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SOLUTIONS:
1. A permutation matrix is a square matrix resulting from permuting the rows of a square identity
matrix. Each row and column has exactly a single non-zero entry, which is 1, and zeros as the
remaining entries.
Use: The transpose of a permutation matrix can be used to factor tensors.
Proof that the product of permutation matrices is a permutation matrix.
Suppose you have two permutation matrices XX and YY.
If the non-zero entry in the first row of XX is in the ii-th column, then the first row of the
matrix X⋅YX⋅Y will be:
(∑nj=1X1, jYj, 1, ∑nj=1X1, jYj, 2… ∑nj=1X1, jYj, n)(∑j=1nX1, jYj, 1, ∑j=1nX1, jYj, 2… ∑j=1nX1, jYj,
n)
We know that:
X1, j=δi, jX1, j=δi, j
(Where δi, jδi, j is the Kronecker delta taking the value 11 when i=ji=j and 00 otherwise).
Hence the first row of X⋅YX⋅Y will be:
(X1, iYi, 1, X1, iYi, 2… X1,iYi,n)(X1,iYi,1,X1,iYi,2,…,X1,iYi,n)
Which since X1, i=1X1, i=1 will be:
(Yi,1,Yi,2,…,Yi,n).(Yi,1,Yi,2,…,Yi,n)
Therefore, first row of X⋅YX⋅Y corresponds to the ii-th row of YY. As we move on, you’ll get to see
that the mm-th row of X⋅YX⋅Y will be the ll-th row of YY, where the 11 in the mm-th row
of XX appears in the ll-th position.
Multiply two permutation matrices, therefore, is basically just rearranging the rows of one of
the matrices resulting in also in a permutation matrix.
Proof that the inverse of a permutation matrix is the transpose of the matrix.
A permutation matrix is the product of a sequence of interchange elementary matrices.
Supposing, P = E1E2 · · · En,
Where P denotes a permutation matrix,
Each Ei interchanges some two rows of the identity matrix. It’s obvious that Ei is symmetric,
So:
ETi = Ei. Also, we have E2i = I because applying the same interchange thrice returns to the
identity.
Therefore,
PPT = (E1E2 · · · En) (E1E2 · · · En) T = (E1E2 · · · En) (ETn ETn−1 · · · ET1)
= (E1E2 · · · En) (EnEn−1 · · · E1) = E1 · · · En−1E2nEn−1 · · · E1
= E1 · · · En−1En−1· · · E1 = · · · = I,
Which implies P−1 = PT
2. Step 1 Define Objectives
Minimize x’Cx x∈Rn such that g’x = r
x x x
x ]T
1 2
n
The asset weight vector is given as
with
portfolio.
Let g be represented by p for individual asset return
xi as the weight of asset i in the
The expected return for each asset in the portfolio is expressed in the vector form
p [ p1 p2
pn ]T with pi as the mean return of the asset i . The portfolio expected return is the
n
weighted average of individual asset return
given as a covariance matrix; C =
11
V
n1
x p pT x xi pi
i 1
. The covariance of an asset is
1n
nn
, where i ,i is the variance of asset i and i , j is the covariance between
n
asset i and asset j . The portfolio variance is
n
p x Vx xi x j i , j
2
T
i 1 j 1
Step 2
Use multi-objective optimization to solve for x:
The Lagrangian multiplier can be used to solve for optimization by:
L( x ) pT x xTVx (1T x 1)
Set
L
0 , it follows that
x
x
To solve, substitute as follows:
1
(V 1 )( p 1)
2
1T V 1 p
2
T 1
T
1
1 V 1 1 V 1
Let a1 1T V 1 1 and a2 1T V 1 p , both of which are scalars, equation 1.3 can be written as:
a2 2
a1 a1
The optimized solution for the portfolio weight vector x is
x*
1 1
V 1 a2 2
V p
( )1
2
2 a1 a1
3. Discuss possible solutions to Ax=0
1. A homogeneous system of m linear equations in n unknowns always has a non–trivial solution if
m < n. Suppose that m < n and that the coefficient matrix of the system is row–equivalent to B, a
matrix in reduced row–echelon form. Let r be the number of non–zero rows in B. Then r ≤ m < n
and hence n − r > 0 and so the number n − r of arbitrary unknowns is in fact positive. Taking one
of these unknowns to be 1 gives a non–trivial solution.
2. The amount of non-zero conditions in the echelon type of the framework is equivalent to the
quantity of leading entries.
3. The number of leading variables is ≤ min (m, n).
4. The quantity of free variables including the amount of leading variables = n which is the size of
columns of A.
5. The homogenous system Ax = 0 has non-trivial solutions if and only if there are free variables.
6. If there are more unknowns than equations, the homogeneous system always has nontrivial
solutions. Why? This is one of only few cases in which we can ...