Adjustment of Temperature Trends In Landstations After
Homogenization ATTILAH
Uriah Heat
Unavoidably Remaining Inaccuracies After Homogenization Heedfully Estimating the Adjustments of Temperature trends
Break and Noise Variance
Homogenization
To 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 proceeding
Random combinations of test breaks are inserted. That one explaining the maximum variance is considered to show the true breaks.
Technical application
Dynamic Programming with Stop criterion
Noise Var Break Var
Dipdoc Seminar – 23.10.2017
Trend bias
Trend bias
If the positive and negative jumps do not cancel out each other, they introduce a trend bias.
Underestimation of trend bias
It 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).
Underestimation of jump height
The 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 detection
x1 and x2 are determined as segment averages.
x1 is nearly correct, but x2 is to high.
Incorrect detection
x1’ and x2’ are determined as segment
averages. x2’ is nearly correct, but x1’ is to low.
In both cases the jump height is underestimated.
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.
Nomenclature
k: 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.)
Statistical Characteristic of the Noise Variance
Beta distributed
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.
Statistic for Noise Variance
We 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:
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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 – 23.10.2017
The mean of a Beta distribution is given by:
Mean explained variance:
Maximum explained variance:
Behavior of Noise
optimum
mean (random)
/ (m-1)
Remember
Statistical Characteristic of the Break Variance
1. Heuristic approach 2. Empirical approach 3. Theoretical approach
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
Second Approach (1/3)
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.
Dipdoc Seminar – 23.10.2017
4 2 4 1
Remember
Second Approach (2/3)
For noise, we had:
Now we have n+k+1
”independents”, thus:
4 2 4 1
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�
�−�
�
�+�= ������������−�
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Second Approach (3/3)
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 1
Effective observation number
If 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?
Sketch of derivation (1/2)
Dipdoc Seminar – 23.10.2017
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�+1�=1�
��
�)
2] = � 12∑
��=1+1 ∑
�+�=11 [ �
��
��
��
�]
� ´ = 1
� ∑
�=1
�+1
�
��
�[��
2����]=¿ 1
�2 ∑
�=1
�+1
[��
2][ ����]=�12 ∑
�=1
�+1
[�� 2]
��� ( ´�)= 1
�2 ∑
�=1
�+1
¿
��� ( ´�)=�+1
�2
1
(
��−1) ∑
�=1�−�
(
��−−�−1 1)
�2=� 1(
��+1) ∑
�=1�−�
(
�−�−�−1 1)
�2The 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 l2
Sketch of derivation (2/2)
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:
∑
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(
��)(
�−�− ��)
=(
��+1+1)
∑
�=1�
�2
(
��−−11−�)
=2(
�+�2)
+(
�+�1)
��� ( �´ )= 2 �−�
�(�+2) ≈
2
�+2
1−��� (�´ )=(�+1)�
�(�+2) ≈
�
�+2
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 series
i = l/m n:
m = l(k+1):
with n* = n/2 +1
Similar to the second approach, but n counts only half.
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1−�=
�
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�+2
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�
1−�= �+2
�+2(�+1)=
� 2 +1
�
2 +�+1
= �∗
�+�∗
�= �
�+�∗
�= �
�+�
Remember
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Remember:
Statistical Characteristic of the Break Variance
1. Heuristic approach 2. Empirical approach 3. Theoretical approach
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.
Stepwise 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).
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�−�=���(�)�
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.
Stepwise Fitting (2/3)
Dipdoc Seminar – 23.10.2017
��−�
� = �
���(�)=�����
with a = 0.629 and b = 1.855
Stepwise Fitting (3/3)
[
� �−� �]
=����(�)=��+���−�
� = �
���(�)
�= �
�+ �
���(�)
�= � , �����∗=�.����+�.���
0.629n + 1.855
solve for v:
and insert an+b:
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:
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�+�∗,�����∗=�.����+�.���
�∗=�.��+�
�∗=�.����+�.���
The mean of Beta distribution is:
The variance of a Beta distribution is:
Which can be solved for a and b:
Method of moments
So far we developed an
empirical equation for v, the mean explained variance.
The same procedure is applied to derive equations for and , the shape
parameters describing the distribution of v. These coefficients are determined by the method of moments.
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(�+ �+�) (�+�)�
�´ = �
�+�
�=
(
�´ (��−���´ ) −�)
�´�=
(
�´ (��−��´ ) −�)
(�−�´ )Empirical values for a and b
Dipdoc Seminar – 23.10.2017
Again 1000 by 1000 permutations for fixed values of n and k are performed.
The explained variance is calculated (for 1,000,000 permutation).
From the mean and the variance of the resulting distribution, a and b are
determined by the method of moments.
This proceeding is repeated for
combinations of k = 1…20 and n = 1…20.
The result is plotted against k.
a is strongly dependent on k, converging obviously to a = k/2 for large n.
b is strongly dependent on n, converging obviously to b = n*/2 for large k.
Alpha/k and Beta/n*
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�+�∗= ��
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� = � �
∗�=�
� = �
�∗
1/f
Dipdoc Seminar – 23.10.2017
f is neither a linear function of k nor of n, it is more promising to depict the reciprocal .
1/f is rather linear in k with a slope reciprocal to n and an incept of 2.
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� =�
� =�
� �+�
Determination of c
For a more detailed determination of c, we solve
for c:
and plot the result against k.
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� = �
� �+�
�=�+�+�.��(�−�)�
�=�� �∗ −2�
�= � �
�∗ −� �
Resulting fits for Alpha and Beta
Dipdoc Seminar – 23.10.2017
�= �
�+�� � �= �
�+�� �∗
�=�+�+�.��(�−�)�
Conclusion
The explained noise variance is Beta distributed with:
The explained break variance is Beta distributed with:
with