Adjustment of Temperature Trends In - PowerPoint Presentation

Slide1

Adjustment of Temperature Trends In

Landstations

After Homogenization

(ATTILAH)

Ralf LindauSlide2

Dipdoc

Seminar – 23.10.2017

Break and Noise Variance

HomogenizationTo homogenize we consider the difference time series between two neighboring stations.The dominating natural variance is cancelled out, because it is very similar at both stations. The relative break variance is increased and we have a realistic chance to detect the breaks.General proceedingRandom combinations of test breaks are inserted. That one explaining the maximum variance is considered to show the true breaks. Technical applicationDynamic Programming with Stop criterion

Noise

Var

Break

VarSlide3

Dipdoc

Seminar – 23.10.2017

Trend bias

Trend biasIf the positive and negative jumps do not cancel out each other, they introduce a trend bias. Underestimation of trend biasIt is impossible to isolate the full break signal from the noise. Thus, only a certain part of it can be corrected. A small fraction remains (which has to be corrected after homogenization).Slide4

Underestimation of jump height

Dipdoc Seminar – 23.10.2017The two fat horizontal lines indicate the true jump height Errors occur when the noise randomly (and erroneously) increases the data above the middle line.

Then, a part of Segment 2 is (erroneously) exchanged to Segment 1

Correct detectionx1 and x2 are determined as segment averages. x1 is nearly correct, but x2 is to high.Incorrect detectionx1’ and x2’ are determined as segment averages. x2’ is nearly correct, but x1’ is to low.In both cases the jump height is underestimated. Slide5

Dipdoc

Seminar – 23.10.2017

Obviously, this systematic underestimation depends on the interaction between noise and break variance.

To quantify this effect, the statistical properties of both break and noise variance has to be known.Nomenclaturek: Number of test breaks (here: 3)n: Number of true breaks (here: 7)m: Total length (here: 100)l: Test segment length (here: 14, 4, etc.)Slide6

Statistical Characteristic of the Noise Variance

Beta distributedSlide7

Dipdoc

Seminar – 23.10.2017

Example for noise variance

We insert k = 3 random test breaks and check the variance they are able to explain.Since we have pure noise, the test segments’ means are very close to zero.However, there is a small random variation: This is the explained variance.Slide8

Statistic for Noise Variance

Dipdoc Seminar – 23.10.2017We insert k = 3 test breaks at random positions into a random noise time series and calculate the explained variance.This procedure is repeated

1000 by 1000 times (1000 time series and 3000 test break positions). Relative explained variance:

 Slide9

Probability density:

with Beta function

For noise the shape parameters are:

when

k

denotes the number of test breaks and

m

the total length

Beta distribution

Dipdoc

Seminar

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23.10.2017 Slide10

The mean of a Beta distribution is given by:

Mean explained variance:

Maximum explained variance:

 Behavior of Noiseoptimum

mean (random)

Dipdoc

Seminar

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23.10.2017

/ (m-1)

RememberSlide11

Statistical Characteristic of the Break

Variance1. Heuristic approach2. Empirical approach 3. Theoretical approachSlide12

For true breaks, constant periods exist. Tested segment averages are the (weighted) means of such (few) constant periods.

This is quite the same situation as for random scatter, only that less independent data is underlying.Obviously, the number of breaks n plays the same role as the time series length m did before for noise.Thus, the first approximation is: First Approach

Dipdoc

Seminar – 23.10.2017  Slide13

Second

Approach (1/2)However, this would lead to

This is obviously only true, when all real breaks are actually matched by the test breaks, (which is not the case for random trials).

Consider k=3, n=7 and count the number of platforms in each test segment. Altogether, there are 11 “independents”, in general n+k+1.Thus, the second approximation is:  Dipdoc Seminar – 23.10.2017 4 2 4 1

RememberSlide14

Second

Approach (2/2)This would lead to

This is rather reasonable, because for

n = k the situation is approximately:Each test segment contains one true break, thus two independents, which are then averaged. This leads to a reduction of the variance by a factor of 2. However, so far we did not take into account that the HSPs have different lengths.The effective number of true breaks must be smaller than the nominal. Dipdoc Seminar – 23.10.2017 4 2 4 1Slide15

E

ffective observation numberIf we generate i = 1…N random time series of length j = 1…m with each element being:

only a fraction of (m-1)/m can be found within the time series (because a fraction of 1/m is “lost” due to the variance of the time series means.

How large is this effect if a step function with n breaks is considered?  Dipdoc Seminar – 23.10.2017 Slide16

Sketch of derivation (1/2)

Dipdoc Seminar – 23.10.2017

The mean of each time series is:

The “lost” variance is:

Which can be reduced to the sum over mean squared lengths:

Which is equal to the weighted sum over l

2

Slide17

Sketch of derivation (2/2)

Dipdoc Seminar – 23.10.2017 The sum of a product of two Binomial coefficients is solvable by the Vandermonde’s identity:Which leads to a solution for the l2 sum:

Inserted into the original expression, we obtain

for the “lost” external variance: The remaining internal variance is then:

 Slide18

Third approach

Dipdoc Seminar – 23.10.2017 The relative unexplained variance of a test segment:with i: number of breaks within a test segment and n: number of breaks within the entire time seriesi = l/m n:

m = l(k+1):

with n* = n/2 +1 Similar to the second approach, but n counts only half. 

Remember

Remember:Slide19

Statistical Characteristic of the Break

Variance1. Heuristic approach2. Empirical approach 3. Theoretical approachSlide20

Empirical

Var(k,n)Dipdoc Seminar – 23.10.2017

Empirical test with 1000 random segmentations

(fixed k) of 1000 time series (fixed n).Calculate the mean relative explained variance v from these 1,000,000 permutations. Repeat this procedure for all combinations of k = 1, …, 20 and n = 1, …, 20.20 functions v(k) for the different n.Slide21

Intelligent Fitting (1/3)

v/(1-v) is proportional to k.The slope is a function of n. (Numbers and lines do not cross). The slope is certainly not proportional, but rather reciprocal to n. (

slp(1) large, slp(20) small).

Thus, better to plot 1/slp(n). Dipdoc Seminar – 23.10.2017 Slide22

We expect horizontal lines, if the reciprocal slope is really independent from k.

This is largely confirmed. Averages over k gives than a value for each n. These seems to be rather linear in n. Thus, plot these averages as a function of n.Intelligent Fitting (2/3)Dipdoc Seminar – 23.10.2017

 Slide23

with

a = 0.629 and b = 1.855 Intelligent Fitting (3/3)Dipdoc Seminar – 23.10.2017

0.629n + 1.855

solve for v:

a

nd insert

an+b

:

Slide24

Application of findings

Dipdoc Seminar – 23.10.2017

Summarizing the stepwise fitting:

The direct fit in the v/k space yields: The best

heuristic approach was:

 Slide25

Conclusion

The explained noise variance is Beta distributed with:The explained break variance is Beta distributed with:

Dipdoc

Seminar – 23.10.2017