Some algebra to seminar exercise 10
Q1 c)
Impact: ∂Ut /∂zt = β u2 < 0 ∂wt /∂zt = β w1 β u2 > 0. Long run: Assume
dynamic stability define Steady state by ∆wt = ∆mc = gmc and Ut = Ut−1 =
U . The long-run version of the model is then:
β w0
(β − 1)
+ w2
gmc
1 − αw
1 − αw
= β u2 z + β u0
β w1
U
1 − αw
−β u1 (w − mc) + (1 − au )U
=
(w − mc) −
Solve this two equation system to obtain:
(w − mc) =
U
=
β w1
z
1 − αw
β
(1 − αu ) − β u1 w1
1 − αw
β u1 β 0w0 + β u2 z
β
(1 − αu ) − β u1 w1
1 − αw
(1 − αu )β 0w0 + β u2
β w0
w2 −1)
where β 0w0 = 1−α
+ (β1−α
gmc , and find the derivatives with respect to z.
w
w
Q1 e)
For simplicity set zt = 0. With αu = 0, the final equation for wt is
wt
= β wo + β w1 β u0 + (αw + β w1 β u1 )wt−1 + β w2 mct
+ [(1 − αw ) − β w2 − β w1 β u1 ] mct−1
Although the stability condition is not required here, it might be noted that
it will be in terms of (αw + β w1 β u1 ) meaning that we can have stability, even
though αw = 1. Impact multiplier is β w2 of course, and additional multipliers
can be obtained from this final equation as well.
For Ut , the following equation can be used (together with the final equation
for wt ):
Ut
= β u0 + β u1 (β wo + β w1 β u0 )
+β u1 (αw + β w1 β u1 )wt−2 + (β u1 β w2 − β u1 )mct−1
+β u1 [(1 − αw ) − β w2 − β w1 β u1 ] mct−2
Long-run. Since αu = 0 is does not damage the stability. The long-run multipliers can be derived from the solution to Q1c).
1