Correction of spurious trends in climate series caused by inhomogeneities Ralf Lindau.
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Transcript of Correction of spurious trends in climate series caused by inhomogeneities Ralf Lindau.
Correction of spurious trends
in climate series caused by inhomogeneities
Ralf Lindau
Dipdoc Seminar – 12. November 2015
Break detectionConsider the differences of one station
compared to a neighbor reference from a surrounding network of stations.
The dominating natural variance is cancelled out, because it is very similar at both stations.
Breaks become visible by abrupt changes in the station-reference time series.
Internal variance (Noise)within the subperiods
External variance (Signal)between the means of different
subperiods
Break criterion:Maximum external (explained) variance
Break-aware ideaBreaks are only detected, but not corrected.
Calculate the mean trend over all homogeneous subperiods (omitting the known breakpoints).
This trend should reflect the true trend.
Dipdoc Seminar – 12. November 2015
𝑡 𝑟𝑒𝑛𝑑=𝑁 2−1𝑁 2 𝑡𝑒𝑥𝑡+
1𝑁2 𝑡 𝑖𝑛𝑡Is not promising, because:
Correction is important
Known break positions alone are not sufficient to derive reliable trends.
The correction part of homogenization determines a fraction of of the trend, when N is the number of subsegments.
Even for only one break this is equal to ¾.
Dipdoc Seminar – 12. November 2015
Station or network trend?
From my point of view, the network-mean (regionally averaged) trend is per se more interesting.
Moreover, trend corrections for individual stations are easy to derive, if the network-mean trend is known.
Therefore, we concentrate on corrections of the network-mean trend.
Dipdoc Seminar – 12. November 2015
Network-mean trend error
Dipdoc Seminar – 12. November 2015
𝑇 𝑖0=𝐶+𝐼𝑖 0Observed = True + Spurious station trend
1𝑛∑𝑖=1
𝑛
(𝑇 𝑖 0−𝑇 𝑖 )=1𝑛∑𝑖=1
𝑛
(𝐶+𝐼 𝑖0−𝐶− 𝐼 𝑖 )=𝐼 𝑖0−1𝑛∑𝑖=1
𝑛
𝐼 𝑖
Observed trend anomaly = Spurious station trend against the network-mean – network-mean trend errortrend
Two ways of correction
1. Break-by-break method:Consider an individual break of one candidate station.
Compare the two homogeneous subperiods before and after this break with homogeneous subperiods of suited
neighbor stations, which preferably long overlap periods with the candidate break.
2. ANOVA method:Minimize the variance of the entire network.Discussed in the following, because it is better defined.
Dipdoc Seminar – 16. June 2014
ANOVA correction scheme (1/3)
Observation b (station, year)Climate signal c (year)Inhomogeneity a (station, year)Noise e (station, year)
Minimize the squared difference between theory and observations.
Derivation with respect to c(j) leads to m equations, one for each c(j).
Dipdoc Seminar – 12. November 2015
𝑏 (𝑖 , 𝑗 )=𝑐 ( 𝑗 )+𝑎 (𝑖 , 𝑗 )+𝜀 (𝑖 , 𝑗 )
∑𝑖=1
𝑛
∑𝑗=1
𝑚
(𝑐 ( 𝑗 )+𝑎 (𝑖 , 𝑗 )−𝑏 (𝑖 , 𝑗 ) )2=min
𝑐 ( 𝑗 )= 1𝑛∑𝑖=1𝑛
(𝑏 (𝑖 , 𝑗 )−𝑎 (𝑖 , 𝑗 ) )=0
ANOVA correction scheme (2/3)
Analogously, we get h equations, one for each of the homogeneous sub-periods.
Altogether we have m+h equations for m+h unknowns.
Thus a (m+h) x (m+h) matrix has to be solved.E.g. 115x115-matrix for 100 years and 15 subperiods.
We can insert the m climate equations for each year into the h equations for each subperiod, so that only a hxh-matrix has to solved.
Dipdoc Seminar – 12. November 2015
ANOVA correction scheme (3/3)
n = 5 stations with data from 100 years.h = 15 homogeneous sub-periods in total.
Consider sub-period no. 8:
It has an overlap of 32 years with a2,of 13 years with a4,of 19 years with a5, etc.
The overlap periods constitutes the non-diagonal matrix elements.
The diagonal is given by (n-1) length(ai)
Dipdoc Seminar – 12. November 2015
Simulated data
Test the performance of the ANOVA correction scheme with simulated data.
The simulated data consist of three superimposed signals:
1. The climate signal, which identical for all stations of a network.
2. Noise, which mimics the difference between the stations, e.g. due to weather.
3. Inhomogeneities inserted at random timings and with random strengths.
Dipdoc Seminar – 12. November 2015
Test under perfect conditions
1000 networks10 stations with 5 breaks each
No mean trend errorNo noiseKnown break positions
Result:
Perfect skill.- We made no programming errors.- The method works perfectly.
Dipdoc Seminar – 19. November 2015
Inserted yearly station inhomogeneity
Det
ecte
d ye
arly
sta
tion
inho
mog
enei
ty
Signal / noise = 1
Still: No mean trend errorPerfectly known break positions
Equal break and noise variance:SNR = 1
Result:
As expected no longer perfect, but r = 0.955
Dipdoc Seminar – 19. November 2015
Inserted yearly station inhomogeneity
Det
ecte
d ye
arly
sta
tion
inho
mog
enei
ty
Network-mean trends
From individual inhomogeneities to network-mean trends.Both regressions are calculated.That taking the x-axis data as independent is in all three cases equal to the 1-to-1 line.
What does this mean?
Dipdoc Seminar – 19. November 2015
Inhomogeneities Station trends Network trends
r = 0.9550 r = 0.9525 r = 0.8092
What does it mean?
It means that:
y is equal to x plus random scatter e.
Because in this case, the following equation chain holds true:
Dipdoc Seminar – 12. November 2015
Remaining trend error
It is convenient to display not the detected (y), but the remaining (y-x) trend error.
As shown the inserted and the remaining quantities are uncorrelated.
The remaining errors are smaller, but comparable in size.
This is valid for SNR = 1
Dipdoc Seminar – 12. November 2015
Inserted network-mean trend error
Rem
aini
ng n
etw
ork-
mea
n tr
end
erro
r
sx2 = 0.265
sy2 = 0.141
Preliminary conclusion I
For perfectly known breakpoints, time series lengths of 100 with 5 breaks, SNR = 1, and no mean trend introduced by the breaks the situation is:
The result after homogenization is slightly better than the original.
The mean trend is zero, the uncertainty before is: 0.265and afterwards: 0.141
Ratio q: (Improvement)
Dipdoc Seminar – 12. November 2015
Improvement ratio (1/2)Ratio q depends on the SNR.
Upper panel:Doubled noise (SNR = ½ ) leads to doubled remaining trend error. The inserted trend error is unchanged.Doubled q
Lower panel:Doubled signal (SNR = 2) leads to doubled inserted trend error. The remaining trend error is unchanged.Halved q
Inserted and remaining errors are independent.The inserted error is determined by the break (signal) variance.The remaining error is determined by the noise variance.
Dipdoc Seminar – 12. November 2015
Inserted
sx2 = 0.265
sy2 = 0.563
Rem
aini
ng
Inserted
Rem
aini
ng
SNR = ½
sx2 = 1.060
sy2 = 0.141
SNR = 2
q = 1.46 (0.73)
q = 0.365 (0.73)
Improvement ratio (2/2)
Further parameters (besides SNR) that may also affect q are: break and station number.
It shows:The ratio does mainly depend on break number and not on station number.
For 6 breaks the ratio is about 1.
Dipdoc Seminar – 12. November 2015
Break number
Stat
ion
num
ber
0.5 1.0 1.5
Preliminary conclusion II Does the correction act neutrally if no correction is necessary?
Yes, depending on SNR,for SNR=1, break number=6, length=100 the data is neither upgraded nor downgraded.
But,the obtained homogenized data is mutually dependent.Standard statistical techniques (using data independency) cannot be applied.All variances are underestimated.
Dipdoc Seminar – 12. November 2015
And IF there is a trend error?
To simulate this, a fourth signal is added:
The heights of the inhomogeneities are shifted by DI upward or downward depending on the middlemost year of the subperiod jm.
Due to boundary effects the mean introduced trend error is decreased from 1 (as naively expected) to 0.873 K/cty
Dipdoc Seminar – 12. November 2015
Year
Mea
n in
sert
ed D
I
Non-zero mean trend error
The scatter is conserved, compared to zero mean trend error.
The data cloud is shifted as a whole to the right.
The uncertainty remains, but the mean trend error is well corrected.
Dipdoc Seminar – 12. November 2015
Inserted network-mean trend error
Rem
aini
ng n
etw
ork-
mea
n tr
end
erro
r
xmean = 0.873ymean = 0.010
Including break position errors
Question:What happens, if the break positions have errors and are not perfectly known (as in reality).
Simulation:Scatter the correct positions by adding noise with standard deviation of 2 years.
Result:Only 80% of the trend error is corrected, 20% remains.
Dipdoc Seminar – 12. November 2015
Inserted network-mean trend error
Rem
aini
ng n
etw
ork-
mea
n tr
end
erro
r
xmean = 0.873ymean = 0.161
Conclusion
If the original data contains no trend error (if the inhomogeneities have by chance no overall effect) the trend error for individual networks is (under realistic conditions) not improved.
However, mutually dependent data results from homogenization. This is a disadvantage.
Mean trend errors (due to inhomogeneities) are corrected perfectly, if the break positions are known perfectly.
For realistic position errors (s = 2 years) the trend error is only partly corrected, 20% remains.
Dipdoc Seminar – 12. November 2015