QUANTITATIVE METHODS for ECONOMIC APPLICATIONS TASK 20/4/2020

QUANTITATIVE METHODS for ECONOMIC APPLICATIONS TASK 20/4/2020

I M 1) After having determined the complex number which is the solution of the equationD

D D

"  3  "  3 œ #3, calculate its cubic roots ^$ D.

D D " " "  3  "  3

"  3  "  3 œ D "  3  "  3 œ D "  3 "  3 œ D # œ #3 Ê D œ #3

     # .

Since #3 œ # cos  3sin we getÀ

# #

 1 1

^$ D œ^$ #cos1 1 sin1 1

'  5 #$  3 '  5 #$ ß ! Ÿ 5 Ÿ # .

For 5 œ ! D œ # # $  3" à

# #

we get _! _$ cos sin  _$  

 1 1

'  3 ' œ

for 5 œ " D œ # #  $  3" à

# #

we get _" _$ cos sin  _$  

 & &  '1  3 '1 œ

for 5 œ # we get D œ_# ^$ #cos$ sin$    ^$ #  3 Þ

#1  3 #1 œ

I M 2) Given the matrix œ , since œ  " is an eigenvalue of , find

" " "

! 5 5  "

! ! 5

 

 

 

 

 

 ^#  -

the value of the real parameter , verify that is diagonalizable and then find a matrix that5  diagonalizes .

From   we get 1 

 

 

 

 

 

 

- ˆ œ ! œ

" " "

! 5 5  "

! ! 5

5 5







   œ !

-

-

-

- - -

#

  # 

For - œ  " we get 1 "5 "5^# " œ ! Ê 5 " œ !Ê 5 œ  ". So the matrix is  œ

" " "

!  "  #

! ! "

 

 

 

 

 

 . We find all the eigenvalues:

T_$ - œ1- "-"-œ !Ê-_" œ-_# œ "ß-_$ œ  ". For 5 œ  " we study the matrix  " œ

! " "

!  #

! ! !

and - œ "    # and since

 

 

 

 

 

 

 ˆ

Rank "ˆœ " Ê 7 œ $  " œ # œ 7¹_" ^#_" and so the matrix is diagonalizable.

Now we go to construct the modal matrix. Firstly we search for two linearly independent eigenvectors corresponding to the eigenvalue - œ ". We must solve the system:

 

     

     

     

     

     

     

   

   

   

   

   

     

 "ˆ †—œ Ê † œ Ê Ê

! " " B !

!  # C !

! ! ! D !

C  D œ ! a B

a B D œ  C

 # .

So eigenvectors corresponding to the eigenvalue - œ " are •œ Bß Cß  C .

Two linearly independent eigenvectors may be •_" œ "ß !ß !  and •_# œ !ß "ß  " .

Now we search for an eigenvector corresponding to the eigenvalue - œ  " solving the sy- stem:  

     

     

     

     

     

     

   

   

   

   

   

    

 "ˆ †—œÊ † œ Ê Ê

# " " B !

!  # C !

! ! # D !

#B  C  D œ !

#D œ !

!

Ê C œ  #B  #B

D œ !

 . Eigenvectors corresponding to - œ  " are •œ Bß ß ! and one of them may be •_$ œ "ß  #ß ! . Finally we construct the modal matrix:

œ  œ ƒ

" ! "

! "  #

!  " !

 

 

 

 

 

 . And we can see that † † Ê

Ê

       

       

       

       

       

       

" " " " ! " " ! " " ! !

!  "  # ! "  # ! "  # ! " !

! ! " !  " ! !  " ! ! !  "

† œ † .

I M 3) Since the linear system has the solution , calculate



B  B  B œ !  

7B  B  B œ ! B  B  5B œ !

"ß !ß  "

" # $

" # $

" # $

the values of the real parameters and and then find a basis for the space of the solutions7 5 of the system.

Substituting we get:





"  !  " œ ! 7  !  " œ !

"  !  5 œ !

Ê 7 œ "

5 œ "

 and we get the system:





B  B  B œ ! B  B  B œ ! B  B  B œ !

B  B  B œ ! B œ  B  B

" # $

" # $

" # $

" # $ $ " #

Ê Ê .

All the solutions of the system are in the form B ß B ß  B  B" # " #.

A basis for the space of the solutions of the system may be "ß !ß  " à !ß "ß  "  .

I M 4) Determine the vectors parallel to the vector — œ "ß  "ß #  and with modulus equal to '.

A vector parallel to the vector — œ "ß  "ß #  is a vector of the type:

˜ œ 5 † "ß  "ß # œ 5ß  5ß #5    . And so we get:

  ˜ œ 5  5  %5 œ^{# # #} '5 œ^# ' Ê 5 œ " Ê 5 œ „"^# . So we get two vectors cor- responding to the request: ˜_" œ "ß  "ß #  and ˜_# œ  "ß "ß  # .

II M 1) Given the equation 0 Bß C œ /  ^{B C# #} /^BC œ ! satisfied at the point  !ß ! , verify that an implicit function B Ä C B  can be defined with it, and then calculate the first order derivative of this function.

The function 0 Bß C  is a differentiable function a Bß C −  ‘^#. It also turns out:

f0 B ß Cœ #B /^{B C# #} /^BCà #C /^{B C# #} /^BC for which f0  !ß ! œ  "à ". Since 0_C^w !ß ! œ " Á ! it is possible to define an implicit function B Ä C B . For its derivative we have: C ! œ   " œ " Þ

"

w 

II M 2) Solve the problem Max min   . s v

Î 0 Bß C œ B  #C  $D Þ Þ B  C  D œ "%^{# # #}

The objective function of the problem is a continuous function, the constraint defines a feasi- ble region which is a compact set since it consists of only boundary points (the surface of aX sphere) and therefore surely maximum and minimum values exist.

For a problem with equality constraints we construct the Lagrangian function and apply to it the first and second order conditions. We have:

ABß Cß Dß-œ B  #C  $D-B  C  D  "%^{# # #} . Applying the first order conditions we have:

f Ê Ê Ê

œ "  # B œ ! œ  #  # C œ ! œ $  # œ !

œ œ  œ

A - 

A -

A -

A -

Bß Cß Dß œ

D B  C  D œ "%

B C D

  œ "%

 

 

 

 

 















wB wC wD

# # #

"

# "

$

" # " *

% %

- - -

-^# -^# -^#

Ê Ê Ê ∪

œ œ  œ

œ œ  œ

œ " œ  "

œ  # œ #

œ $ œ  $



 

 

   

   

   

   

   

   

   

 

 







 

 

B C D

œ "% œ

B C D

B B

C C

D D

œ œ 

Þ

"

# "

$

"% # "

%

"

# "

$

# " "

- - - -

- - -

# - - -

# % # #

We have only two stationary points: T œ "ß  #ß $_"   and T œ  "ß #ß  $_#  .

By Weierstrass' theorem, since 0 "ß  #ß $ œ "%  and 0  "ß #ß  $ œ  "%  it obviously turns out that is the maximum point while is the minimum point.T_" T_#

If we want to apply the second order conditions we have to built the bordered Hessian matrix:

‡ 

 

 

 

 

 

 

 

 

Bß Cß Dß œ

! #B #C #D

#B  # ! !

#C !  # !

#D ! !  #

- -

-

-

to calculate then, in the points and , theT" T#

minors ‡_$ and ‡_%. Since ‡ 

 

 

 

 

 

 

 

 

T_" œ À

! #  % '

#  " ! !

 % !  " !

' ! !  "

it is

‡_$ T_" œ #!  ! and ‡_% T_" œ  &'  ! and so is the maximum point;T_"

Since ‡ 

 

 

 

 

 

 

 

 

T_# œ À

!  # %  '

 # " ! !

% ! " !

 ' ! ! "

it is

‡$ T# œ  #!  ! and ‡% T# œ  &'  ! and so is the minimum point.T#

II M 3) Given 0 Bß C œ B C  and 1 Bß C œ B  C  ^{# #} determine for which values of parameter α it is W_@0 "ß " œ  W_@1 "ß " , with @œcosαßsinα.

The functions 0 Bß C œ B C  and 1 Bß C œ B  C  ^{# #} are polynomials and therefore are diffe- rentiable in any order a Bß C −  ‘^#.

So H 0_@  "ß " œ f0 "ß " † @ and H 1_@  "ß " œ f1 "ß " † @. Since:

f0 B ß C œ Cà BÊf0   "ß " œ "ß " and f1 B ß C œ #Bà #CÊf1 "ß " œ #à # , to get W_@0 "ß " œ  W_@1 "ß "  it will be:

 "ß " cosαßsinαœ # à # cosαßsinαÊcosαsinαœ ! Êcosαœ sin and so:α αœ $1 αœ (1

% or % .

II M 4) Given the function 0 Bß Cß D œ B CD   ^{# $} logC  B  / ^CD, determine the gradient vector of the function at T œ "ß #ß #_!  .

It is f BCD   " à B D  "  / à $B CD  /

C  B C  B

0 Bß Cß D œ  # ^{$ # $ CD # # CD} and so:

f "ß #ß #   " à )  "  / à #%  /

#  " #  "

0  œ $# ^{## ##} œ $$à )à #$.