A Symmetry Theorem for Dynamic Stochastic Programming
Miles Kimball1
University of Michigan
April 2010
Define:
vt  (Kt , Xt , vt1)
K t 1    K t , X t , t 1 
V t (Kt ) 
max
Xt ;( Kt , Xt )W


 Kt , Xt ,V t 1  Kt , Xt ,t 1 


Assume S KT ,V T (KT )  V T TK (KT ) for some terminal time T .
Theorem: Given the following conditions:
A) v t11 FSD v t21 (Kt , Xt , vt11 )  (Kt , Xt , vt21 ) (RFSD)
B) (Kt , X t ) W  TK (Kt ), TX (Kt , X t )  W
(SCC)
C) Kt1  Kt , Xt ,t1  TK (Kt 1) TK (Kt ),TX (Kt , Xt ),t 1  (SIC)

D) S ( K t , t )   TK ( K t ), TX ( K t , X t ), S    K t , X t , t 1  , t 1 

Eii) S  K

)   V T

(SLU)
t 1
t 1
Ei) S Kt 1 ,V ( Kt 1 )  V TK ( K t 1 )  (IH i)
t 1
,V t 1 ( Kt 1
t 1
K
( K t 1 )  (IH ii)
Then:
S  Kt ,V t (Kt )  V t TK (Kt ) Kt
To demonstrate this result, we prove the following two steps:
1
I am deeply grateful to students in my “Advanced Mathematical Methods for
Macroeconomics” course at the University of Michigan, particularly Thomas Bridges and Noah
Smith, for helping me work out and write out carefully succeeding iterations of this theorem.
The text of this iteration is the solution to a final exam problem in Winter 2010 generously
provided by Noah Smith. Compare to Thomas Bridges’ lecture notes in “symmetrytheorem.pdf” to see in what sense this exam problem was a generalization.


i) V t TK (Kt )  S Kt ,V t (Kt )  0 Kt


ii) S Kt ,V (Kt ) V TK (Kt )  0 Kt
t
t
Step (i):
Define:

X *  arg max  Kt , X t ,V t 1    Kt , X t , t 1  
X t ; Kt , X t W

Lemma (L1):
u(x)  v(x)x  u( X ) FSD v ( X )
Proof:
a, v( X )  a  u ( X )  a
a,  X ; v( X )  a   X ; u( X )  a
a, P  u ( X )  a   P  v( X )  a 
u ( X ) FSD v ( X )
Now we write:
V t TK ( K t )   S  K t ,V t ( K t ) 

max
X t ;TK  K t , X t W

 TK ( K t ), X t ,V t 1   TK ( K t ), X t , t 1  



 S  K t , max  K t , X t ,V t 1    K t , X t , t 1   
X t ; K t , X t W




  TK ( Kt ), TX ( Kt , X * ), V t 1  TK ( Kt ), TX ( Kt , X * ), t 1 



S Kt ,  Kt , X * ,V t 1   Kt , X * , t 1 


 

 using (SCC)

 using (SIC) and (SLU)
T

K
,
X
,


  
  TK ( Kt ), TX ( K t , X * ),V t 1 TK   Kt , X * , t 1 


 TK ( Kt ), TX ( K t , X * ), S   K t , X * , t 1  ,V t 1
 0  using (IH i), (L1), and (RFSD)
Step (ii):
To prove (ii), it is sufficient to show that


S TK 1  Kt  ,V t TK 1  Kt   V t  Kt   0 Kt
*
K
t
t 1
Assume that both TK 1 and TX 1 exist.
Define:

X *  arg max  Kt , X t ,V t 1    Kt , X t , t 1  
X t ; Kt , X t W

Lemma (L2):
TK  Kt 1    TK ( Kt ), TX ( Kt , X t ), t 1 
 TK 1  Kt 1    TK 1 (Kt ), TX 1 (Kt , X t ), t 1 
Proof:
Taking TK 1 of both sides of the RHS gives us
 Kt , Xt ,t 1   TK 1 TK (Kt ),TX (Kt , Xt ),t 1 
Now take Kt  TK 1 ( Kt ) and (Kt , Xt ) TX 1(Kt , Xt ) . This yields:
 TK 1 (Kt ), TX 1 (Kt , Xt ),t 1   TK 1  Kt 1 
Now we write:


S TK 1  Kt  ,V t TK 1  Kt    V t  Kt 




 S  TK 1  Kt  ,
max
 TK 1  Kt  , X t ,V t 1  TK 1  Kt  , X t , t 1  
1
X t ;TK  Kt , X t W




 Kt , X * ,V t 1   Kt , X * , t 1 




 
 S TK 1  K t  ,  TK 1  K t  , TX 1  K t , X *  ,V t 1  TK 1  K t  , TX 1  K t , X *  , t 1

 Kt , X ,V
*
t 1
  K , X
*
t

, t 1 


 
 S TK 1  Kt  ,  TK 1  Kt  , TX 1  Kt , X *  , V t 1 TK 1   Kt , X * , t 1 

 Kt , X ,V
*

t 1
  K , X ,  
*
t
 

 

 Kt , X ,V
t 1
   K , X ,  
*
t
t 1
 0  by (IH ii), (L1), and (RFSD)
This completes the proof of the Generalized Symmetry Theorem.

 using (SCC)
 using (SIC) and (L2)
t 1
  Kt , X * , S TK 1   Kt , X * , t 1  ,V t 1 TK 1   Kt , X * , t 1 
*


 using (SLU)