7:43 p.m.
Hello,
I programmed the following loop and wanted to count how often the null
is rejected for the ADF test. Unfortunately I do not know whether there
is a simple function for this.
loop 1000 --progressive --quiet
genr a = 10*uniform()
genr y = 1*a + normal()
alpha = 0.99
genr y = alpha*y(-1) + normal()
adf 1 y --nc --quiet
genr pv = $pvalue
print pv
store coeffs.gdt pv
endloop
Do you know a solution for this?
Thanks in advance.
Artur
8:40 p.m.
Hi,
Here is my proposal (please correct any errors), assuming that adf
does not produce variables for your purpose. Note that I am a beginner
at this :).
<script>
nulldata 1000
confidence_lvl=0.05
series rejected=0
i=1
loop 1000 --progressive --quiet
genr a = 10*uniform()
genr y = 1*a + normal()
alpha = 0.99
genr y = alpha*y(-1) + normal()
adf 1 y --nc --quiet
genr pv = $pvalue
print pv
genr rejected[i]=(pv>confidence_lvl)?1:0
i++
store coeffs.gdt pv
endloop
printf "Count of rejected null hypothesis: %d\n", sum(rejected)
</script>
Bye,
Hélio
On Sat, Jan 2, 2010 at 7:43 PM, Artur T. <artur.tarassow(a)googlemail.com> wrote:
Hello,
I programmed the following loop and wanted to count how often the null
is rejected for the ADF test. Unfortunately I do not know whether there
is a simple function for this.
loop 1000 --progressive --quiet
genr a = 10*uniform()
genr y = 1*a + normal()
alpha = 0.99
genr y = alpha*y(-1) + normal()
adf 1 y --nc --quiet
genr pv = $pvalue
print pv
store coeffs.gdt pv
endloop
Do you know a solution for this?
Thanks in advance.
Artur
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8:50 p.m.
Hélio Guilherme schrieb:
nulldata 1000
confidence_lvl=0.05
series rejected=0
i=1
loop 1000 --progressive --quiet
genr a = 10*uniform()
genr y = 1*a + normal()
alpha = 0.99
genr y = alpha*y(-1) + normal()
adf 1 y --nc --quiet
genr pv = $pvalue
print pv
I suggest to move the print command out of the loop, it will slow things
down.
genr rejected[i]=(pv>confidence_lvl)?1:0
I guess this should be < instead of >?
i++
or loop for i=1..1000
store coeffs.gdt pv
again, better process the results of the loop after the loop is finished
cheers,
sven
8:59 p.m.
Sven Schreiber schrieb:
Hélio Guilherme schrieb:
> nulldata 1000
> confidence_lvl=0.05
> series rejected=0
> i=1
> loop 1000 --progressive --quiet
> genr a = 10*uniform()
> genr y = 1*a + normal()
> alpha = 0.99
> genr y = alpha*y(-1) + normal()
> adf 1 y --nc --quiet
> genr pv = $pvalue
> print pv
>
I suggest to move the print command out of the loop, it will slow things
down.
> genr rejected[i]=(pv>confidence_lvl)?1:0
>
I guess this should be < instead of >?
> i++
>
or loop for i=1..1000
> store coeffs.gdt pv
>
again, better process the results of the loop after the loop is finished
cheers,
sven
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Thank you for the script and
hints, Helio and Sven.
I've got two questions left:
1. What is "?1:0" exactly doing?
2. Why is it not possible to set a sample smaller than the number of
loops any more? If I set nulldata 100 for loop 1000 I get the following
output: "Observation number out of bounds
'[i]': bad observation specifier
> genr rejected[i]=(pv<=confidence_lvl)?1:0"
Since I
want to study the sample properties of the ADF test this is not
quite useful :)
Cheers,
Artur
9:10 p.m.
Artur T. schrieb:
I've got two questions left:
1. What is "?1:0" exactly doing?
=> is the expression before the '?' true? then assign the value before
the colon (1); otherwise assign the value after the colon (0). It's like
condensend if-elif-endif block.
2. Why is it not possible to set a sample smaller than the number of
loops any more? If I set nulldata 100 for loop 1000 I get the following
output: "Observation number out of bounds
'[i]': bad observation specifier
>> genr rejected[i]=(pv<=confidence_lvl)?1:0"
Since I want to study the sample properties of the ADF test this is not
quite useful :)
I'm not sure but maybe it's because of the --progressive loop option. I
would actually remove all dependences on nulldata, samples, series, and
so on and do it with scalars (and in general matrices, but probably not
needed here) only.
something like:
numberofruns = 1000
siglevel = 0.05
counter = 0
loop numberofruns --quiet
.. blahblah...
adf ...blah
pv = $pvalue
if pv<siglevel
counter += 1
endif
...blah
endloop
printf "Got %d rejections out of %d trials\n", counter, numberofruns
good night,
sven
9:18 p.m.
On 02-01-2010, at 20:43, Artur T. wrote:
Hello,
I programmed the following loop and wanted to count how often the null
is rejected for the ADF test. Unfortunately I do not know whether there
is a simple function for this.
loop 1000 --progressive --quiet
genr a = 10*uniform()
genr y = 1*a + normal()
alpha = 0.99
genr y = alpha*y(-1) + normal()
adf 1 y --nc --quiet
genr pv = $pvalue
print pv
store coeffs.gdt pv
endloop
Do you know a solution for this?
How about
<script>
scalar rcount = 0
scalar cfdlev = 0.05
loop 1000 --progressive --quiet
genr a = 10*uniform()
genr y = 1*a + normal()
alpha = 0.99
genr y = alpha*y(-1) + normal()
adf 1 y --nc --quiet
genr pv = $pvalue
if pv <= cfdlev
rcount = rcount + 1
endif
store coeffs.gdt pv
endloop
print rcount
</script>
rcount is the number of rejections
Berend
4:29 a.m.
On Sat, 2 Jan 2010, Artur T. wrote:
I programmed the following loop and wanted to count how often the
null
is rejected for the ADF test. Unfortunately I do not know whether there
is a simple function for this.
loop 1000 --progressive --quiet
genr a = 10*uniform()
genr y = 1*a + normal()
alpha = 0.99
genr y = alpha*y(-1) + normal()
adf 1 y --nc --quiet
genr pv = $pvalue
print pv
store coeffs.gdt pv
endloop
<script>
nulldata 100 # or whatever
scalar alpha = 0.99
scalar level = 0.05 # or whatever
scalar reject = 0
scalar replics = 1000
loop replics --progressive --quiet
genr a = 10*uniform()
genr y = 1*a + normal()
genr y = alpha*y(-1) + normal()
adf 1 y --nc --quiet
genr pv = $pvalue
reject += (pv < level)
print pv
store pvals.gdt pv
endloop
printf "rejections = %d (%.2f%%)\n", reject, 100*reject/replics
</script>
From running this a few times, it looks as if this test is
over-rejecting quite seriously (if I'm interpreting the results
correctly). I'll look into this further.
Allin.
9:35 a.m.
Allin Cottrell schrieb:
On Sat, 2 Jan 2010, Artur T. wrote:
> I programmed the following loop and wanted to count how often the null
> is rejected for the ADF test. Unfortunately I do not know whether there
> is a simple function for this.
>
> loop 1000 --progressive --quiet
> genr a = 10*uniform()
> genr y = 1*a + normal()
> alpha = 0.99
> genr y = alpha*y(-1) + normal()
> adf 1 y --nc --quiet
> genr pv = $pvalue
> print pv
> store coeffs.gdt pv
> endloop
>
<script>
nulldata 100 # or whatever
scalar alpha = 0.99
scalar level = 0.05 # or whatever
scalar reject = 0
scalar replics = 1000
loop replics --progressive --quiet
genr a = 10*uniform()
genr y = 1*a + normal()
genr y = alpha*y(-1) + normal()
adf 1 y --nc --quiet
genr pv = $pvalue
reject += (pv < level)
print pv
store pvals.gdt pv
endloop
printf "rejections = %d (%.2f%%)\n", reject, 100*reject/replics
</script>
>From running this a few times, it looks as if this test is
over-rejecting quite seriously (if I'm interpreting the results
correctly). I'll look into this further.
Allin.
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Thanks for all the versions - they all do their job properly.
@Allin
I was also surprised how often the null is rejected. But actually, as is
known, the result depends heavily on the number of observations and how
close %alpha is to the unit.
Artur
9:58 a.m.
Artur T. schrieb:
Allin Cottrell schrieb:
> On Sat, 2 Jan 2010, Artur T. wrote:
>
>
>> I programmed the following loop and wanted to count how often the null
>> is rejected for the ADF test. Unfortunately I do not know whether there
>> is a simple function for this.
>>
>> loop 1000 --progressive --quiet
>> genr a = 10*uniform()
>> genr y = 1*a + normal()
>> alpha = 0.99
>> genr y = alpha*y(-1) + normal()
>> adf 1 y --nc --quiet
>> genr pv = $pvalue
>> print pv
>> store coeffs.gdt pv
>> endloop
>>
> <script>
> nulldata 100 # or whatever
> scalar alpha = 0.99
> scalar level = 0.05 # or whatever
> scalar reject = 0
> scalar replics = 1000
> loop replics --progressive --quiet
> genr a = 10*uniform()
> genr y = 1*a + normal()
> genr y = alpha*y(-1) + normal()
> adf 1 y --nc --quiet
> genr pv = $pvalue
> reject += (pv < level)
> print pv
> store pvals.gdt pv
> endloop
>
> printf "rejections = %d (%.2f%%)\n", reject, 100*reject/replics
> </script>
>
> >From running this a few times, it looks as if this test is
> over-rejecting quite seriously (if I'm interpreting the results
> correctly). I'll look into this further.
>
> Allin.
> _______________________________________________
> Gretl-users mailing list
> Gretl-users(a)lists.wfu.edu
> http://lists.wfu.edu/mailman/listinfo/gretl-users
>
Thanks for all the versions - they all do their job properly.
@Allin
I was also surprised how often the null is rejected. But actually, as is
known, the result depends heavily on the number of observations and how
close %alpha is to the unit.
What actually are you trying to analyze? Are you aware that the adf test
with --nc is mis-specified, or is this the whole point?
cheers,
sven
11:20 a.m.
Sven Schreiber schrieb:
Artur T. schrieb:
> Allin Cottrell schrieb:
>
>> On Sat, 2 Jan 2010, Artur T. wrote:
>>
>>
>>
>>> I programmed the following loop and wanted to count how often the null
>>> is rejected for the ADF test. Unfortunately I do not know whether there
>>> is a simple function for this.
>>>
>>> loop 1000 --progressive --quiet
>>> genr a = 10*uniform()
>>> genr y = 1*a + normal()
>>> alpha = 0.99
>>> genr y = alpha*y(-1) + normal()
>>> adf 1 y --nc --quiet
>>> genr pv = $pvalue
>>> print pv
>>> store coeffs.gdt pv
>>> endloop
>>>
>>>
>> <script>
>> nulldata 100 # or whatever
>> scalar alpha = 0.99
>> scalar level = 0.05 # or whatever
>> scalar reject = 0
>> scalar replics = 1000
>> loop replics --progressive --quiet
>> genr a = 10*uniform()
>> genr y = 1*a + normal()
>> genr y = alpha*y(-1) + normal()
>> adf 1 y --nc --quiet
>> genr pv = $pvalue
>> reject += (pv < level)
>> print pv
>> store pvals.gdt pv
>> endloop
>>
>> printf "rejections = %d (%.2f%%)\n", reject, 100*reject/replics
>> </script>
>>
>> >From running this a few times, it looks as if this test is
>> over-rejecting quite seriously (if I'm interpreting the results
>> correctly). I'll look into this further.
>>
>> Allin.
>> _______________________________________________
>> Gretl-users mailing list
>> Gretl-users(a)lists.wfu.edu
>> http://lists.wfu.edu/mailman/listinfo/gretl-users
>>
>>
> Thanks for all the versions - they all do their job properly.
>
> @Allin
> I was also surprised how often the null is rejected. But actually, as is
> known, the result depends heavily on the number of observations and how
> close %alpha is to the unit.
>
>
What actually are you trying to analyze? Are you aware that the adf test
with --nc is mis-specified, or is this the whole point?
cheers,
sven
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Yes, actually I wanted to test the validity of the test under
misspecification, sample size and structural breaks. The code was only
one part of the total story.
Artur
9:27 p.m.
Artur T. schrieb:
Sven Schreiber schrieb:
> Artur T. schrieb:
>
>> Allin Cottrell schrieb:
>>
>>>>
>>> <script>
>>> nulldata 100 # or whatever
>>> scalar alpha = 0.99
>>> scalar level = 0.05 # or whatever
>>> scalar reject = 0
>>> scalar replics = 1000
>>> loop replics --progressive --quiet
>>> genr a = 10*uniform()
>>> genr y = 1*a + normal()
>>> genr y = alpha*y(-1) + normal()
>>> adf 1 y --nc --quiet
>>> genr pv = $pvalue
>>> reject += (pv < level)
>>> print pv
>>> store pvals.gdt pv
>>> endloop
>>>
>>> printf "rejections = %d (%.2f%%)\n", reject, 100*reject/replics
>>> </script>
>>>
>>> >From running this a few times, it looks as if this test is
>>> over-rejecting quite seriously (if I'm interpreting the results
>>> correctly). I'll look into this further.
>>>
>>
>> @Allin
>> I was also surprised how often the null is rejected. But actually, as is
>> known, the result depends heavily on the number of observations and how
>> close %alpha is to the unit.
>>
>>
> What actually are you trying to analyze? Are you aware that the adf test
> with --nc is mis-specified, or is this the whole point?
>
>
Yes, actually I wanted to test the validity of the test under
misspecification, sample size and structural breaks. The code was only
one part of the total story.
still it's a strange setup;
AFAICS the lines
>>> genr a = 10*uniform()
>>> genr y = 1*a + normal()
only determine the initial condition of y
at t=0, because you're
overwriting all other values of y in the following line anyway. The
influence of "unusual" initial conditions on unit root tests is known
and there is a literature about that. IMO that would explain the
"overrejection" that you're seeing. BTW, here the DGP does not have a
unit root so we all should be happy that the test is rejecting :-)
cheers,
sven
4:14 p.m.
On 03-01-2010, at 22:27, Sven Schreiber wrote:
still it's a strange setup;
AFAICS the lines
>>>> genr a = 10*uniform()
>>>> genr y = 1*a + normal()
only determine the initial condition of y at t=0, because you're
overwriting all other values of y in the following line anyway. The
influence of "unusual" initial conditions on unit root tests is known
and there is a literature about that. IMO that would explain the
"overrejection" that you're seeing. BTW, here the DGP does not have a
unit root so we all should be happy that the test is rejecting :-)
You beat me to it.
I was getting some doubts about the genr's.
So I decided to initialise y to 0.
The modified script then is
<script>
nulldata 100
# directory of this script and where to store data
string Rundir = "adfcount"
scalar alpha = 0.99
scalar reject = 0
scalar rlevel = 0.05
scalar N = 1000
set stopwatch
loop N --progressive --quiet
series y = 0
genr y = alpha*y(-1) + normal()
adf 1 y --c --quiet
genr pv = $pvalue
reject += (pv<rlevel)
store @Rundir/coeffs.gdt pv
print pv
endloop
time=$stopwatch
printf "Rejections = %d (%.2f%%)\n", reject,100*(reject/N)
printf "CPU seconds used %.2f\n",time
</script>
I've run this several times. The rejection rate still seems to reach 6 or 7 %.
Berend
5:11 p.m.
On Mon, 4 Jan 2010, Berend Hasselman wrote:
You beat me to it.
I was getting some doubts about the genr's.
So I decided to initialise y to 0.
[...]
I've run this several times. The rejection rate still seems to
reach 6 or 7 %.
Slightly modified version:
<script>
set echo off
set messages off
nulldata 100
scalar n = 9
matrix alphas = 0.9 + 0.1/(n+1)*seq(0,n)
scalar rlevel = 0.05
scalar N = 1000
matrix rjs = zeros(1, cols(alphas))
set stopwatch
loop for i=0..n --quiet
scalar reject = 0
alpha = alphas[i+1]
loop N --quiet
series y = 0
genr y = alpha*y(-1) + normal()
adf 1 y --c --quiet
genr pv = $pvalue
reject += (pv<rlevel)
endloop
rjs[i+1] = reject
end loop
time=$stopwatch
results = (alphas | (100/N * rjs))'
colnames(results, "alpha rej_rate")
print results
printf "CPU seconds used %.2f\n",time
</script>
which of course illustrates the well known lack of power of the ADF test:
if you use "adf --gls" things change quite a bit.
Riccardo (Jack) Lucchetti
Dipartimento di Economia
Università Politecnica delle Marche
r.lucchetti(a)univpm.it
http://www.econ.univpm.it/lucchetti
1:13 p.m.
Riccardo (Jack) Lucchetti schrieb:
On Mon, 4 Jan 2010, Berend Hasselman wrote:
> You beat me to it.
> I was getting some doubts about the genr's.
> So I decided to initialise y to 0.
[...]
> I've run this several times. The rejection rate still seems to reach
> 6 or 7 %.
Slightly modified version:
<script>
set echo off
set messages off
nulldata 100
scalar n = 9
matrix alphas = 0.9 + 0.1/(n+1)*seq(0,n)
scalar rlevel = 0.05
scalar N = 1000
matrix rjs = zeros(1, cols(alphas))
set stopwatch
loop for i=0..n --quiet
scalar reject = 0
alpha = alphas[i+1]
loop N --quiet
series y = 0
genr y = alpha*y(-1) + normal()
adf 1 y --c --quiet
genr pv = $pvalue
reject += (pv<rlevel)
endloop
rjs[i+1] = reject
end loop
time=$stopwatch
results = (alphas | (100/N * rjs))'
colnames(results, "alpha rej_rate")
print results
printf "CPU seconds used %.2f\n",time
</script>
which of course illustrates the well known lack of power of the ADF
test: if you use "adf --gls" things change quite a bit.
Riccardo (Jack) Lucchetti
Dipartimento di Economia
Università Politecnica delle Marche
r.lucchetti(a)univpm.it
http://www.econ.univpm.it/lucchetti
------------------------------------------------------------------------
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That's a nice script. I am
surprised that the power of the gls variant
is that higher.
Thanks for the script!
artur
5:06 p.m.
Ricardo, I copied and paste your code and then run it; this message appeared
gretl version 1.8.6
Current session: 2010-01-05 11:04
? <script>
Syntax error in command line
Error executing script: halting
<script>
What could be causing the problem?
Thanks!
--
Charles Koss
http://charlesonnet.blogspot.com/
On Tue, Jan 5, 2010 at 7:13 AM, Artur T. <artur.tarassow(a)googlemail.com> wrote:
> Riccardo (Jack) Lucchetti schrieb:
>> On Mon, 4 Jan 2010, Berend Hasselman wrote:
>>
>>> You beat me to it.
>>> I was getting some doubts about the genr's.
>>> So I decided to initialise y to 0.
>> [...]
>>> I've run this several times. The rejection rate still seems to reach
>>> 6 or 7 %.
>>
>> Slightly modified version:
> <script>
>> set echo off
>> set messages off
>> nulldata 100
>>
>> scalar n = 9
>> matrix alphas = 0.9 + 0.1/(n+1)*seq(0,n)
>> scalar rlevel = 0.05
>> scalar N = 1000
>> matrix rjs = zeros(1, cols(alphas))
>> set stopwatch
>>
>> loop for i=0..n --quiet
>> scalar reject = 0
>> alpha = alphas[i+1]
>> loop N --quiet
>> series y = 0
>> genr y = alpha*y(-1) + normal()
>> adf 1 y --c --quiet
>> genr pv = $pvalue
>> reject += (pv<rlevel)
>> endloop
>> rjs[i+1] = reject
>> end loop
>>
>> time=$stopwatch
>>
>> results = (alphas | (100/N * rjs))'
>> colnames(results, "alpha rej_rate")
>> print results
>> printf "CPU seconds used %.2f\n",time
>> </script>
>>
>> which of course illustrates the well known lack of power of the ADF
>> test: if you use "adf --gls" things change quite a bit.
>>
>> Riccardo (Jack) Lucchetti
>> Dipartimento di Economia
>> Università Politecnica delle Marche
>>
>> r.lucchetti(a)univpm.it
>> http://www.econ.univpm.it/lucchetti
>> ------------------------------------------------------------------------
>>
>> _______________________________________________
>> Gretl-users mailing list
>> Gretl-users(a)lists.wfu.edu
>> http://lists.wfu.edu/mailman/listinfo/gretl-users
> That's a nice script. I am surprised that the power of the gls variant
> is that higher.
>
> Thanks for the script!
>
> artur
>
> _______________________________________________
> Gretl-users mailing list
> Gretl-users(a)lists.wfu.edu
> http://lists.wfu.edu/mailman/listinfo/gretl-users
>
5:18 p.m.
Em 5 de janeiro de 2010 Charles Koss <hqtiger(a)gmail.com> escreveu:
Ricardo, I copied and paste your code and then run it; this message
appeared
gretl version 1.8.6
Current session: 2010-01-05 11:04
? <script>
Syntax error in command line
Error executing script: halting
> <script>
What could be causing the problem?
Charles, try to run Riccardo's script without the lines <script> and
</script>.
Best,
Henrique
5:38 p.m.
Enrique, it worked as you said!!! Thanks
--
Charles Koss
http://charlesonnet.blogspot.com/
On Tue, Jan 5, 2010 at 11:18 AM, Henrique <henrique.coelho(a)gmail.com> wrote:
Em 5 de janeiro de 2010 Charles Koss <hqtiger(a)gmail.com>
escreveu:
> Ricardo, I copied and paste your code and then run it; this message appeared
>
> gretl version 1.8.6
> Current session: 2010-01-05 11:04
> ? <script>
> Syntax error in command line
>
> Error executing script: halting
>> <script>
>
> What could be causing the problem?
Charles, try to run Riccardo's script without the lines <script> and
</script>.
Best,
Henrique
_______________________________________________
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Gretl-users(a)lists.wfu.edu
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5:38 p.m.
On Tue, 5 Jan 2010, Charles Koss wrote:
Ricardo, I copied and paste your code and then run it; this message
appeared
gretl version 1.8.6
Current session: 2010-01-05 11:04
? <script>
Syntax error in command line
Error executing script: halting
> <script>
What could be causing the problem?
Thanks!
Nothing serious: the <script> and </script> lines are just used to mark
the beginning and the end of the script in the mail message. Just delete
those two lines and you should be ok.
Riccardo (Jack) Lucchetti
Dipartimento di Economia
Università Politecnica delle Marche
r.lucchetti(a)univpm.it
http://www.econ.univpm.it/lucchetti
5:41 p.m.
Thank you Ricardo, it worked. Now, I am learning from the script!!!
Regards,
--
Charles Koss
http://charlesonnet.blogspot.com/
On Tue, Jan 5, 2010 at 11:38 AM, Riccardo (Jack) Lucchetti
<r.lucchetti(a)univpm.it> wrote:
On Tue, 5 Jan 2010, Charles Koss wrote:
> Ricardo, I copied and paste your code and then run it; this message
> appeared
>
> gretl version 1.8.6
> Current session: 2010-01-05 11:04
> ? <script>
> Syntax error in command line
>
> Error executing script: halting
>>
>> <script>
>
> What could be causing the problem?
>
> Thanks!
Nothing serious: the <script> and </script> lines are just used to mark the
beginning and the end of the script in the mail message. Just delete those
two lines and you should be ok.
Riccardo (Jack) Lucchetti
Dipartimento di Economia
Università Politecnica delle Marche
r.lucchetti(a)univpm.it
http://www.econ.univpm.it/lucchetti
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participants (8)
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Allin Cottrell -
Artur T. -
Berend Hasselman -
Charles Koss -
Henrique -
Hélio Guilherme -
Riccardo (Jack) Lucchetti -
Sven Schreiber