advanced microeconomics – assignment 1

Advanced  Microeconomics  –  Assignment  1    

Problem  1:  Axioms  on  preferences  

!! ≿ !!,  !! ≻ !!  

By  definition,  !! ≻ !!  means  !! ≿ !!  but  not  !! ≿ !!  

So,  !! ≿ !! ≿ !!  

By  transitivity,  this  means  !! ≿ !!  

Now,  suppose  instead  we  have  !! ≿ !!.  Since  !! ≿ !!,  by  transitivity  !! ≿ !!  

But  there  is  a  contradiction  here  because  we  know  that  !! ≿ !!  

So,  we  have  shown  that  by  transitivity  !! ≿ !!  is  true,  and  by  contradiction  !! ≿ !!  cannot  hold.  

Therefore,  we  conclude  !! ≻ !!  

Problem  2:  Utility  Functions  

1. The  binary  relation  ≿  on  the  consumption  set  X  is  called  a  preference  relation  if  it  satisfies  Axioms  1  and  2.    Axiom  1  Completeness:  For  all  !!  and  !!  in  X,  either  !! ≿ !!  or  !! ≿ !!.    

a. Since  bundle  !!  is  better  than  !!  whenever  !! − !! ! ≥ !! −!! !,  it  is  possible  to  describe  all  the  relationships  between  bundles.  This  satisfies  axiom  1.    

Axiom  2  Transitivity:  For  any  three  elements  !!,  !!  and  !!  in  X,  if  !! ≿ !!  and  !! ≿ !!,  then  !! ≿ !!.  

a. Suppose  there  was  a  third  bundle  !!,  given  !!  is  better  than  !!  whenever  !! − !! ! ≥ !! − !! !,  by  transitivity  this  means  that  !! ≿ !!  if  !! − !! ! ≥ !! − !! !.  Therefore  transitivity  is  satisfied.    

Thus,  ≿  represents  preference  relations.  

2. Axiom  3  Continuity:  For  all  ! ∈ ℝ!!,  the  ‘at  least  as  good  as’  set,  ≿ (!),  

and  the  ‘no  better  than’  set,  ≾ (!),  are  closed  in  ℝ!!.  

 Axiom  4  Strict  Monotonicity:  For  all  !!, !! ∈ ℝ!

!,  if  !! ≥ !!,  then  !! ≿ !!,  while  if  !! ≫ !!,  then  !! ≻ !!.        

a. For  Axiom  3,      ≿ !! = {!! ∈ !|!! ≿ !!}  when  !! − !! ! ≤ !! − !! !  

≾ !! = {!! ∈ !|!! ≾ !!}  when  !! − !! ! ≥ !! − !! !    This  reads  as;  the  upper  contour  set  (!!)  that  is  weakly  preferred  to  set  !!  is  closed  when  !! − !! ! ≤ !! − !! !  is  satisfied,  and  the  lower  contour  set  (!!)  that  is  no  better  than  to  set  !!  is  closed  when  !! − !! ! ≥ !! − !! !  is  satisfied.  Based  on  this  definition,  the  continuity  axiom  is  met.    

b. For  Axiom  4,    given  ! ∈ [0,1]  and  that  bundle  !!  is  weakly  preferred  to  !!  whenever  !! − !! ! ≥ !! − !! !,  if  !!  contains  0.7  units  and  !!  contains  0.6  units,  !! ≿ !!,  despite    !!  >  !!.  So  in  this  case,  monotonicity  is  not  satisfied  since  more  is  not  necessarily  better.      For  axiom  4  to  be  satisfied,  the  set  must  be  restricted  to  the  region  ! ∈ [0,0.5],  so  that  the  more  of  one  bundle  (e.g.  !! >  !!)  necessitates  in  the  preference  of  !!  ≿  !!.    

3. Theorem  1.1:  If  the  binary  relation  ≿  is  complete,  transitive,  continuous,  and  strictly  monotonic,  there  exists  a  continuous  real-­‐valued  function,  !:  ℝ!

! → ℝ,  which  represents  ≿.    

a. From  part  2.2,  it  has  been  established  that  over  the  region  ! ∈ [0,1],  axiom  4  is  violated.  However,  theorem  1.1  only  states  sufficient  conditions  for  the  existence  of  a  !,  which  does  not  necessarily  mean  that  the  theorem  does  not  still  allow  for  a  !  to  exist.      

4. In  proving  theorem  1.1,  sets  A  and  B  are  defined  as:    ! ≡ ! ≥ 0 !" ≿ !                ! ≡ {! ≥ 0|!" ≾ !}      where  ! ≡ (1),  and  !∗ ∈ ! ∩ !  will  define  ! ! !~!  when  ! ! = !∗  such  that  ! ∈ [0,1],  and  !!  ≿  !!  whenever  !! − !! ! ≥ !! − !! !.      The  proof  fails  in  determining  the  uniqueness  of  ! ! .  Suppose  there  is  only  one  number  such  that  !!!~!~!!!,  by  transitivity  this  means  !!!~!!!.  But  it  is  no  longer  true  that  holds  when  strict  monotonicity  is  violated  because  !! = !!,  !! < !!  and  !! > !!  all  can  represent  !!!~!!!.                  

5. From  part  2.4,  a  utility  function  can  be  proven  to  exist.    In  the  region  of  ! ∈ [0,1],  since  strict  monotonicity  is  violated,  set  A  is  true  when  ! > !  is  in  the  interval  of  [0,0.5]  and  when  ! < !  is  in  the  interval  of  [0.5,1].      Similarly  for  set  B,  it  is  true  when  ! < !  in  the  interval  of  [0,0.5]  and  when  ! > !  is  in  the  interval  of  [0.5,1],  with  all  sets  closed  given  the  assumption  of  continuity.      Suppose  there  is  a  !  that  doesn’t  exist  in  the  set  of  A  and  B.  From  completeness,  it  must  be  the  case  that  either  !" ≿ !  or  !" ≾ !.  But  this  must  mean  that  ! ∈ ! ∪ !  for  either  case  to  hold.  Thus,  by  proof  of  contradiction,  there  is  at  least  one  !∗  that  represents  ! ! .