FRE6083_HW3_solutions

FRE6083_HW3_solutions - Solutions Question 1 Proceed term...

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Unformatted text preview: Solutions Question 1 Proceed term by term: (10 − 0.5 Chain rule: ln(1 + Chain rule again (ln(1 + which gives ln(1 + Last term: 4 ln( ) exp( ) 2) 2 2) 2 )=− 2 = 1 (ln(1 + 2 ))−1/2 × 2 −1 (ln(1 + 2 )) )) = 2 (1 + ) = (ln(1 + 2 ))−1/2 (1 + −1 2 −1 ) =4 ln( ) exp( ) − exp( ) exp( )2 =4 ln( ) − −1 exp( ) Putting the pieces together yields − + (ln(1 + 2 ))−1/2 (1 + 2 −1 ) +4 ln( ) − −1 exp( ) Question 2 Gradient is the vector of first partial derivatves g= 2 2 −2 −4 Stationary points are found where g = 0, so we solve the two equations = 1, for two stationary points (2, 0.4) and (−2, −0.5). The Hessian is the matrix of second partial derivatives, H= Evaluating H at each point gives 1 4 4 0 , −1 −4 −4 0 2 2 2 0 2 =4 These matrices are not positive definite or negative definite...so the function must be evaluated to determine what the nature of the stationary points. SInce 2 = 4 at the stationary ponts, ( , ) = −2 , so the function is constant at 4 for all when = −2 and −4 for all when = 2. When = −2 there is a minimum in the direction at (2, 0.5) and maximum at (−2, −0.5). Question 3 Row and column operations (interchanging rows or columns of a matrix) change the determinant of a matrix by a factor of (−1) where = + and and are the row (or column) indices. Here we can interchange rows 1 and 3 to simplify the determinant to just two 3 × 3 cofactors: 1 0 1 2 21 1 −1 00 21 0 10 2 01 = 1 12 1 22 0 −1 1 1 1 2 2 =1 0 2 1 3 10 12 −1 2 − 0 1 11 22 1 −1 1 The first component has just two elements, 2 1 2 10 −1 2 12 −1 2 = 2 − 11 21 11 In the second component we can interchange rows 1 and 2 and work with just two elements, 1 0 2 Hence the result, ∣ ∣ = −6 + 3 − 3 + 6 = 0 Since + = 1 + 3 = 4 is even, the determinant is not affected. In this example the determinant is 0 so it is immaterial what rows or columns are interchanged. Question 4 Write the system in matrix form, = , where ⎛ ⎞ ⎛⎞ 1 −2 3 = ⎝2 4 0 ⎠, = ⎝ ⎠, 0 2 −1 Now = −1 21 1 −1 2 1 −1 = −2 21 1 21 1 −1 ⎛⎞ 1 = ⎝3⎠ 0 if is non-singular. Here, −1 ⎞ −1 1 −3 = ⎝ 0.5 −0.25 1.5 ⎠ 1 − 0 .5 2 and vector . ⎛ which gives the solution ( , , ) = (2, −0.25, −0.5). In R you can use the command Question 5 Part(a) 1 0 .5 0 .5 1 ( , ) for appropriately defined matrix Part(b) Variance is the square of volatility, so the asset annual variances are 0.12 and 0.22 . Since the correlation is 0.5 the annual asset covariance matrix is 2 1 12 12 2 2 = = = 1 0 2 1 12 12 1 0 2 0 1 0 0 .1 0 0 0 .2 0.01 0.01 0.01 0.04 1 0 .5 0 .5 1 0 .1 0 0 0 .2 Part(c) The porfolio is a linear combination of assets with weight vector portfolio return is 0.01 0.01 ′ = ( 0 .4 0 .6 ) 0.01 0.04 = 2.08 × 10−2 4 ′ = (0.4, 0.6) so the annyal variance of 0 .4 0 .6 Question 6 Part a Eigenvalues 1 + and 1 − , so the matrix is positive definite unless the correlation has magnitude 1. The 1 1 eigenvectors are and respectively. Typically eigenvectors are normallized to have length 1, so 1 −1 √ these would be scaled by 1/ 2. Part b V= 2 1 12 12 2 2 so 1 = 2 = means that V = 2 C and hence both matrices have the same eigenvectors, the eigenvalues of V being 2 times the eigenvalues of C. Part c V= 0.04 0.025 0.025 0.0625 Eigenvalues are 0.0787 and 0.0238. Normalized eigenvectors are (0.543, 0.840)′ and (−0.840, 0.543)′. In general the eigen structure of correlation and covariance matrices are different because the relationship between the two is nonlinear. Question 7 By independence the joint pdf is the product of the marginal pdfs, ( , , )= Now, ( ≥ )= ≥ 1 1 0 1 1 0 1 1 () () ( )=1 0≤ , , ≤1 (,,) = 0 = 0 (1 − 2 ) = 0 1− = 3/4 Question 8 Part (a) Conditional on 1 the other two random variables are independent, so the joint conditional pdf is the product of the marginal conditional pdfs, 23 ( 2 , 3∣ 1) = 2− 1 1( 2 + 3 ) 0 for 2 > 0 and otherwise 1, 2, 3) 3 >0 > 0, > 0 and > 0. The joint pdf of 1 , For any such point, Now for 2 2 and ( 3 is positive only for points ( 2, 3) such that 1 2 3 1, = 1 ( 1 ) 23 ( 2 , 23 ( 2 , 3) = 2− 1 1 (1+ 2 + 3 ) > 0 and 3 > 0 the marginal joint pdf ∞ 23 ( 2 , 3) 3) follows by integration over = 2 (1 + 2 3 3) 1, = 0 ( 1, 2, 3) 1 + 5 Part (b) 4 4− 0 3 Pr( 2 + 3 < 4) = 0 23 ( 2 , 3) 2 3 = 16 25 Part (c) From the definition of conditional probability, 1( 1∣ 2, 3) = ( 1, 2, 2, 1 3) 3) 23 ( Substituting the pdf forms previously derived we obtain for 1( 1∣ 2, 3) > 0, 1 (1+ 2 + 3 ) = 1 (1 + 2 2 + 32− 3) 1 For 1 ≤ 0, the density is zero. 6 ...
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This note was uploaded on 02/16/2011 for the course FE 6083 taught by Professor Neveskyi during the Spring '11 term at NYU Poly.

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