7. homogenization seminar budapest – 24. – 27. october 2011 what is the correct number of break...

7. Homogenization Seminar Budapest – 24. – 27. October 2011

What is the correct numberof break points

hidden in a climate record?

Ralf LindauVictor VenemaBonn University


Defining breaks

Consider the differences of one station compared to a reference. (Kriged ensemble of surrounding stations)

Breaks are defined by abrupt changes in the station-reference time series.

Internal variancewithin the subperiods

External variancebetween the means of different

subperiods

Criterion:Maximum external variance attained bya minimum number of breaks


Decomposition of Variance

n total number of yearsN subperiodsni years within a subperiod

The sum of external and internal variance is constant.


Two questions

Titel of this talk asks: How many breaks?

Where are they situated?

Testing of all permutions is not feasible.

The best solution for a fixed number of breaks can be found by Dynamical Programming

131010987654321

90919293949596979899

10

991

k

n


Dynamical Programming (1)

Find the optimum positions for a fixed number of breaks.

Consider not only the complete time series, but all possible truncated variants.





Find the first break by simply testing all permutions.






Fill up all truncated variants. The internal variance consists now of two parts: that of the truncated variant plus that of the rest.






Fill up all truncated variants. The internal variance consists of two parts: that of the truncated variant plus that of the rest.

Search the minimum out of n.



The 2-breaks optimum for the full length is found.

To begin the search for 3 breaks, we need as before the previous solutions for all, also shorter length.

This needs n2/2 searches, which is for larger numbers of breaks k much less than all permutations (n over k).


Position & Number

Solved:

The optimum positions for a fixed number of breaks are known by Dynamical Programming.

Left:

Find the optimum number of breaks.

The external variance increase in any case with increasing number of breaks.

Use as benchmark the behaviour of a random time series.


Segment averages

with stddev = 1

Segment averages xi scatter randomly

mean : 0

stddev: 1/

Because any deviation from zero can beseen as inaccuracy due to the limited number of members.

in


External Variance

The external varianceis equal to the mean square sumof a random normal distributed variable.

Weighted measure for thevariability of the subperiods‘means


2-distribution

n: Length of time series (Number of years)k: Number of breaksN = k+1: Number of subsegments[ ]: Mean of several break position permutations

[varext] = (N-1)/n = k/nIn average, the external variance increases linearly with k.However, we consider the best member as found by DP.

varext ~ N2 The external variance is chi2-distributed.

Def.:

Take N values out of N (0,1), square and add them up.

By repeating a N2-distribution is obtained.


21-years random data (1)

1000 random time series are created.

Only 21-years long, so that explicite tests of all permutations are possible.

The mean increases linearly.

However, the maximum is relevant

(the best solution as found by DP)

Can we describe this function?

First guess: 4*11 kv


21-years random data (2)

Above, we expected the datafor a fixed number of breaksbeing chi2-distributed.


The random data does not fit exactly to a chi2-distribution.

The reason is that chi2 has no upper bounds.

But varext cannot exceed 1.

A kind of confined chi2 is the beta distribution.

From 2 to distribution

n = 21 yearsk = 7 breaks

data



n = 21 yearsk = 7 breaks

data

X ~ 2(a) and Y ~ 2(b)

X / (X+Y) ~ (a/2, b/2)

If we normalize a chi2-distributed variable by the sum of itself and another chi2-distributed variable, the result will be -distributed.

The -distribution fits well to the data and is the theoretical distribution for the external variance of all break position permutations.



11

15

17

7

)(

)()(),(

ba

babaB

2

1,2

1)(

12

112

knkB

vvvp

knk

with

We are interested in the best solution,with the highest external variance, as provided by DP.

We need the exceeding probability forhigh varext


Incomplete Beta Function

2

1,2

1)(

12

112

knkB

vvvp

knk

External variance v is -distributedand depends on n (years) and k (breaks):

The exceeding probability P gives thebest (maximum) solution for v

Incomplete Beta Function

1

0

1)(i

l

lml vvl

mvP

Solvable for even k and odd n:

2

ki

2

3n

m

v

pdvvP0

1)(


Example 21 years, 4 breaks

1

0

1)(i

l

lml vvl

mvP

k = 4 i = 2n = 21 m = 9

89 191)( vvvvP

2

ki

2

3n

m


Theory and Data

89 191)( vvvvP

Theory (Curve):

Random data (hached) fits well.


Nominal Combination Number

48454

201

k

n

For n = 21 and k = 4 there are

break combinations.

If they all were independent wecould read the maximum externalvariance at (4845)-1 ≈ 0.0002 being 0.7350

However, we suspect that thebreak combinations are notindependent. And we know thecorrect value of varext.


Effective and Nominal

Remember: varext= 0.5876 for k=4

The reverse reading leads to an 23 times higher exceeding probability.

This shows that the break permutationsare strongly dependent and the effectivenumber of combinations is smaller than the nominal.

However, the theorectical function is correct.


From 21 years to 101 years

As we now know the theoretical function, we quit the explicit check by random data.

And skip from unrealistic short time series (n=21) to more realistic (n=101).

Again the numerical values of the external variance is known and we can conclude the effective combination numbers.

Can we give a formula for in order to derive v(k) ?

220

breaks

dk

dv


dv/dk sketch

Increasing the break number from k to k+1 has two consequences:

1. The probability function changes.

2. The number combinations increase.

Both increase the external variance.

k breaks

k+1 breaks


Using the Slope

P(v) is a complicated function and hard to invert into v(P).

Thus, dv is concluded from dP / slope.

We just derived P(v) by integrating p(v), so that the slope p(v) is known.

k breaks

k+1 breaks

1

0

1)(i

l

lml vvl

mvP


The Slope

)(

)())(ln(

vP

vpvP

dv

d

1

0

1

1

111

))(ln(i

l

lml

imi

vvl

m

i

mimvv

vPdv

d

11

1

11

111

))(ln(

imi

imi

vvi

m

i

mimvv

vPdv

d

v

imvP

dv

d

1

1))(ln(

vkn

vPdv

d

12

1))(ln(

Insert the known functions:

The last summand dominates:

Reduce and replace m and i:


Distance between the Curves

1

0

01

1

1

lnlnlni

l

lml

i

l

lml

ii

vvl

m

vvl

m

PP

11

1

11

1

lnlnlnimi

imi

ii

vvi

m

vvi

m

PP

vi

vimPP ii 1

1lnlnln 1

vk

vknPP kk 1

1ln2

1lnln 1

The last summand dominates:

Reduce and replace m and i:


Effective combination growth

Nominal Growth Rate

-2 ln ( (n-1- k) / k) Ln: Logarithmic sketch minus: Number of combinations is reciprocal to Exceeding Probability

2: Exceeding Probability only known for even break numbers

k

n 1

1

1

k

n(n-1-k) / k

However, break combinations are not independentand we know the effective number of combinations


Ratio: nominal / effective

k1 k2 k nominal effectiv c=nom/eff

2 4 3 -2.552 -7.784 0.328

4 6 5 -2.186 -6.952 0.315

6 8 7 -1.963 -6.356 0.309

8 10 9 -1.765 -5.889 0.300

10 12 11 -1.645 -5.503 0.299

12 14 13 -1.514 -5.173 0.293

14 16 15 -1.435 -4.885 0.294

16 18 17 -1.363 -4.627 0.295

18 20 19 -1.292 -4.394 0.295

The ratio of nominal / effective is approximatly constant with c = 0.3


Approximative Solution

vk

vkn

k

knc

kn

v

dk

dv

1

1ln2

11ln

1

12

1*

n

kkNormalisation

for small k*

)4ln()100ln(3.02

:

4

)1(

)1(ln

1ln2

1

1*

*

*

*

*

*

vk

vk

n

k

kc

dk

dv

v

k

15.439.176.2

for n = 100


Exact Solution

5ln21ln

2

1

1

1*

***

k

kk

dk

dv

v

k

***

*

1

5ln21ln

2

1

1

1dk

kk

kdv

v

*

2

1

*

*

2

1)5ln(2* 1

11k

k

kkv


Constance of Solution

101 ye

ars21 yea

rs

The solution for the exponent is constant for different length oftime series (21 and 101 years).


Conclusion

We have found a general mathematical formulation how the external variance of a random time series is increasing when more and more breaks as given by Dynamical Programming are inserted.

This can be used as benchmark to define the optimum number of breaks.


Integrated result

How does the found function look like after integration?

Crosses: Test data

Line: Theory

Error bars: 90 and 95 percentile


Appendix (1)

vl

m

vl

m

f

i

i

1

1

vl

vlmf

i

i

1

1

vl

vlnf

k

k

1

1

vk

vknf

1

1

Consider the individual summands of the sum as defined in The factor of change f between a certain summand and its successor is:

m and i can be replaced by n and k:

inserting k instead of lk is a lower limit for f because (n-1-lk)/lk, the rate of change of the binomial coefficients, is decreasing monotonously with k:

where li runs from zero to i. The ratio of consecutive binomial coefficients can be replaced and it follows:

normalised by 1/(n-1):

vk

vkf

1

1*

*

1

0

1)(i

l

lml vvl

mxP


Appendix (2)

4**

4**

1

111

kk

kkf

3**

4*

1

11

kk

kf

4

31

4

31

4

31

411***

*

**

*

kkk

k

kk

kf

1

011*

4*3*

k

kkf

the approximate solution is known with 1-v = (1- k*)4

0k

1k

We can conclude that each element of the sum givenabove is by a factor f larger than the prior element.For small k* the factor f is greater than about 4 and grows to infinity for large k*. Consequently, we canapproximate the sum by its last summand according to:

111

11

1)(

imi

i

il

lml vvi

mvv

l

mxP


Application (1)

Insert in each of 1000 random time series 5 breaks of variance 1.

The change of external variance for low break numbers (1, 2, 3 up to about 10) increase.

Lying above the theoretical function for random time series without any break (arrow).

Variances of break numbers higher than 5 increase, because the inserted 5 breaks are not always the biggest.


Application (2)Stop break search, when the growth rate for

the external variance drops firstly below the theoretical one for zero breaks.

1 Example of 1000 test time series

Crosses: Observations

Thin line: Inserted breaks

Fat line: Detected breaks

In average over 1000 samples:

Added variance: 86%

(theoretically 5/6)

Remaining after correction: 27%

Average detected break number 5.48

7. homogenization seminar budapest – 24. – 27. october 2011 what is the correct number of break...

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