320 Economic Cycles
such incomplete observations are dropped from the sample. For the sub-sample of
expansions, we consider instead 1−St, so that( 1 −St)=1 marks the turning point
of an economic expansion.
An economic expansion or contraction is aphaseof the business cycle. Definedt
as the number of months in a given phase. The above table demonstrates values
for laggeddt. Now drop those observations from the sample ifdt− 1 =0, and define
mas the number of remainingStvalues. For our example,m=8 since we drop two
observations becausedt− 1 =0 and we drop the last contraction due to censoring. A
straightforward test for duration dependence in contractions is obtained by testing
the null hypothesis,H 0 :β 1 =0, in the simple regression equation:
St=β 0 +β 1 dt− 1 + (^) t, (7.6)
whereEt− 1 (
t)=0.
For constant hazards,β 1 =0 andβ 0 =p 01 , wherep 01 =P(St= 1 |St− 1 = 0 ).
The termdt− 1 captures autonomous changes in the hazard function. The resulting
model can be written as:
St=p 01 +vt. (7.7)
Hamilton (1994, p. 684) shows how to write such an equation if one is considering
complete cycles rather than half-cycles. For the purpose of a duration analysis,
however, it is only necessary to consider half-cycles: expansions or contractions,
bull or bear markets, upswings or downswings.
For non-constant hazards,β 1 =0, and the termination probability depends on
dt− 1 , the length of time in the specified phase. Ohn, Taylor and Pagan (2004)
show that the SBt-test is appropriate for testingH 0 :β 1 =0. We argue here the
same point, but from a different approach. In particular, by the definition of the
binary indicator,St, the duration of an expansion must be at least one month. The
geometric density is thus left-censored at unity. The censored probability function
isP(T=t)=( 1 −p)t−^1 pfor 1≤t≤∞, withE(T)= 1 /pandV(T)=( 1 −p)/p^2. Letn
be the number of turning points. ThenT=( 1 /n)
n
i= 1
Tiis a consistent estimator for
E(T), ̃σ^2 =( 1 /n)
n
i= 1
(Ti−T)^2 is a consistent estimator forV(T), andSis a consistent
estimator forp.
The least squares estimator ofβ 1 is:^10
β̂ 1 =
1
m
m
t= 1
(St−S)dt− 1
1
m
m
t= 1
(dt− 1 −d)^2
, (7.8)
whereS=m^1
m
t= 1
St=n/mandd=m^1
m
t= 1
dt− 1.
Our goal is to show thatplim̂β 1 =0 under the null hypothesis that dura-
tions follow the geometric density. To do so, consider the numerator of the least