Fractional Cointegration- LM test Robinson 1994

[arvo]

Dear all,
Does anyone have the procedure code for Robinsons LM test in his 1994 paper " Efficient tests of nonstationary hypotheses"
Thanks very much,

[TomDoan]

This looks like it does the Q=0 test for first differencing.

Code: Select all

*
* Robinson, JASA, December 1994
* Series B (Box-Jenkins IBM stock data)
*
open data "ibm-stock.txt"
data(format=free,org=var) 1 369 ibmstock
*
* Make the hypothesized difference of the data (1-L)^1 to create U-tilde
*
diff ibmstock / dseries
*
* Compute the variance of U-tilde
*
stats dseries
compute sigsq=%variance
*
compute endl=369,startl=2
*
compute n = endl-startl+1
*
freq 1 n
rtoc startl endl 1
# dseries
# 1
*
* Compute the periodogram and the psi function
*
fft 1
cmult(scale=1.0/(2*%pi*n)) 1 1
set pgram 1 n = %real(%z(t,1))
set psi   1 n = log(2 * abs(sin(-%pi*(t-1)/n)))
*
* Compute the sums of lambda to make small a-tilde and big A-tilde.
* Avoid the undefined value for psi at t=1 (zero frequency)
*
sstats 2 n  2.0*psi^2>>bigA
sstats 2 n -2.0*%pi*psi*pgram>>smallA
compute bigA=bigA/n,smallA=smallA/n
*
* This is closed form value rather than the value approximated off the sum
*
*compute bigA=%pi^2/6.0
*
* This is the chi-squared statistic
*
compute r=n*smalla^2/(bigA*sigsq^2)
*
* This is the one-tailed equivalent
*
compute onetail=sqrt(n)*smalla/(sqrt(bigA)*sigsq)
disp "Chi-Squared" r "One-tail" onetail

[arvo]

Dear Tom,
Thank you very much for the fast reply!
Now I just have to figure out how to include intercept and intercept and trend when computing.
I am assuming this code above is for no regressors? Or have I read it incorrectly.
Once again thank you!

[TomDoan]

The regression variant is y(t)=z(t)beta+x(t), phi(L)x(t)=u(t), where phi(L) is typically (1-L)^(1+theta) so you're testing a fractional alternative to first differencing. Under the null, that would mean estimating

(1-L)y=((1-L)z(t))beta+u(t)

If z is 1 and t, you would first difference the data and regress on a constant only, since the 1 drops out when differenced and the t converts to a constant.

[arvo]

Once again thank you for the quick reply.
I am looking to present 3 cases - no regressors, constant, constant and trend for different values of the differencing parameter.
With the code you have shown me, I am unsure how to adjust for the other 2 cases.
Your help is much appreciated, thank you.

[TomDoan]

Just detrend the data first, and fractionally difference the results using DIFF with the D option with D=.5 or whatever is your differencing parameter under the null. Fractional differencing can't eliminate a trend, so you have to get it out of the data first.

[mskare69]

Dear Tom,

Could you provide a rats code for replicating the results in TESTING THE ORDER OF INTEGRATION OF THE U.K. UNEMPLOYMENT - Gil Alana or guide me how to do it in rats to replicate the results so I can do it on myself for the data I have.

Thanks,

Fondly,

Marinko

[arvo]

Dear Tom,
Marinkos request would also be of significant use.

In addition when I said 3 cases,I meant on u(t), so u(t) would have no constant no trend, constant, constant and trend. Like in the paper Marinko states.
So given a process how would I consider these cases.Sorry if this is just what you have stated , I just obtained odd results before.

Thank you,

[TomDoan]

If you read the paper, in all cases, you simply apply the difference operator under the null to the dependent variable and the regressors, run a regression to get the residuals, and apply the calculation in my original post to the resulting series, with the residuals replacing my "dseries". If d=1, the constant would disappear and the trend would downgrade to a constant. If d<1, the constant would remain as is. I have no idea how Prof. Gil-Alana handles the model with constant and trend with a hypothesized d<1 since a fractional difference applied to a trend (theoretically) generates a non-convergent sum.

[vtsa]

Dear Tom,
Could you provide above mentioned Robinson(1994) LM test with autocorrelated Bloomfield errors (q=1, q=2, q=3, etc)...
I appreciate your responce.
Thank you in advance