Professor Erdinç
ECO 402
Handout on Rational Expectations Model
Consider the following model.
1.
y s t  y n   ( pt  E( pt )) Lucas Supply
2.
y d t  brt  z t
3.
mt  pt  rt  yt  t
Aggregate Demand
Money Market Equilibrium
where  , b ,  are positive constants. In this model, AD is subject to real shocks with known mean and
2
variance, i.e. z t ~ N (0,  z ). At the same time, the money demand in equation (3) is subjected to velocity
shocks, t ~ N (0,   ) . y n is full employment (natural level of output) output.
2
1. Derive the solution values for the endogenous variables, y , r and p under rational expectations.
2. Show that in the presence of velocity (nominal) and demand (real) shocks, it is possible for the
output to deviate from its natural level (normalized here to zero for simplicity).
3. If the policy makers wish to stabilize output around its natural level via a constant money supply
rule, what is the variance of output under this policy? (Hint: Set mt = m = E (mt ) ).
4. If the policy makers wish to stabilize output around its natural level via interest rate targeting, what
is the variance of output under this policy? (Hint: Set rt = r and take it exogenous. Now you should
treat mt as endogenous.)
5. Which of the two policies, fixed money supply or fixed interest policy regime, will be optimal in
stabilizing he output around its natural level? Does your answer depend on the relative magnitude of
these shocks?
Solutions: Rational Expectations Model
1.) yts    pt  E  pt 
y t  pt  E  pt 

y td  brt  z t
y t  brt  z t
 yt  pt  rt  mt  t
mt  pt  rt  yt  t
The above equations correspond to the following matrix equation:
 1   0   yt   E  pt 
1
0 b   pt    zt 

 1  1    rt   t  mt 
Let
1

J  1
0
0
b 
1 1 
0
b
1 
   
1
b
1 
 b    b
It follows that
 E  pt   
zt
y t 
t  mt
1
b
1
zt
b
1
t  mt

1 
J
, p t 
J
, and rt 
1

 E  p t 
1
0
zt
t  mt
1 1
J
 E  pt  0
1
pt 
0
 E  p t  0
0
1
zt
b
1
 t  mt

J
zt

b
t  mt 
  E  pt 
1
b
1 
J

z t  bt  bmt  E  pt   b 
b    b

Assuming rational expectations, we can substitute pt with E  p t  in the above equation. We also
assume that random disturbance terms are equal to zero. We get:
 
 
z t  bt  bmt  E pt   b  bmt  E pt   b
E p 

b    b
b    b
 

t

bmt

   b  
E pt 1 


 b    b  b    b
 
 

E pt b    b    b  bmt

E pt 
 
bmt
 mt
b
Then we have:
z t  bt  bmt  E  pt   b  z t  bt  bmt  mt   b  z t  bt  bmt  mt  mt b



b    b
b    b
b    b
z   bt  mt b    b 
z   b t
 t
 t
 mt
b    b
b    b
pt 
 E  pt   
zt
y t 

 t  mt
b
1 
J
 E  pt 

0
b
1 
   
zt
b
 t  mt 
b    b

 E  pt b  z t  bt  bmt  bmt  z t  bt  bmt  z t  bt 


b    b
b    b
b    b
rt 

0
0
1

 E  pt 
1
0
zt
 t  mt
1 1
J
0

zt
 1  t  mt
   
1
zt
 1  t  mt
  E  pt 
1
0
1 1
b    b

z t   t  mt  z t   E  pt  z t  t  mt  z t  mt z t  t  z t


b    b
b    b
b    b
2.) Let the natural level of output be y n  0 . In the absence of velocity and demand shocks:
y t 
 z t  bt   0    b  0

 0  y n , i.e. the output is at its natural level equal to zero.
b    b
b    b
When either velocity or demand shocks t and z t are present (this means they are not equal to zero),
y t 
 z t  bt 
b
 0  y n , unless we have the unlikely event that z t  t , in which case
b    b

y t 
 z t  bt 
 0  y n as well.
b    b
3.) Let mt  m  E mt  . This would mean that the random disturbance term t =0. Then
y t 
 z t  bt 
z t

varies only according to the value of the demand shock term z t .
b    b b    b
 y2
 E yt 
 
2
m
 2 2
 z2
2
b    b
4.) Let rt  r . Then m will be endogenous in the model and we would have the following system of
equations:
yts    pt  E  pt 
y ts  pt  E  pt 

y td  br  z t
y td  z t  br
 yt  pt  mt  r  t
mt  pt  r  yt  t
In matrix form, the above equations are equivalent to:
 1   0  yt   E  pt 
1
0 0  pt    z t  br 

 1  1 1 mt   t  r 
1

J  1
0
0
0
1 1 1
0
1 1
 E  pt   
yt 

zt  br
t  r
0
1
0
1 1

0
0 0
1 1
J
   

   zt  br 

 br  zt
 
 y2 r  E yt   z2
2
5.) In the previous two parts we saw that neither a fixed money supply nor a fixed interest rate
 2 2
 z2 , therefore a constant
policy totally eliminates the variance of output. But  >
2
b    b 
2
z
money rate policy is optimal as the coefficient is less than 1.