assign02soln

# assign02soln - (2.1 To prove β 2 = 0 if and only if μ is...

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Unformatted text preview: (2.1) To prove : β 2 = 0 if and only if μ is a multiple of l . Proof: Note that β 2 = h 1 Σ h 1 , where h 1 = Σ- 1 μ- l Σ- 1 μ l Σ- 1 l ! Σ- 1 l . ( ⇒ ) Suppose β 2 = h 1 Σ h 1 = 0. Since Σ is positive definite, we must have h 1 = 0. Therefore, Σ- 1 μ = l Σ- 1 μ l Σ- 1 l ! Σ- 1 l . Left-multiplying both sides of the equation by Σ we get μ = l Σ- 1 μ l Σ- 1 l ! l, so we see that μ is a multiple of l . ( ⇐ ) Suppose μ = cl . Then h 1 = c Σ- 1 l- c l Σ- 1 l l Σ- 1 l ! Σ- 1 l = c Σ- 1 l- c Σ- 1 l = 0 , which implies that β 2 = h 1 Σ h 1 = 0. (2.2) To prove : β 1 = 0. Proof: Note that β 1 = h Σ h 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. Then, β 1 = h Σ h 1 = l Σ- 1 l Σ- 1 l Σ h 1 = l l Σ- 1 l h 1 = l l Σ- 1 l Σ- 1 μ- l Σ- 1 μ l Σ- 1 l ! Σ- 1 l ! = l Σ- 1 μ l Σ- 1 l- ( l Σ- 1 μ ) l Σ- 1 l ( l Σ- 1 l ) 2 = l Σ- 1 μ l Σ- 1 l- l Σ- 1 μ l Σ- 1 l = 0 as required. 1 (2.3) To prove : β 2 = α 1 . Proof: Note that β 2 = h 1 Σ h 1 and α 1 = μ h 1 , where h 1 = Σ- 1 μ- l Σ- 1 μ l Σ- 1 l ! Σ- 1 l . Then, β 2 = h 1 Σ h 1 = h 1 μ- l Σ- 1 μ l Σ- 1 l ! l ! = h 1 μ- l Σ- 1 μ l Σ- 1 l ! h 1 l = α 1- l Σ- 1 μ l Σ- 1 l ! l h 1 . Therefore, α 1- β 2 = l Σ- 1 μ l Σ- 1 l l Σ- 1 μ- l Σ- 1 μ l Σ- 1 l ! l Σ- 1 l ! = l Σ- 1 μ l Σ- 1 l l Σ- 1 μ- l Σ- 1 μ = 0 , so β 2 = α 1 . Remark . Note that if μ is a scalar multiple of l , then h 1 is a zero vector, and consequently α 1 = β 2 = 0. (2.4) Question : Write out the optimality conditions for min { x Σ x | μ x = μ p , l x = 1 } , (1) and solve them in terms of Σ- 1 . Show that these efficient portfolios are identical to the portfolios given by...
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assign02soln - (2.1 To prove β 2 = 0 if and only if μ is...

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