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7/31/2019 Serial Correlation Coefficient
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Investigating a new estimator of the
serial correlation coefficient
Manfred Mudelsee
Institute of Mathematics and StatisticsUniversity of Kent at Canterbury
CanterburyKent CT2 7NF
United Kingdom
8 May 1998
Abstract
A new estimator of the lag-1 serial correlation coefficient ,n1i=1
3i i+1 /
n1i=1
4i , is compared with the old estimator,
n1i=1
i
i+1
/ n1i=1
2i
, for a stationary AR(1) process with known mean. The
mean and the variance of both estimators are calculated using the second or-
der Taylor expansion of a ratio. No further approximation is used. In case of
the mean of the old estimator, we derive Marriott and Popes (1954) formula,
with (n 1)1 instead of (n)1, and an additional term (n 1)2. In caseof the variance of the old estimator, Bartletts (1946) formula results, with
(n 1)1 instead of (n)1. The theoretical expressions are corroborated withsimulation experiments. The main results are as follows. (1) The new esti-
mator has a larger negative bias and a larger variance than the old estimator.
(2) The theoretical results for the mean and the variance of the old estimator
describe the principal behaviours over the entire range of, in particular, the
decline to zero negative bias as approaches unity.
Mudelsee M (1998) Investigating a new estimator of the serial correlation coefficient.
Institute of Mathematics and Statistics, University of Kent, Canterbury, IMS Technical
Report UKC/IMS/98/15, Canterbury, United Kingdom, 22 pp.
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1 Introduction
Consider the following estimators of the serial correlation coefficient oflag 1 from a time series i (i = 1, . . . , n) sampled from a process Ei withknown (zero) mean:
=
n1i=1 ii+1n1
i=1 2i
, (1)
=
n1i=1
3i i+1n1
i=1 4i
. (2)
Estimators of type (1) are well-known (e. g. Bartlett 1946, Marriott andPope 1954), whereas (2) is new to my best knowledge.
Let Ei be the stationary AR(1) process,
E1
N(0, 1),
Ei = Ei1 + Ui, i = 2, . . . , n , (3)with Ui i. i. d. N(0, 2) and 2 = 1 2. The following misjudgement ofmine led to the present study. minimises
n1i=1
(i+1 i)2 2i ,
which is a weighted sum of squares. However, since Ei has constant varianceunity, no weighting should be necessary.
Nevertheless, we compare estimators (1) and (2) for process (3). We re-strict ourselves to 0 < < 1. We first calculate their variances, respectivelymeans, up to the second order of the Taylor expansion of a ratio, similarlyto Bartlett (1946), respectively Marriott and Pope (1954). Since we concen-trate on the lag-1 estimators and process (3), no further approximation isnecessary. These theoretical expressions and also those of White (1961) forestimator (1) are then examined with simulation experiments.
2 Series expansions
2.1 Old estimator2.1.1 Variance
We write (1) as
=1n
n1i=1 ii+1
1n
n1i=1
2i
(4)
=c
v, say,
1
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and have, up to the second order of deviations from the means of c and v,the variance
var() = var(c/v) var(c)E2(v)
2 E(c)cov(v, c)E3(v)
+ E2(c)var(v)
E4(v). (5)
We now apply a standard result for quadravariate standard Gaussian distri-butions with serial correlations j (e. g. Priestley 1981:325),
cov(aa+s, bb+s+t) = ba ba+t + ba+s+t bas,
from which we derive, for i following process (3):
cov
1
n
n1a=1
aa+s,1
n
n1b=1
bb+s+t
=
=
1
n2
n1
a,b=1(
|ba|
|ba+t|
+
|ba+s+t|
|bas|
). (6)
Without further approximation we can directly derive the various constituentsof the right-hand side of (5) by the means of (6).
var(c) = var
1
n
n1i=1
ii+1
= (put s = 1 and t = 0 in (6))
=1
n2
n1a,b=1
2|ba| +n1
a,b=1
|ba+1| |ba1|
.
We find, by summing over the points of the a-b lattice in a diagonal manner,
n1a,b=1
2|ba| = (n 1) + 2n2i=1
i 2(ni1)
= (n 1)
2
1 2 1
+ 2(2n4 2)
(1 2)2 (7)
(at the last step we have used the arithmetic-geometric progression), and
n1a,b=1
|ba+1|
|ba1|
= (n 1) 2
+ 2
n2i=1 i
2(ni1)
= (n 1)
2
1 2 + 2 2
+ 2
(2n4 2)(1 2)2 . (8)
(7) and (8) give var(c). Further,
cov(v, c) = cov
1
n
n1i=1
2i ,1
n
n1i=1
ii+1
2
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= (put s = 0 and t = 1 in (6))
=2
n2
n1a,b=1
|ba| |ba+1|.
We findn1
a,b=1
|ba| |ba+1| = 2n3 +n2i=1
(2 i + 1) 2(ni)3
= (n 1) 2
1
1 2 1
+2(2n5 3)
(1 2)2 +(2n3 1)
(1 2) . (9)
(At the last step we have used the arithmetic-geometric and also the geomet-ric progression.) (9) gives cov(v, c). Further,
var(v) = var
1
n
n1i=1
2i
= (put s = t = 0 in (6))
=2
n2
n1a,b=1
2|ba|,
which is given from (7). We further need
E(v) =n 1
n(10)
and
E(c) =(n 1)
n. (11)
All three terms contributing to var() have a part (n 1)2. However,these parts cancel out:
var() (1 2)(n 1) . (12)
Bartlett (1946) already gave, up to the order (n)1,
var() (1 2)
n, (13)
which cannot be distinguished from our result.
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2.1.2 Mean
We have, up to the second order of deviations from the means of c andv, the mean
E() = E(c/v) E(c)
E(v) cov(v, c)
E2(v) + E(c)
var(v)
E3(v) . (14)
As above, we derive cov(v, c) and var(v) from (9) and (7), respectively, andhave E(v) and E(c) given by (10) and (11), respectively, yielding
E() 2 (n 1) +
2
(n 1)2( 2n1)
(1 2) . (15)
Marriott and Pope (1954) investigated the estimator
=1
n1 n1i=1 ii+1
1
nn1
i=1 2i
and gave, up to the order (n)1,
E() 2 n
,
which cannot be distinguished from the first two terms of our result.
2.1.3 Unknown mean of the process
I have also tried to calculate up to the same order of approximation the
mean and the variance of the estimatorn1i=1
i 1n1
n1j=1 j
i+1 1n1
n1j=1 j+1
n1
i=1
i 1n1
n1j=1 j
2 ,for the case when the mean of the process is unknown. That would requireto calculate the following sums over the points of a cubic lattice:
n1a,b,c=1
|ba| |cas|
for s = 0 and 1. However, I was unable to obtain exact formulas.
2.2 New estimator
2.2.1 Variance
We write (2) as
=1n
n1i=1
3i i+1
1n
n1i=1
4i
=d
w, say.
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var(), up to the second order of the Taylor expansion, is given by (5),substituting d for c and w for v, respectively. We need the following resultfor octavariate standard Gaussian distributions with serial correlations j(Appendix),
cov(
3
aa+s,
3
b b+s+t) = 9 ba ba+t + 9 ba+s+t bas
+ 18 s ba ba+s+t + 18 ba bas s+t
+ 18 s 2ba s+t + 18
2ba ba+s+t bas
+ 6 3ba ba+t, (16)
from which we derive, for i following process (3):
cov
1
n
n1a=1
3aa+s,1
n
n1b=1
3bb+s+t
=
= 1n2
n1a,b=1
( 9 |ba| |ba+t| + 9 |ba+s+t| |bas|
+ 18 |s| |ba| |ba+s+t| + 18 |ba| |bas| |s+t|
+ 18 |s| 2|ba| |s+t| + 18 2|ba| |ba+s+t| |bas|
+ 6 3|ba| |ba+t| ) . (17)
Without further simplification we can directly obtain the various constituentsof var() by the means of (17).
var(d) = var1
n
n1i=1
3i i+1
= (put s = 1 and t = 0 in (17))
=1
n2
9 n1
a,b=1
2|ba| + 9n1
a,b=1
|ba+1| |ba1|
+ 18 n1
a,b=1
|ba| |ba+1| + 18 n1
a,b=1
|ba| |ba1|
+ 18 2n1
a,b=1
2|ba| + 18n1
a,b=1
2|ba| |ba+1| |ba1|
+ 6n1
a,b=1
4|ba|
.
The first and the fifth sum is given by (7), the second sum by (8) and thethird by (9). We find,
n1a,b=1
|ba| |ba1| =n1
a,b=1
|ba| |ba+1|,
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which is given by (9),
n1a,b=1
2|ba| |ba+1| |ba1| = (n 1) 2 + 2n2i=1
i 4(ni1)
= (n 1)
21 4 +
2 2
+ 2 (4n8 4)(1 4)2 (18)
and
n1a,b=1
4|ba| = (n 1) + 2n2i=1
i 4(ni1)
= (n 1)
2
1 4 1
+ 2(4n8 4)
(1 4)2 . (19)
(7), (8), (9), (18) and (19) give var(d). Further,
cov(w, d) = cov
1
n
n1i=1
4i ,1
n
n1i=1
3i i+1
= (put s = 0 and t = 1 in (17))
=1
n2
36 n1
a,b=1
|ba| |ba+1| + 36 n1
a,b=1
2|ba|
+ 24n1
a,b=1
3|ba| |ba+1| .The last sum of this expression,
n1a,b=1
3|ba| |ba+1| = (n 1) +n2i=1
i 4(ni)3 +n2i=1
i 4(ni)5
= (n 1)
1
1 3 +1
5 1
+
(4n7 + 4n9) (1
4n+4)
(1 4)2 . (20)(7), (9) and (20) give cov(w, d). Further,
var(w) = var
1
n
n1i=1
4i
= (put s = t = 0 in (17))
=1
n2
72 n1
a,b=1
2|ba| + 24n1
a,b=1
4|ba|
,
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which is given from (7) and (19), respectively. We further need
E(w) =3 (n 1)
n(21)
and
E(d) =3 (n 1)
n. (22)
As for the old estimator, the terms (n1)2 which contribute to var()cancel out:
var() 53
(1 2)(n 1) . (23)
2.2.2 Mean
We use (14) with d instead of c and w instead of v, respectively. Wederive cov(w, d) from (7), (9) and (20), further, var(w) from (7) and (19),and we have E(w) and E(d) given by (21) and (22), respectively. This yields,up to the second order of deviations from the means of d and w,
E() 1(n 1)
4
3
5 2
1 + 2
+1
(n 1)2
4( 2n1)
(1 2) +8
3
(3 4n1)(1 2) (1 + 2)2
. (24)
2.3 Remark
is found to have a larger negative bias and also a larger variance than .In the simulation experiment, we intend not only to prove that. We are alsointerested to examine the significance of the term (n 1)2 in (15). In thecomparison we include the theoretical results of White (1961) who studiedthe old estimator (1) for process (3). He calculated the kth moment of (c/v)via the joint moment generating function m(c, v), with c and v defined as in(4) without the factors 1/n,
E(
k
) =0
vk . . .
v2
km(c, v)
ck
c=0 dv1 dv2 . . . d vk.
He expanded the integrand to terms of order n3 and 4 and gave the fol-lowing results (the index W refers to his study):
var()W
1
n 1
n2+
5
n3
1
n 9
n2+
53
n3
2 12
n34, (25)
E()W
1 2n
+4
n2 2
n3
+
2
n23 +
2
n25. (26)
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3 Simulation experiments
In the first experiment, for each combination of n and listed in Table1, we generated a set of 250 000 time series from process (3). For every timeseries, and have been calculated. The sample means, sim and sim,
and the sample standard deviations, sim and sim, over the simulationsare listed in Table 1. Therein, our theoretical valuesfrom (12), (15), (23)and (24)are included. In case of the means, these are additionally splittedinto terms (n 1)2 and terms of lower order. Also the theoretical valuesfrom Whites (1961) formula, herein (25) and (26), are listed.
We further investigated whether the distribution of approaches Gaus-sianity as fast as that of . Fig. 1 shows histograms of , respectively ,from the first simulation experiment in comparison with Gaussian distribu-tions N( sim,
2 sim), respectively N( sim,
2 sim).
In the second simulation experiment (Fig. 2), formulas (12) versus (25)
and (15) versus (26) are examined over the entire range of , with particularinterest on large. We plot the negative bias, E(), and var(). Foreach combination of and n, the number of simulated time series after (3)is 250 000.
The error due to the limited number of simulations is small enough fornot influencing any intended comparison.
4 Results and discussion
The first simulation experiment (Table 1) confirms, over a broad range of
and n, that has a larger bias and a larger variance, respectively, than .In general, the deviations between the theoretical results and the simula-
tion results decrease with n increased, as is to be expected.For any combination of and n listed in Table 1, our terms (n1)2 in
(15), respectively (24), bring these theoretical results closer to the simulationresults, with the exception of E() for = 0.9 and n = 800 where thesimulation noise prevents that comparison. For large, respectively n small,these second order terms contribute heavier.
The frequency distribution of (Fig. 1) has a similar shape as that of. It is shifted to smaller values against that and also broader, reflecting the
larger negative bias, respectively the larger variance. The functional formof the distribution of is approximately the Leipnik distribution, which isheavier skewed for large and tends to Gaussianity with n increased. Bothestimators seem to approach Gaussianity equally fast (Fig. 1).
The second simulation experiment (Fig. 2) compares Whites (1961) the-oretical results for , (25) and (26), with ours, (12) and (15). It shows thathis are better performing up to a certain value of .
Above, our formulas describe better the simulation, particularly, the de-
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cline to zero negative bias, respectively zero variance, as approaches unity.The decline to zero negative bias is caused by the term (n 1)2 in (15).Those declines are reasonable, since for = 1 all time series points haveequal value and c = v in (4).
For small , his formulas perform better since they are more accurate
with respect to powers of (1/n) than ours. For larger his approximationbecomes less accurate. In particular, for n 3 and > 0, E()W cannotbecome zero. For n 10, d
d( E()W) cannot become negative. For n 8,
var()W cannot become zero. That means, for those cases Whites (1961)formulas cannot produce the decline to zero negative bias and zero variance,respectively.
The fact that Bartletts (1946) formula for the variance, (13), describesthe simulation better than (12) for 0, is regarded as spurious.
These results mean that the expansion formulas for var() and E(), (5)and (14), respectively, are sufficient to derive the principal behaviours for
1. In case of the mean, however, no further approximation is allowedwhich would lead to the term (n 1)2 in (15) be neglected.
5 Conclusions
1. In case of the stationary AR(1) process (3) with known mean, the newestimator, , has a larger negative bias and a larger variance than theold estimator, . Its distribution tends to Gaussianity about equallyfast.
2. Our formula (15) for E() describes the expected decline to zero nega-tive bias for 1.
3. The formula (12) for var() describes the expected decline to zero vari-ance for 1.
4. The second order Taylor expansion of a ratio is sufficient to derivethese two principal behaviours for 1, if no further approximationsare made. This condition could be fulfilled since we had concentratedourselves on the lag-1 estimator and process (3).
5. For unknown mean of the process, it is more complicated to derive theequations with the same accuracy.
Acknowledgements
Q. Yao is appreciated for comments on the manuscript. The present studywas carried out while the author was Marie Curie Research Fellow (EU-TMRgrant ERBFMBICT971919).
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References
Bartlett, M. S., 1946, On the theoretical specification and sampling prop-erties of autocorrelated time-series: Journal of the Royal Statistical SocietySupplement, v. 8, p. 27-41 (Corrigenda: 1948, Journal of the Royal Statis-
tical Society Series B, v. 10, no. 1).
Marriott, F. H. C., and Pope, J. A., 1954, Bias in the estimation of au-tocorrelations: Biometrika, v. 41, p. 390-402.
Priestley, M. B., 1981, Spectral Analysis and Time Series: London, Aca-demic Press, 890 p.
Scott, D. W., 1979, On optimal and data-based histograms: Biometrika,v. 66, p. 605-610.
White, J. S., 1961, Asymptotic expansions for the mean and variance ofthe serial correlation coefficient: Biometrika, v. 48, p. 85-94.
Appendix
We assume that i is drawn from a standard Gaussian distribution withserial correlations j. We derive (16) by means of the moment generatingfunction. Write
cov(3aa+s,
3b b+s+t) = E(
3aa+s
3b b+s+t) E(
3aa+s) E(
3b b+s+t)
= E(3aa+s3b b+s+t) 9 s s+t.
Now, E(3aa+s3b b+s+t) is the coefficient of (t
31 t2 t
33 t4)/(3!1!3!1!) in the mo-
ment generating function
m(t1, t2, t3, t4) = exp
1
2
4i=1
4j=1
ijtitj
,
with
11 = 22 = 33 = 44 = 1,
12 = 21 = s,
13 = 31 = ba,
14 = 41 = ba+s+t,
23 = 32 = bas,
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24 = 42 = ba+t,
34 = 43 = s+t.
Terms (t31 t2 t33 t4) in m(t1, t2, t3, t4) can only be within the fourth order ofthe series expansion of the exponential function, i. e., within
1
4!
1
2
4i=1
4j=1
ijtitj
4 ,
further restricting, within
1
384(t21 + t
23 + 2 12t1t2 + 2 13t1t3
+ 2 14t1t4 + 2 23t2t3 + 2 24t2t4 + 2 34t3t4)4,
further restricting, within
1
384(4 213t
21t
23 + 2 t
21t
23 + 4 12t
31t2 + 4 13t
31t3 + 4 14t
31t4 + 4 23t
21t2t3
+ 4 24t21t2t4 + 4 34t
21t3t4 + 4 12t1t2t
23 + 4 13t1t
33 + 4 14t1t
23t4
+ 4 23t2t33 + 4 24t2t
23t4 + 4 34t
33t4 + 8 1213t
21t2t3 + 8 1214t
21t2t4
+ 8 1234t1t2t3t4 + 8 1314t21t3t4 + 8 1323t1t2t
23 + 8 1324t1t2t3t4
+ 8 1334t1t23t4 + 8 1423t1t2t3t4 + 8 2334t2t
23t4)
2.
Finally, these terms are
1
384(2 4 213t21t23 8 1234t1t2t3t4 + 2 4 213t21t23 8 1324t1t2t3t4
+ 2 4 213t21t23 8 1423t1t2t3t4 + 2 2 t21t23 8 1234t1t2t3t4+ 2 2 t21t23 8 1324t1t2t3t4 + 2 2 t21t23 8 1423t1t2t3t4+ 2 4 12t31t2 4 34t33t4 + 2 4 13t31t3 4 24t2t23t4+ 2
4 13t
31t3
8 2334t2t
23t4 + 2
4 14t
31t4
4 23t2t
33
+ 2 4 23t21t2t3 4 14t1t23t4 + 2 4 23t21t2t3 8 1334t1t23t4+ 2 4 24t21t2t4 4 13t1t33 + 2 4 34t21t3t4 4 12t1t2t23+ 2 4 34t21t3t4 8 1323t1t2t23 + 2 4 12t1t2t23 8 1314t21t3t4+ 2 4 13t1t33 8 1214t21t2t4 + 2 4 14t1t23t4 8 1213t21t2t3+ 2 8 1213t21t2t3 8 1334t1t23t4 + 2 8 1314t21t3t4 8 1323t1t2t23).
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Thus, we find
E(3aa+s3bb+s+t) =
3!1!3!1!
384(96 1234 + 96 1324 + 96 1423
+ 192 121314 + 192 132334
+ 192 1221334 + 192
213
21423
+ 64 31324).
This gives the final result
cov(3aa+s, 3b b+s+t) = 9 ba ba+t + 9 ba+s+t bas
+ 18 s ba ba+s+t + 18 ba bas s+t
+ 18 s
2ba
s+t
+ 18 2ba
ba+s+t
bas
+ 6 3ba ba+t.
It should be noted that this result has been checked using the joint cumulantof order eight, in my assessment without less effort.
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Table1:
(Continued)
Sim
Sim
T(24)1
st
T(24)2nd
T(24,
23)
T(15)1st
T(15)2nd
T(15,1
2)
T(26,
25)
=
.900
Mean
.7
94824765
.831981030
.6539982
55
.060176382
.714174637
.805263157
.025764012
.83102717
0
.825372450
n
=
20
S.d.
.1
52898617
.141649998
.129099444
.10000000
0
.139640968
S.d.
/500
.0
00305797
.000283299
=
.900
Mean
.8
67598466
.883239327
.8527875
43
.002251858
.855039402
.881818181
.000966603
.88278478
5
.882622098
n
=
100
S.d.
.0
56144769
.049552189
.056556637
.04380858
2
.049831684
S.d.
/500
.0
00112289
.000099104
=
.900
Mean
.8
94646055
.897751699
.8941501
46
.000034571
.894184717
.897747183
.000014839
.89776202
3
.897759744
n
=
800
S.d.
.0
19176153
.015732535
.019908007
.01542067
5
.015723824
S.d.
/500
.0
00038352
.000031465
=
.980
Mean
.9
03670968
.929367351
.7063014
70
.182799711
.889101182
.876842105
.073477667
.95031977
2
.900780563
n
=
20
S.d.
.1
17565805
.106515873
.058937969
.04565315
4
.118185438
S.d.
/500
.0
00235131
.000213031
=
.980
Mean
.9
64341773
.971335719
.9538679
79
.002915320
.956783299
.970150753
.001249438
.97140019
2
.970390010
n
=
200
S.d.
.0
21591051
.019022430
.018211487
.01410655
7
.019544022
S.d.
/500
.0
00043182
.000038044
=
.980
Mean
.9
77759602
.979033841
.9773985
63
.000028899
.977427462
.979019509
.000012386
.97903189
5
.979021902
n
=
2000
S.d.
.0
05526193
.004652735
.005745999
.00445083
1
.004658731
S.d.
/500
.0
00011052
.000009305
=
.999
Mean
.9
95105991
.996604567
.9883253
15
.006226665
.994551981
.994995991
.002535138
.99753113
0
.995035904
n
=
500
S.d.
.0
05699764
.004845877
.002583928
.00200150
2
.005953760
S.d.
/500
.0
00011399
.000009691
=
.999
Mean
.9
97587240
.998180155
.9963353
34
.000574434
.996909768
.998000500
.000245543
.99824604
4
.998002994
n
=
2000
S.d.
.0
01870816
.001622354
.001290994
.00100000
0
.001728444
S.d.
/500
.0
00003741
.000003244
=
.999
Mean
.9
98907925
.998960630
.9988934
64
.000000932
.998894397
.998960039
.000000399
.99896043
9
.998960043
n
=
50000
S.d.
.0
00247577
.000207948
.000258136
.00019995
1
.000207779
S.d.
/500
.0
00000495
.000000415
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-0.5 0.0 0.5 1.0
= 0.20n= 10
-0.5 0.0 0.5
= 0.20
n= 20
-0.2 0.0 0.2 0.4 0.6
= 0.20
n= 50
Figure 1: First simulation experiment (cf. Table 1). Histograms of
(thick line), respectively (thin line), compared with Gaussian distributionsN( sim,
2 sim) (heavy line), respectively N( sim,
2 sim) (light line). The
number of histogram classes follows Scott (1979).
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= 0.50n= 10
= 0.50
n= 50
= 0.50
n= 150
-0.5 0.0 0.5 1.0
0.0 0.5
0.3 0.4 0.5 0.6 0.7
Figure 1: (Continued)
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= 0.90n= 20
= 0.90
n= 100
= 0.90
n= 800
0.5 1.0
0.7 0.8 0.9 1.0
0.85 0.90 0.95
Figure 1: (Continued)
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= 0.98n= 20
= 0.98
n= 200
= 0.98
n= 2000
0.6 0.8 1.0 1.2
0.90 0.95 1.00
0.96 0.97 0.98 0.99
Figure 1: (Continued)
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20/23
= 0.999n= 500
= 0.999
n= 2000
= 0.999
n= 50000
0.98 0.99 1.00 1.01
0.995 1.000
0.9985 0.9990 0.9995
Figure 1: (Continued)
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21/23
0.0 0.2 0.4 0.6 0.8 1.0
0.00
0.05
0.10
- E(^)
n = 10
SQRT[var(^)]
n = 10
0.0 0.2 0.4 0.6 0.8 1.0
0.00
0.10
0.20
0.30
Figure 2: Second simulation experiment. Above: negative bias, below:standard deviation. Simulation results (dots). Theoretical result, our for-mula, (12) and (15), respectively, (heavy line). Theoretical result, Whites(1961) formula, (25) and (26), respectively, (light line).
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22/23
0.00
0.04
0.08
0.0 0.2 0.4 0.6 0.8 1.0
- E(^)
n = 20
0.0 0.2 0.4 0.6 0.8 1.0
0.00
0.10
0.20
SQRT[var(^)]
n = 20
Figure 2: (Continued)
21
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23/23
0.0 0.2 0.4 0.6 0.8 1.0
0.00
0.01
0.02
- E(^)
n = 100
SQRT[var(^)]
n = 100
0.0 0.2 0.4 0.6 0.8 1.0
0.00
0.05
0.10
Figure 2: (Continued)
22