# Cointegration

So now the fun starts. We have established the integration order of the variables in the RadF file, we impose the rule that only variables of the same order can be combined, and in particular that they cannot be cointegrated with temperature which is I(1). In this case all the anthropogenic variables in RadF are I(2) — W-M_GHGs, O3, StratH2O, LandUse, SnowAlb, BC, ReflAer, AIE — while Solar and StratAer are I(1) or I(0).

Adding these AGW variables together would be one way, as they are all I(2) and are in the same units of Watts/m2. However, the result is still I(2), as it takes two differencings before the ADF test rejects.

[1,] 0 0.9662216 -1.088566 0.9208762
[2,] 1 0.5309850 -2.192266 0.4966766
[3,] 2 -0.4976567 -4.875666 0.0100000

What we can do is see if the variables cointegrate. To do this, we fit a linear regression model to the variables and test the residuals for integration order. If the residual is a lower order I(1) then the variables are all related and can be treated as a common trend. It turns out the residuals are I(1).

[1,] 0 0.9988194 -3.087874 0.1245392
[2,] 1 0.4176963 -4.468787 0.0100000
[3,] 2 -0.4492167 -6.912500 0.0100000

Some of the variables are not significant however, so we restrict the cointegration vector to the following:

W-M_GHGs = -4.51*O3 + 54.56*StratH2O + 11.21*LandUse – 0.35*AIE, R2=0.9984

The plot of the residuals of the full (black) and restricted (red) model is below. Interestingly, htere is a clear 20 year cycle not apparent in the original data.

Finally we have the information we need to develop our full model. We have Temperature and Solar as I(1), and we will leave out StratAer (volcanics) as it turns out to be non-significant. The sum of the I(2) variables is AGW=I(2) so we use the first difference (or rate of change) deltaAGW as a variable which will be I(1). We also use the residuals of the cointegration g=I(1).

The resulting linear model turns out to have marginal cointegration on the ADF test (as also found by Beenstock) but the PP test clearly rejects, so the result is I(0) and the variables cointegrate.

[1,] 0 0.7316112 -3.132276 0.1060901
[2,] 1 -0.1910926 -6.674352 0.0100000
[3,] 2 -0.9669920 -8.286269 0.0100000

Phillips-Perron Unit Root Test
data: l\$residuals
Dickey-Fuller Z(alpha) = -33.0459, Truncation lag parameter = 4, p-value = 0.01

T = -0.36007 + Solar*1.22403 + deltaAGW*5.31922 + g*1.12901, R2=0.629

Below I plot the temperature (black), and the prediction from the model (red).

The plot also explores the impact of different levels of CO2. The upper (green) is what would have happened to temperature if CO2 had increased at twice the rate. The lower (blue) is the result if CO2 had been zero throughout.

So far the reproduction of the Beenstock analysis on an alternate data set resembles his analysis, suggesting some robustness to the result. The message of this analysis is that higher emissions of CO2 make no difference to the eventual temperature levels, although they change faster, arriving at the equilibrium levels sooner.

## 0 thoughts on “Cointegration”

1. dewittpayne says:

I still don't buy it. It's far more likely that there's something biasing the test rather than some magical mechanism that somehow balances the forcing due to ghg's in less than a year when the system obviously has time constants longer than that. Ocean heat content, in spite of the current pause, has been going up for as long as it's been measured. That's a direct measure of the radiative imbalance. In fact, that may be the problem. The forcing going into OHC is only reflected in the temperature over the span of decades to centuries.

2. davids99us says:

You could test your idea by seeing if the OHC is integratingtemperature: eg is OHC I(2), and maybe sea level too, as good datagoes back to before 1880.

3. Anonymous says:

I still don’t buy it. It’s far more likely that there’s something biasing the test rather than some magical mechanism that somehow balances the forcing due to ghg’s in less than a year when the system obviously has time constants longer than that. Ocean heat content, in spite of the current pause, has been going up for as long as it’s been measured. That’s a direct measure of the radiative imbalance. In fact, that may be the problem. The forcing going into OHC is only reflected in the temperature over the span of decades to centuries.

• Anonymous says:

You could test your idea by seeing if the OHC is integrating
temperature: eg is OHC I(2), and maybe sea level too, as good data
goes back to before 1880.

• Anonymous says:

I think this is the first major objection to the paper. There is a ‘closed system’ assumption. You need to include all the major sinks. You can’t support the conclusions when a major sink like the ocean is left out.

4. davids99us says:

I think this is the first major objection to the paper. You need to include all the major sinks. You can't support the conclusions when a major sink like the ocean is left out.

5. dewittpayne says:

I think I may have found the real problem. The ghg column in RADF.txt may not be real data, or it may have been smoothed or otherwise manipulated. I found the OHC data and it was I(1). But it was only from 1955-2009. So I thought maybe if I restrict the ghg forcing to the last 55 points, that might make a difference. No, it's still I(2). So what about real data? I took the annual Muana Loa CO2 data from 1958-2009 and converted it to forcing using 5.3*ln(CO2(t)/CO2(to)). That's I(1). The p value for the first difference of the data is 0.01874. That makes me suspicious of all the data in that file and it means determining what data B&R actually used is critical.

6. davids99us says:

Oh I don't really trust the RadF data much, its just convenient. AndBeenstock's reporting of the data used is weak too. Checking whetherCO2 is really I(2) is a study in itself.

7. Anonymous says:

I think I may have found the real problem. The ghg column in RADF.txt may not be real data, or it may have been smoothed or otherwise manipulated. I found the OHC data and it was I(1). But it was only from 1955-2009. So I thought maybe if I restrict the ghg forcing to the last 55 points, that might make a difference. No, it’s still I(2). So what about real data? I took the annual Muana Loa CO2 data from 1958-2009 and converted it to forcing using 5.3*ln(CO2(t)/CO2(to)). That’s I(1). The p value for the first difference of the data is 0.01874. That makes me suspicious of all the data in that file and it means determining what data B&R actually used is critical.

• Anonymous says:

Oh I don’t really trust the RadF data much, its just convenient. And
Beenstock’s reporting of the data used is weak too. Checking whether
CO2 is really I(2) is a study in itself.

8. David,I may sound like a broken record, but this is not statistics at all. You can just keep making unqualified statements that this vble is I(1), This is I(2) etc. You have a finite set of data with noise. You have to be able to quantify your level of confidence in those statements if you want to make inferences. So far neither you nor B&R seem to have a clue how to do that.

9. davids99us says:

Sorry Nick, I keep getting distracted on other things. So tell me whyI(n) isnt rejection of the unit root (ie stationary) after ndifferences with certain level of confidence, and how that is not aquantification of confidence. If you think you need a different test,what?

10. Firstly, you need to state the number. And if you want to use the ADF rejection statistic, you need to say why it allows you to use that number for some other proposition. And exactly what is the proposition that your quantified confidence applies to.The cointegration argument says that something couldn't happen because A is I(1) and B is I(2). OK, how confident are you that A is I(1) and couldn't be I(2). And conversely for B. This is statistics – you have to be able to quote those figures. Your statements are meaningless without them.

11. David,
I may sound like a broken record, but this is not statistics at all. You can just keep making unqualified statements that this vble is I(1), This is I(2) etc. You have a finite set of data with noise. You have to be able to quantify your level of confidence in those statements if you want to make inferences. So far neither you nor B&R seem to have a clue how to do that.

• Anonymous says:

Sorry Nick, I keep getting distracted on other things. So tell me why
I(n) isnt rejection of the unit root (ie stationary) after n
differences with certain level of confidence, and how that is not a
quantification of confidence. If you think you need a different test,
what?

• Firstly, you need to state the number. And if you want to use the ADF rejection statistic, you need to say why it allows you to use that number for some other proposition. And exactly what is the proposition that your quantified confidence applies to.

The cointegration argument says that something couldn’t happen because A is I(1) and B is I(2). OK, how confident are you that A is I(1) and couldn’t be I(2). And conversely for B. This is statistics – you have to be able to quote those figures. Your statements are meaningless without them.

• Anonymous says:

You wont get an argument from me that the null hypothesis should be
stated precisely. I could say that the probability of falsely
accepting the alternate hypothesis that the series is a level or trend
stationary first-order autoregressive (AR(1)) model plus additive
constant is less than 0.05. But I am trying to write for the layman.

• Anonymous says:

Nick, I understand that you are saying that the statement S=I(n) is TRUE is a composite of statements, something like the (n-1)th difference is not stationary and the nth difference is stationary. I suppose I think it would not make much difference, except to reduce the certainty, and if that was a concern then the increase the confidence in some way. Beenstock uses 4 tests.

“Test 1 shows that according to all three test statistics rfCO2 is not trend stationary. Test 2 shows that according to the PP statistic rfCO2 is marginally difference stationary, but the KPSS and ADF statistics clearly reject this hypothesis. Test 3 establishes that the 2nd difference of rfCO2 is stationary according to all three test statistics. Finally, a direct estimate of d using the fractional unit root method gives a value of 1.56 to d which provides independent evidence that rfCO2 is I(2).”

Now I suppose you would say that he did not rigorously determine the confidence of a composite hypothesis, and that’s fair enough.

• Yes indeed. But more to the point, he said nothing about the confidence attached to any statement about temperature. Yet AGW is refuted!!! 100%!

• Anonymous says:

Maybe he’s going for the cash prize.

• Anonymous says:

Nick,
For data with any noise at all, the best you can say is that the integration order is at least whatever level that the test fails to reject. You can’t say that it’s less than some value if it does reject because the test is very sensitive to the noise level. So the finding that null hypothesis is rejected for the second difference in temperature does not mean that temperature is not I(2). A pure quadratic function with no noise does have a unit root for the second difference. But start adding noise and the ADF test will first fail to find a unit root for the second difference, then the first difference and finally for original series.

• fail to find a unit root.
I think you mean fail to reject a unit root.

The way I think of it is this. You model the series as something for which a linear combination
(a0 + a1*D +a2*D^2+…)y_t=ε_t
RHS has mean zero etc. D is the forward difference operator.
Then saying it is I(1) means a0 is zero; I(2) means a1 is also zero. Rejecting the possibility of a unit root at I(1) level means saying that a0 is significantly different from zero.

That’s my complaint about these assertions that, say, temperature is I(1). All that means is that you have failed to prove that a0 is nonzero. It’s a weak assertion. And if you have failed to prove it at 95% confidence, that doesn’t mean that you are 95% confident of anything.

• Anonymous says:

No, we’re both wrong, it will reject a unit root, not fail to reject or fail to find. But I think we agree that rejecting a unit root for the second difference is weak. Failing to reject tells you something, rejecting tells you much less. In your equation, if only one coefficient is equal to 1 and all the rest are zero, the tests will always work correctly. But I don’t think the data actually look like that. The measurement process adds noise and I doubt that the various quasi-periodic oscillations like ENSO are correctly modeled by your equation. As I remember, those things are approximately AR(1) so maybe that’s all the tests see for temperature. I need to check on what applying a moving average does to the test results. A three point average didn’t make stationary white noise non-stationary. The KPSS test seems to have some advantages over ADF, although it looks like the null hypothesis for the KPSS test is that there isn’t a unit root. So for that test you keep differencing until it fails to reject.

• it will reject a unit root, not fail to reject or fail to find.
It’s a test, so it has to have alternative possible outcomes (else it’s not much of a test). The alternative is fail to reject.

oscillations like ENSO are correctly modeled by your equation
Well, it’s not really my equation. But actually it does embrace oscillations – think of it as a general autonomous homogeneous linear ODE with constant coefficients, Nonzero a0, a2=1, all else 0 gives oscillation (SHM).

I’ve put up a blog post on this stuff.

12. davids99us says:

You wont get an argument from me that the null hypothesis should bestated precisely. I could say that the probability of falselyaccepting the alternate hypothesis that the series is a level or trendstationary first-order autoregressive (AR(1)) model plus additiveconstant is less than 0.05. But I am trying to write for the layman.

13. davids99us says:

Nick, I understand that you are saying that the statement S=I(n) is TRUE is a composite of statements, something like the (n-1)th difference is not stationary and the nth difference is stationary. I suppose I think it would not make much difference, except to reduce the certainty, and if that was a concern then the increase the confidence in some way. Beenstock uses 4 tests.”Test 1 shows that according to all three test statistics rfCO2 is not trend stationary. Test 2 shows that according to the PP statistic rfCO2 is marginally difference stationary, but the KPSS and ADF statistics clearly reject this hypothesis. Test 3 establishes that the 2nd difference of rfCO2 is stationary according to all three test statistics. Finally, a direct estimate of d using the fractional unit root method gives a value of 1.56 to d which provides independent evidence that rfCO2 is I(2).”Now I suppose you would say that he did not rigorously determine the confidence of a composite hypothesis, and that's fair enough.

14. Yes indeed. But more to the point, he said nothing about the confidence attached to any statement about temperature. Yet AGW is refuted!!! 100%!

15. davids99us says:

I suppose that's why its still a draft.

16. dewittpayne says:

Nick,For data with any noise at all, the best you can say is that the integration order is at least whatever level that the test fails to reject. You can't say that it's less than some value if it does reject because the test is very sensitive to the noise level. So the finding that null hypothesis is rejected for the second difference in temperature does not mean that temperature is not I(2). A pure quadratic function with no noise does have a unit root for the second difference. But start adding noise and the ADF test will first fail to find a unit root for the second difference, then the first difference and finally for original series.

17. fail to find a unit root.I think you mean fail to reject a unit root. The way I think of it is this. You model the series as something for which a linear combination (a0 + a1*D +a2*D^2+…)y_t=Îµ_tRHS has mean zero etc. D is the forward difference operator.Then saying it is I(1) means a0 is zero; I(2) means a1 is also zero. Rejecting the possibility of a unit root at I(1) level means saying that a0 is significantly different from zero. That's my complaint about these assertions that, say, temperature is I(1). All that means is that you have failed to prove that a0 is nonzero. It's a weak assertion. And if you have failed to prove it at 95% confidence, that doesn't mean that you are 95% confident of anything.

18. dewittpayne says:

No, we're both wrong, it will reject a unit root, not fail to reject or fail to find. But I think we agree that rejecting a unit root for the second difference is weak. Failing to reject tells you something, rejecting tells you much less. In your equation, if only one coefficient is equal to 1 and all the rest are zero, the tests will always work correctly. But I don't think the data actually look like that. The measurement process adds noise and I doubt that the various quasi-periodic oscillations like ENSO are correctly modeled by your equation. As I remember, those things are approximately AR(1) so maybe that's all the tests see for temperature. I need to check on what applying a moving average does to the test results. A three point average didn't make stationary white noise non-stationary. The KPSS test seems to have some advantages over ADF, although it looks like the null hypothesis for the KPSS test is that there isn't a unit root. So for that test you keep differencing until it fails to reject.

19. it will reject a unit root, not fail to reject or fail to find.It's a test, so it has to have alternative possible outcomes (else it's not much of a test). The alternative is fail to reject.oscillations like ENSO are correctly modeled by your equationWell, it's not really my equation. But actually it does embrace oscillations – think of it as a general autonomous homogeneous linear ODE with constant coefficients, Nonzero a0, a2=1, all else 0 gives oscillation (SHM).I've put up a blog post on this stuff.

20. dewittpayne says:

No, we're both wrong, it will reject a unit root, not fail to reject or fail to find. But I think we agree that rejecting a unit root for the second difference is weak. Failing to reject tells you something, rejecting tells you much less. In your equation, if only one coefficient is equal to 1 and all the rest are zero, the tests will always work correctly. But I don't think the data actually look like that. The measurement process adds noise and I doubt that the various quasi-periodic oscillations like ENSO are correctly modeled by your equation. As I remember, those things are approximately AR(1) so maybe that's all the tests see for temperature. I need to check on what applying a moving average does to the test results. A three point average didn't make stationary white noise non-stationary. The KPSS test seems to have some advantages over ADF, although it looks like the null hypothesis for the KPSS test is that there isn't a unit root. So for that test you keep differencing until it fails to reject.

21. it will reject a unit root, not fail to reject or fail to find.It's a test, so it has to have alternative possible outcomes (else it's not much of a test). The alternative is fail to reject.oscillations like ENSO are correctly modeled by your equationWell, it's not really my equation. But actually it does embrace oscillations – think of it as a general autonomous homogeneous linear ODE with constant coefficients, Nonzero a0, a2=1, all else 0 gives oscillation (SHM).I've put up a blog post on this stuff.

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