Here we generate and test a number of series with different integration order I(n) and polynomial order O(n). The test is the Augmented Dickey Fuller test, one of the most well known of the unit root tests. Beenstock used three tests, because the tests for unit roots are known to have low power.

The six series shown above are as follows (black, red, green):

In the first group is a random series I(0), the integrated random series I(1), and the twice integrated random series I(2).

The second group is a linear trend plus noise O(1), a quadratic trend plus noise O(2), and a cubic trend plus noise O(3).

The way I diagnose order is like the code below returning the number of differences before the adf.test rejects (adf.test is in R package tseries).

for (n in 0:2)

{

d<-adf.test(x,k=2)$p.value

if (d<0.05) return n

x<-diff(x)

}

The result for the 6 series: [1] 0 1 2 0 0 1

The first 5 were guessed right, but the cubic series is a false positive. Note that even when I use a highly curved function such as the sixth power or exponential I still get I(1).

This reaction to curvature is I guess the reason that Beenstock examine the rfCO2 for breaks. They found that rfCO2 was I(2), but also showed signs of being I(1) with a break. On the basis of another test they reject that possibility.

Next, some real data.

### Like this:

Like Loading...

David, the problem with this test, and with B&R, is there is no quantification of the strength of the conclusion. You eventually want to say – these two series have incompatible I(n) status so AGW is falsified. But you need to answer, if they have compatible status, what is the probability that what you observed arose by chance? You'd want to show that that is small.I just don't see that here. You could have two almost identical series, one just squeaking through the ADF test for I(1), one not. So you'd say they were incompatible?

Nick, I have no doubt such a test is possible. I am not too sure thatis not contained in the rejection levels as stated. I am havingenough trouble understanding the implications of the existing work asit is.This idea if incompatability by order of integration is prettyprofound, and novel in system identification AFAIK. It would seem tome that it is also indicative of causal direction: ie from low tohigh. Perhaps they do not pay much attention to causation ineconomics. Anyway, if order incompatability is so important, then thetest should be more watertight. But I haven't read all theliterature.I read some of the earlier Kaufmann papers and found them fairly opaque.

I am not too sure that is not contained in the rejection levels as stated.I am sure. Your adf test isif (d<0.05) return nSo if with n=1, Temp returns 0.0499 and CO2 returns 0.0501, you'd say AGW is refuted?I mean the way Beenstock does it, not me.

David, the problem with this test, and with B&R, is there is no quantification of the strength of the conclusion. You eventually want to say – these two series have incompatible I(n) status so AGW is falsified. But you need to answer, if they have compatible status, what is the probability that what you observed arose by chance? You’d want to show that that is small.

I just don’t see that here. You could have two almost identical series, one just squeaking through the ADF test for I(1), one not. So you’d say they were incompatible?

Nick, I have no doubt such a test is possible. I am not too sure that

is not contained in the rejection levels as stated. I am having

enough trouble understanding the implications of the existing work as

it is.

This idea if incompatability by order of integration is pretty

profound, and novel in system identification AFAIK. It would seem to

me that it is also indicative of causal direction: ie from low to

high. Perhaps they do not pay much attention to causation in

economics. Anyway, if order incompatability is so important, then the

test should be more watertight. But I haven’t read all the

literature.

I read some of the earlier Kaufmann papers and found them fairly opaque.

I am not too sure that is not contained in the rejection levels as stated.I am sure. Your adf test is

if (d<0.05) return nSo if with n=1, Temp returns 0.0499 and CO2 returns 0.0501, you’d say AGW is refuted?

I mean the way Beenstock does it, not me.

Nick, If 5% of Temp is wrongly diagnosed, and 5% of CO2, the 95% of

each are correctly diagnostd. The probability of getting them both

right is 95×95 or 90%.If you are worried about a 10% error rate, just

test them at a higher level. 99% will give a probability of being

right of 98%.

If 5% of Temp is wrongly diagnosed, and 5% of CO2,No, that’s not what the result says. One is a rejection at 5%, the other is a non-rejection. If you go to 99% they will both pass (I(2) on your test), at 90% they are both I(1).

That’s what’s lacking – a quantitative measure of incompatibility.

Both have 5% rejections, just at different levels.

The I(2) level test is irrelevant, as it only determines whether CO2 is I(2) or I(3) – both would be incompatible. It’s the n=1 test that decides the compatibility issue.

Nick, If 5% of Temp is wrongly diagnosed, and 5% of CO2, the 95% ofeach are correctly diagnostd. The probability of getting them bothright is 95×95 or 90%.If you are worried about a 10% error rate, justtest them at a higher level. 99% will give a probability of beingright of 98%.

If 5% of Temp is wrongly diagnosed, and 5% of CO2,No, that's not what the result says. One is a rejection at 5%, the other is a non-rejection. If you go to 99% they will both pass (I(2) on your test), at 90% they are both I(1).That's what's lacking – a quantitative measure of incompatibility.Both have 5% rejections, just at different levels.

The I(2) level test is irrelevant, as it only determines whether CO2 is I(2) or I(3) – both would be incompatible. It's the n=1 test that decides the compatibility issue.

Your noise to signal is too high in your polynomial examples. I created a series where y=0.01x^2 +Z(x) where x=1 to 100 and a vector Z created by rnorm(100) with the default mean of 0 and an s.d.of 1. That series tests I(1) by adf.test. The p value for the original series is 0.99 and less than 0.01 for the first difference. OTOH, if the series is 0.001x^2 with the same range for Z, the p value for the test of the original series is 0.14. For the series y=0.0001x^2 plus the same Z, the adf.test p value is less than 0.01. Yet more evidence that the integration order test used by B&R is showing the difference in signal to noise for the temperature anomaly compared to rfCO2, not the real integration order, whatever it is. I'll try to write all this up. How do I send it and in what form do you want it? I can do Word with either separate or embedded figures or both.

However suits you. You should check you are doing trend stationary tests and run Phillips-Perron (and KPSS) tests too.

Your noise to signal is too high in your polynomial examples. I created a series where y=0.01x^2 +Z(x) where x=1 to 100 and a vector Z created by rnorm(100) with the default mean of 0 and an s.d.of 1. That series tests I(1) by adf.test. The p value for the original series is 0.99 and less than 0.01 for the first difference. OTOH, if the series is 0.001x^2 with the same range for Z, the p value for the test of the original series is 0.14. For the series y=0.0001x^2 plus the same Z, the adf.test p value is less than 0.01. Yet more evidence that the integration order test used by B&R is showing the difference in signal to noise for the temperature anomaly compared to rfCO2, not the real integration order, whatever it is. I’ll try to write all this up. How do I send it and in what form do you want it? I can do Word with either separate or embedded figures or both.

However suits you. You should check you are doing trend stationary tests and run Phillips-Perron (and KPSS) tests too.

Your noise to signal is too high in your polynomial examples. I created a series where y=0.01x^2 +Z(x) where x=1 to 100 and a vector Z created by rnorm(100) with the default mean of 0 and an s.d.of 1. That series tests I(1) by adf.test. The p value for the original series is 0.99 and less than 0.01 for the first difference. OTOH, if the series is 0.001x^2 with the same range for Z, the p value for the test of the original series is 0.14. For the series y=0.0001x^2 plus the same Z, the adf.test p value is less than 0.01. Yet more evidence that the integration order test used by B&R is showing the difference in signal to noise for the temperature anomaly compared to rfCO2, not the real integration order, whatever it is. I'll try to write all this up. How do I send it and in what form do you want it? I can do Word with either separate or embedded figures or both.

However suits you. You should check you are doing trend stationary tests and run Phillips-Perron (and KPSS) tests too.

Pingback: cat memes

Pingback: Poemas de Octavio Paz

Pingback: hearthstone accounts

Pingback: bal sylwestrowy

Pingback: link do strony

Pingback: zobacz oferte

Pingback: pomoc pielegniarska wroclaw

Pingback: oferta

Pingback: oferta

Pingback: pity2015program.pl

Pingback: zobacz