Question

10. The following modified output from PcGive presents infomation concerning unit root tests on the logarithm of Y (LY), where a constant and trend is included. The sample size is 100. t-adf 2.7256 0.93930 2.5568 0.94380 2.2990 0.94949 2.0632 0.95425 2.0181 0.95546 2.0442 0.95518 lag -AY LY LY LY LY LY LY 0.049070 5 0,0491504 0.049613 3 0.050322 20.72567 0.050230 017339 0.8626 0,050044 0 1.1881 0.2370 0,0665 0.0318 0.4694 2.1707 lag AY ALY 3.8343 0.29629 ALY 3.6152 0.36690 ALY ALY ALY ALY 0.050268 5 0.050382 0.78918 0.4315 0.050308 3 0.050585 2 0.051159 0.05105 0 0.2111 4.1618 0.31845 5.3150 0.20919 7.6599 0.043753 .3150 -0.015670 1.9914 -0.66 0.1218 0,0486 743 0.5057 i) How many lags should be used for the ADF test on LY? a) 0 c) 2 d) 3 i) What is the order of integration for LY? b) 1 c) 2 d) 3 iii) How many lags should be used for the ADF test on ALY? b) 1 c) 2 d) 3 iv Which ADF for ALY should be reported? a) -3.8343 b) -3.6152 c) -4.1618 d) 5.3150 e)-7.6599 What does Δ1nV represent? c) (AlnYF d) AlmY. / Almy,' e) None of them

Answer 1

i) Null Hypothesis of Unit Root test (ADF test) is that THERE IS PRESENCE OF UNIT ROOT i.e NON STATIONARY.

We reject Null Hypothesis when P-value of any test is less than 0.05 for 95% Confidence Interval. Only at lag 3, P-value is less than 0.05 (0.0318<0.05). So, lag 3 should be used for LY i.e. option (d).

ii) Order of integration is the minimum number of differences required to obtain stationary time series. Following from i), it should be option (d) i.e 3

iii) Using the same logic as (i) , number of lags for $\Delta$LY in ADF test should be 2.( P-value of 0.0486< 0.05). i.e option (c)

iv) Following from (iii), ADF of -5.3150 should be reported as it has corresponding lag 2 which makes the series stationary. i.e option (d)

v) $\Delta$2 means second order differencing. It is calculated as follows:

Series First Order Second Order

lnY(t) 0 0

lnY(t-1) lnY(t-1)-lnY(t) 0

lnY(t-2) lnY(t-2)-ln(t-1) [ lnY(t-2) - ln(t-1) ] - [ lnY(t-1)- lnY(t) ]

= lnY(t-2) - lnY(t-1) -  lnY(t-1) + lnY(t)

= lnY(t) - 2lnY(t-1) + lnY(t-2).

So, it is option (a).