Palgrave Handbook of Econometrics: Applied Econometrics

664 Panel Methods to Test for Unit Roots and Cointegration

(1999). For pooling to lead to tests with power requires all the idiosyncratic pro-
cesses to be integrated under the null hypothesis, while under the alternative a
strictly positive fraction of these processes is stationary even asN→∞.

13.2.3.1 Break dates known

Two models are considered, given by the specification of the deterministic
processes.

Model 1: Di,t=μi+

∑li

j= 1

θi,jDUi,j,t,i=1, 2,...,N.

Model 2: Di,t=μi+δit+

∑li

j= 1

θi,jDUi,j,t+

∑mi

k= 1

γi,kDTi∗,k,t,i=1, 2,...,N.

Thus in Model 1 only the intercept of the process is broken, with theith unit being
characterized bylibreaks at fractions (of the time-span of the sample) given by
Tai,j
T =γi,j, which are fractions that remain constant asT→∞. In Model 2, by
contrast, both the intercept and trend are broken (the latter at fractions given by
Tbi,k
T =λi,k,k=1, 2,...,mi). The break dates can be positioned heterogeneously
acrossi, the breaks can be of different magnitudes, each unit may have different
numbers of breaks, and the breaks in the intercept can be located at different time
periods from the breaks in trend.
To presage the results somewhat, it is simple to note that in Model 1, since the
factors are extracted from a model for the differenced observations, the breaks in
intercept reduce to impulse dummies which do not, in fact, have any impact on
the asymptotic distribution of the MSB statistics. Thus whether or not the breaks
are known or unknown does not make any difference, as long as these breaks are
restricted to the intercept. By contrast, for Model 2, where the breaks in trend
reduce upon differencing to changes in the intercept of the differenced process,
the statistics do depend upon the nuisance parametersλi,kand critical values need
to be computed to take this into account. In addition, if the break dates are not
knowna priori, consistent estimates of the break fractions are needed.

Analysis of Model 1
Differencing Model 1 yields:

yi,t=πi′Ft+e∗i,t, (13.14)

where:

ei∗,t=ei,t+

∑li

j= 1

θi,jD(Tai,j)t,

andD(Tai,j)tare the impulse dummies withD(Tai,j)t=1ift=Tai,j+1 and 0
elsewhere.