 l + t Σ - 1 μ and 0 = l 0 x = - ul 0 Σ - 1 l + tl 0 Σ - 1 μ. Therefore, u = t l 0 Σ - 1 μ l 0 Σ - 1 l and x = - t l 0 Σ - 1 μ l 0 Σ - 1 l Σ - 1 l + t Σ - 1 μ = th 1 , which has μ p = μ 0 x = 0 h 1 = 1 , σ 2 p = x 0 Σ x = t 2 h 0 1 Σ h 1 = t 2 β 2 = t 2 α 1 , so t = ± σ p / α 1 . This gives us two straight lines in ( σ p , μ p ) space: μ p = σ p α 1 and μ p = - σ p α 1 . . . σ p 0 . . . . μ p 1
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(2.13) Recall the two optimization problems: (2 . 1) min x Σ x s.t. μ x = μ p , l x = 1 (2 . 3) min - x + 1 2 x Σ x s.t. l x = 1 (where t is allowed to be negative). Since both problems have convex quadratic objective functions and linear equality constraints, the necessary and sufficient conditions for the two problems are respectively given by (2 . 1 * ) braceleftBigg μ x = μ p , l x = 1 - x = u 1 μ + u 2 l for some u 1 , u 2 R (2 . 3 * ) braceleftBigg l x = 1 - Σ x = wl for some w R If x is optimal for (2.1), x must satisfy (2.1*); taking t = - u 1 / 2 and w = u 2 / 2, we have that x satisfies (2.3*). Thus x is optimal for (2.3) with parameter t = - u 1 / 2. Conversely, if x is optimal for (2.3), x must satisfy (2.3*); taking μ p = μ x , u 1 = - 2 t and u 2 = 2 w , we have that x satisfies (2.1*). Thus x is optimal for (2.1) with parameter μ p = μ x . (2.14a) From equation (2.8), the parametric-efficient portfolio for t 0 is x ( t ) = h 0 + th 1 , where h 0 = Σ 1 l l Σ 1 l and h 1 = Σ 1 μ - l Σ 1 μ l Σ 1 l Σ 1 l . Consider the linear inequality system e 1 x = e 1 h 0 + te 1 h 1 0 e 2 x = e 2 h 0 + te 2 h 1 0 . . . e n x = e n h 0 + te n h 1 0 in terms of t 0, where e i is the i -th row of the n -by- n identity matrix. Note that e i x ( t ) corresponds to the i -th entry of x ( t ); solvability of the above system for some t 0 is equivalent to existence of non-negative portfolio. 1
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First note that if there exists i s.t. e i h 0 < 0 and e i h 1 = 0, the above system does not have any solution t 0. Thus there are no non-negative efficient portfolios.
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