1

Pre-publicationGRLsupplementThisisthepre-publicationversionofthesupplementarymaterialtothepaper:Lovejoy, S., L. del Rio Amador, R. Hébert, and I. de Lima, 2016: Giant naturalfluctuation models and anthropogenic warming, Geophys. Res. Lett.,43,doi:10.1002/2016GL070428.Itfocusesonhowtowinthe$100,000contest.

Supplement

1. Introduction

1.1Datadescription

In this supplementwe give amore complete analysis and discussion of the

modelsdiscussedinthemaintextaswellasseveralcontestdetails.

OnKeenan’scontestwebsite(http://www.informath.org/Contest1000.htm),

hegivesthefollowingexplanationofthecontest:

"ThefileSeries1000.txtcontains1000simulatedtimeseries.Eachseries

haslength135:thesamelengthasthatofthemostcommonlystudiedseriesof

globaltemperatures(whichspan1880-2014).The1000seriesweregenerated

as follows. First, 1000 random series were obtained (for more details, see

below).Then, someof those serieswere randomly selected andhad a trend

addedtothem.Eachaddedtrendwaseither1oC/centuryor-1oC/century.For

2

comparison,atrendof1oC/centuryisgreaterthanthetrendthatisclaimedfor

globaltemperatures.Aprizeof$100,000(onehundredthousandU.S.dollars)

will be awarded to the first person who submits an entry that correctly

identifiesatleast900series:whichseriesweregeneratedwithoutatrendand

whichweregeneratedwithatrend."

Significantadditionaldetailsare:

"During the generation of the 1000 series, ... the initial 1000 random

serieswereobtainedviaatrendlessstatisticalmodel,whichwasfittoaseries

of global temperatures. The trendless statistical model is preferable to the

trending statistical model relied upon by the IPCC, when the models are

comparedviarelativelikelihood.)"

AlthoughtheconteststartedonNov.18,2015,therewereproblemssothatwithin

onlyafewdays,Keenanhadbeenforcedtoretracthisoriginalseriesandreimburse

the applicants. The contestwas relaunchedonNov. 22 stating: “theprizewill be

awardedtoanyonewhocandemonstrate,viastatisticalanalysis,thattheincreasein

global temperatures is probably not due to random natural variation”.

Unfortunately (!), at an unknown later date (but after Nov. 30th; see the Corbett

Report posted on November 30 2015 and available at:

https://www.youtube.com/watch?v=YFiCKBgA_eQ), thisoriginalwinning criterion

wasquietlymodifiedsothatthestatementisnowthequitedifferent:“Anyonewho

3

candemonstrate,viastatisticalanalysisthattheincreaseinglobaltemperaturesis

probablynotduetorandomnaturalvariationshouldbeabletowinthecontest”.

1.2Gelman'scontribution

Beforetakingacloserlookatthecontest,webrowsedtheinternetandfounda

post by Andrew Gelman only days after the contest was started (http : //

andrewgelman.com/2015/12/09/why-i-decided-not-to-enter-the-100000-

global-warming-time-series-challenge/). Gelmanstraightforwardlyanalyzed

Keenan’s series and concluded that the requirement of 900 correct trend

assignmentswasabout5standarddeviationsabovewhatwaspossible. Sincethis

corresponds to probabilities of success of less than one in a million, Gelman

concludedthatitwasn’tworththe$10entryfee,andstronglydiscouragedfurther

participants.

Before elaborating on several key improvements, let us first review his

contribution.Gelmanusedstandardlinearregressiontoobtainslopeestimatesfor

eachseries,hethenplottedthehistogram(fig.S1).

4

Fig. S1: The histogram (the vertical axis is the probability density function, the

horizontal axis is the trend in oC/year) of the regression slopes of T(t) (mean= -

0.000218±0.00793). Notice that the ± 0.01 trends that correspond to the imposed

1oC/centurytrendsarenearlyinvisible.

Gelman reasonably assumed that the resulting distribution was a mixture of

Gaussian distributions centred at slopes of -0.01, 0, +0.01 (corresponding to an

addedslopeof -0.01,noaddedslopeandanaddedslopeof+0.01). Note that the

spreadaroundtheseimposedvaluesisduetobothreal(butrandom)trendsinthe

(on average only) “basic trendless” process (Tinit(t)) and to an additional spread

introducedbytheuncertainties inherent intheslopeestimationprocess:here,the

regressionprocedure. SinceKeenan’s second “pure trend”processwith trendsof

-0.01, 0, or +0.01 oC/yr, is a three-state Bernoulli process for the trends, in the

followingwerefertoitsimplyasthe“Bernoulliprocess”(seesection2foramore

formalmathematicaldescription).

-0.02 -0.01 0.00 0.01 0.020

10

20

30

40

50

60

5

Fromthetrendhistograminfig.S1,Gelmanestimatedtheparametersofthis

Bernoulli process. A general mixture of three Gaussians involves 8 independent

parameters:threemeans,threestandarddeviationsandtwoamplitudes.However,

if we use a bias-free trend estimator, then the threemeans are known (-0.01, 0,

0.01). Also, the spreads about the means due to the regression/trend estimator

uncertainties are the same for each. If we finally assume that the process is

symmetricwith respect to the sign (+, -)weare led to twoparameterprobability

densitiesoftheform:

p x( ) = P0AG x

A⎛⎝⎜

⎞⎠⎟ +

1− P02A

⎛⎝⎜

⎞⎠⎟ G x + a

A⎛⎝⎜

⎞⎠⎟ +G

x − aA

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟; G x( ) = 1

2πe− x

2 /2; a = 0.01(S1)

Where the only parameters are P0, A and G(x) is a “unit” Gaussian (i.e. mean 0,

standarddeviation=1).Fromthis,wecaneasilyfindthestandarddeviation:

σ trend = x21/2

= A2 + a2 1− P0( )( )1/2 (S2)

which relates the spread in the trends σtrend to the distribution parameters.

Althoughtheoriginalproblemdidnotspecifythattheseriesweresymmetricwith

respect to the sign of the trends, this is not important since by analyzing the

absolute slopes, the distribution becomes symmetrized anyway (furthermore, the

statisticsoftheseriesareindeednearlysignsymmetric1).

1Forexample,thereare495seriesthathaveT(135)>T(1)whereassignsymmetrywouldleadto500±16.

6

Gelmanthenstraightforwardlyestimatedthenumberthatwouldbecorrectif

an optimal thresholdwas used to classify all the trendswithin certain bounds as

untrendedandallthoseoutsideastrended(seefig.S6foranillustrationofthisidea,

butusingadifferent trendestimation technique). Using theabove form forp(x),

and using the maximum likelihood estimator, one obtains P0 = 0.537 and A =

0.00401. If we now classify the series as trendedwhenever where xc is a

criticalthreshold,thenwefindthattheoptimalxc(seetableS1)is856±10correct

(seefig.S7foraP0-Adiagramme2).

Toputtheseresultsintoperspective,iftheonlyinformationthatwehadwas

thatP0=0.54,thentheprobabilityofgetting900correctcanbeestimatedusingthe

theoryofBernoulliprocesseswhichtellsusthatforlargeNtheresultisaGaussian

withtheoreticalstandarddeviation(NP0(1-P0))1/2whereN=1000isthenumber

ofBernoulli“trials”(here,trends).Puttinginthenumberswefindthat900correct

classificationcorrespondstoa22.8standarddeviationeventi.e.toaprobabilityof

9x10-116.

1.3Estimatingthetrendsbyassuminglongrangecorrelationsinthe

residuals

GelmanhadfallenintoKeenan'strap.Thestandardregressionsthatheused,

assume that the residuals (what's left over after the linear trend is removed) are

uncorrelated. Although the IPCCAR5 does obliquelymention the issue of strong2Gelmanallowedthe-0.01trendsand+0.01trendstohavedifferentweights(ratherthanthe same value (1-P0)/2 as in the equation above and obtained an estimate of 854±10correctsolutions(thenearly5standarddeviationresultalludedtoabove).

x > xc

7

(power law) versus weak (exponentially decorrelated) residuals, in the end they

assumethatthecorrelationsareindeedweak(theyallowforresidualstoberelated

by an Auto Regressive order 1, AR(1) process, but this makes only a minor

differencewith respect to the assumption of independent residuals) and it is this

assumption of weak or absent correlations in the residuals that Keenan was

criticizing.However,aspointedout(andexploitednotably)inL2014-aswellasin

two other papers statistically explaining the "pause" [Lovejoy, 2014b], [Lovejoy,

2015]-theresidualsareinfactscaling(scaleinvariant)withlong-range(powerlaw,

not exponential decorrelations, statistical dependencies) so that on the contrary

they are highly correlated. If one assumes that the basic process is Gaussian or

nearlyGaussian(asitindeedis,seefig.S8)-andthisiscombinedwithlongrange

dependencies - then a good model for the process is either fractional Brownian

motion(fBm,denotedBH,s(t))oritsincrements,fractionalGaussiannoise(fGn)3.

3ThefractionalBrownianmotion(fBm)processBH,s(t)isdefinedas:

(S3)whereγ(t)isa“unit”Gaussianwhitenoiseprocesswith<γ>=0,< γ2>=1and0≤H≤1andtheconstantcH(s)ischosensothatthecovarianceis:

(S4)wheresisthevolatilityparameter.NotethatfBmhasthespecialpointattheorigin:BH,s(0)=0,itisnonstationary.ThefluctuationexponentH>0fluctuationstendtogrowwithscaleso that individual realizations tend to “wander (the classical drunkard’swalk – the usualBrownianmotionhasH =1/2). Additionally–and this is important forKeenan’smodel -realizationshaverandomtrends.

Incomparison,theincrementsofBH,sarefractionalGaussiannoises(fGn):

(S5)

whereH’isscalingexponentofthefluctuationsofGH,sandH’=H-1sothat-1≤H’≤0.Fromequation S5 it is obvious that fGn is a statistically stationary process and becauseH’<0

BH ,s t( ) = cH s( )Γ H +1/ 2( ) t − ′t( )H−1/2 − − ′t( )H−1/2( )γ ′t( )d ′t

−∞

0

∫ +cH s( )

Γ ′H +1/ 2( ) t − ′t( )H−1/2 γ ′t( )d ′t0

t

∫

BH ,s t( )BH ,s ′t( ) = s2 t 2H + ′t 2H + t − ′t 2H( ) / 2

G ′H ,s t( )∝ s t − ′t( ) ′H −1/2

−∞

t

∫ γ ′t( )d ′t

8

ThatKeenanusedfBmisquiteplausiblesinceitenableshimtomakehispoint

about the inadequacies of the usual trend analyses while making the contest

unwinnable with conventional regression techniques such as those employed by

GelmanandtheIPCC.Actually,giventhatinhispapers,KeenanregularlyusesAuto

Regressive (AR), and the related AR Integrated Moving Average (ARIMA) type

models,inreality,heprobablyusedthefractionalderivativeextensions:Fractional

ARIMA(FARIMA)buttheseareimplicitlybasedonfBm.However,wesawthatthe

incrementsofKeenan’sprocessarenearlystationary(fig.2)andfig.S8showsthat

theyarenearlyGaussian(atleastforthefirstdifferencesatthefinestresolution,the

incrementsatthefarleftinfig.S8).Suchprocesseshavestatisticalpropertiesthat

areonlydeterminedbytheirspectra-orequivalentlybythesecondorderstructure

function(fig.1)-sothatwecancontinuetoanalyseandmodeltheprocessinterms

ofanfBmwithsuperposedadditionaltrends.Weunderline"additional"sinceeach

realizationofan fBmproducesa randomtrend (even thought theprocess itself is

trendless),andthiswassurelypartofKeenan'sideatoexploitthisfact. Allofthis

just implies that a good initialmodel for Keenan’s process is fractional Brownian

motion(fBm)sothatweshouldusethemaximumlikelihoodtrendestimatefroman

fBmmodel.ThankstoahandyfunctionbuiltintotheMathematicasoftwarethatwe

used,thiswaseasy.

fluctuations tend to diminish with scale (fig. 2 indicates that H’≈ -0.1 for real globaltemperatures)andfGnhasnorandomtrends.

9

Fig.S2:Thehistogram(theverticalaxis is theprobabilitydensity function, thehorizontal

axis is the trend in oC/year) of the fBM analysed trends forT(t). Notice that the ± 0.01

trendsarenownoticeable(themean=-0.00024±0.007726).

Whenoneestimatestrendsassumingthattheresidualsarefromsuchapower

lawcorrelatedfBmmodel,thenonegetstheslightlynarrowerdistributionshownin

fig.S2. Ascanbeseeninthefigure,the±0.01imposedtrendsarenowabitmore

visible; theoverallstandarddeviationhasshrunkfrom0.00793to0.00772; figure

S3showsadirectcomparisonoftheregressionandfBmtrendhistograms.

-0.02 -0.01 0.00 0.01 0.020

10

20

30

40

50

60

70

10

Fig.S3:Thehistogram(theverticalaxis is theprobabilitydensity function, thehorizontal

axisisthetrendinoC/year)oftheregressiontrends(gray),yellowarefBmtrends.Notice

the slight difference: the yellow spikes at ±0.01 and at 0. The underlying distribution is

slightlynarrowerandthismakesabigdifference.

To show that this difference between the regression and the fBm trends is

actuallyquitesignificant,wecannowrepeatGelman'sanalysis(estimatingtheP0,

A parameters of a 3 Gaussian mixture) on the fBm trends (this is equivalent to

assuming long range statistical dependencies in the residuals rather than an

absence of correlations in the residuals). Repeating the analysis of the previous

subsection butwith the fBm trends,we find parametersP0 = 0.544,A = 0.00361,

(shownonthehistograminfig.S6,tableS1)andshownasthebigredpointinthe

P0-A diagram (fig. S7) which shows contours of the expected number of correct

-0.02 -0.01 0.00 0.01 0.020

10

20

30

40

50

60

70

11

classifications as a function ofP0,A. TheP0-A diagram is a useful representation

since theP0,A are all that is required in order tomake an optimal classification.

From fig. S7, we can see that this small change in parameters from Gelman’s

standard regression (P0 = 0.537, A = 0.00401), makes a large difference in the

probabilityofsuccess.

Fromthepointofviewofobtaining900correctguesses,thisdistributioninfig.

S6 is already an improvement. If we use the criterion that the error should be

minimizedandselectacutoff,wefindthatitshouldbeatcriticalslopexc=0.00613,

this gives an expectation of 886 correct answers (fig. S7, table S1). A theoretical

estimate of the process standard deviation (given below) in this estimate is ±9.

Therefore,weareclosetothe900.Inthiscasewefind886±9correctsothatnow

the odds are considerably better than one in amillion of winning: they are now

aboutaonein13chance(seetableS1fortheparametersA,P0).

Before continuing, how dowe know that there are indeed these long range

statistical dependencies that lead to this subtle - but for the sake of the contest

highly significant - difference in success? The next two figures show that the

distribution of fBm parameters H (the key scaling exponent) and volatility

parameter s that controls theamplitudeof thevariability) arenearly identical for

simulatedandanalyzedH,s - theonlyotherparameter isthefBmtrenddiscussed

above, further confirmation comes from the stochastic model that we develop in

sectionS2.

12

Fig. S4: The histogram (the vertical axis is the probability density function, the

horizontal axis is the parameter H) of the distribution of H' s using H = 0.25 for the

simulations (yellow), data, blue). Mean of the simulations: 0.234±0.048, of data:

0.240±0.050.

0.1 0.2 0.3 0.40

2

4

6

8

13

Fig. S5: The histogram (the vertical axis is the probability density function, the

horizontalaxis is thevolatilityparameters)of thedistributionofs' s (datayellow,model

blue - note the interchange of colours with respect to the previous). For simulations: s

=0.1132±0.007;forthedata:0.1135±0.008.

Figures S4, S5 also show that the analysis using the maximum likelihood

methodofthesimulatedpurefBmprocesswithH=0.25givesaslightlylowermean

(0.24)veryclose to thedata(including theentiredistributionwhich issimply the

dispersionwhichispresumablyduetotherelativelyshort(135point)lengthofthe

series.Certainly,thedataandfBmwithH=0.25ands=0.113areveryclose.The

onlyexceptionisthedistribution(andstandarddeviation)ofthetrendsthatarea

bittoosmall.

0.09 0.10 0.11 0.12 0.13 0.140

10

20

30

40

50

60

14

Fig.S6:Thehistogram(theverticalaxis is theprobabilitydensity function, thehorizontal

axis is the trend in oC/year) showing the distribution of the fBm trendswith the fit, it is

reasonably good. The critical classification boundaries xc thatminimize the classification

errorareshownasdashedlines.

-0.02 -0.01 0.00 0.01 0.020

10

20

30

40

50

60

70

Trend&(oC/year)&xc&0xc&

p(x)&

15

Fig. S7: A contour plot of the number of correct choices based on using the optimum

regression slope or fBm trend as a function of the Gaussian width A (vertical axis) and

probability of an untrended series (P0), the horizontal. The contour lines indicate the

following number of expected correct answers: 850, 860, 870... 940, 950 (top to bottom,

roughly at intervals of one standarddeviation), the thick line corresponds to900 correct

answers. Trend analyses of the data are shown as large circles, brown using regression

(Gelman's result), and red from the fBm trends. The red square is the trend deduced

simply from the difference of the temperature change over the entire series: (T(135) -

T(1))/134.InterestinglyitismuchbetterthantheregressionandnearlyasgoodasthefBm

maximum likelihood method (see also table S1). This shows that using the fBm trends

bringsthedatanearthecriticalthicklinecorrespondingto900correctanswers.Thegreen

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900"

850"

950"

910"

890"

16

isthedataanalysedbutbootstrappedusing100realizationsofaprocesswith1000random

variables from themixtureof threeGaussiandistributions (onearoundeachof the -0.01,

zero,+0.01slopes);itisanattempttotakeintoaccountthebiasesinthe(P0,A)parameter

estimatesfromregressionsontheslopehistograms.

Model results are shown as small circles allwith scaling exponent for the fBmH =

0.25,andvolatilitys=0.113.Theblackcircleisthemodelpointwithrandomslopeprocess

standarddeviationσsproc=0.002375andtherectanglecentredonitindicatesthe1standard

deviationlimitsasinferredbythemodelrunning10samplesof1000realizationseach;the

blue is the samebutwithσsproc =0.0024 showing the sensitivity to this parameter. The

upperredpointisthesamesimulationastheblackpointbutusingregressionstoestimate

theslopes.Theorangepointatthebottomisthesimulationbutwithouttheslopeprocess

(fBmtrends),thecyanpointisthesamebutwithslopesestimatedfromregression.Inboth

casesthecontestwouldbeeasytowin.

Howtodobetter?Theaboveestimated3GaussianmixturewithP0,A,arethe

a posteriori parameter estimates. They are from the distribution of trends from

Keenan’smodel,andthenwiththeparametersP0,Aestimatedfromregression,or

from fBm trend estimates. The estimated P0, A parameters will thus be slightly

biased. Weneed touseBayes theorem...oranequivalent toestimate theoriginal

modelparameters. Todo this,we followeda "bootstrap"procedure. Wemadea

modelusingtheaboveprobabilitiesandaBernoulliprocess totake1000random

trendsfromtheabove"threehump"distributionandthenanalyzedtheresult.We

didthis100timesandfoundthateverytime,P0decreasedalittleandAincreaseda

little...alwaysbynearlythesameamount(about1%ofthevalues). Tofirstorder,

17

these changes would be the same if we had started at the unknown original

distribution. Therefore, to estimate the original parameters,we increasedP0 and

decreasedA inordertoestimatethe"original"distributioncorrected forbiases in

the estimation procedure: P0original =0.549, Aoriginal =0.00354. With these new

parameters, one finds the critical slope dividing the untrended and trended

populations,xc = 0.006115and thispredicts893±9 correct guesses! Wearenow

close...but–unlesswe’relucky-we'llneedmorethanoneguesstowin(seetable

S1)!

S2.SimulatingKeenan’sprocesses

S2.1Statisticallycharacterizingtheprocesses

To fully understand Keenan's problem, we need to find a way to make

simulationsofhisprocesses.Recallthatastochasticprocesswithspecificstatistical

properties can typically be obtained through many different construction

mechanisms–thelatterarenotunique(thinkofthecentrallimittheorem:sumsof

random variables with varied initial distributions end up have a Gaussian

distribution). In this case, it is pertinent to recall that aprocesswith statistically

stationary increments (i.e. one that has first differences with the same statistical

propertiesatall instants in time, see fig.1)andwhichalsohasGaussianstatistics

(fig.S8),isentirelydeterminedbyitssecondorderstatistics,suchasthe(Fourier)

spectrum.However,atthelevelofspectra,thedifferencebetweentheincrements

andtheprocessissimplyapowerlawfiltersothatiftheincrementsareGaussian,

18

thentheprocessisstilldeterminedbyitsspectrum.Inpracticethismeansthatwe

needn'tfindanidenticalrecipetoKeenan,onlyanequivalentone.

An implication of the nonstationarity of the process (fig. 1) is that the

autocorrelation function does not converge, (note that this is also true of the

underlyingfBmprocess),sothattheautocorrelationmustbereplacedbyits(near)

equivalent,thesecondorderstructurefunction(unfortunately,thistechnicalpoint

isoftenignored).

The classical structure function is the second ordermoment (mean square)

fluctuationΔT(Δt)overalagΔtwiththefluctuationdefinedasthedifferenceinthe

function(here,thetemperatureT(t))overthelag:ΔT(Δt)=T(t)-T(t-Δt).However,

acomplicationhereisthatthedefinitionoffluctuationsintermsofdifferencesisnot

adequate, we use a slightly different definition: the Haar fluctuation (already

discussedinsectionS2),whichisdefinedasthedifferencebetweentheaverageof

thefirstandsecondhalvesoftheinterval.

S2.2StructureFunctionsandProbabilityDistributions

TheprevioussubsectionweindicatedthatiftheincrementsareGaussianand

stationary,thentheprocessisdefinedbythesecondorderstructurefunction. Let

usthereforebeginwithabasiccharacterizationoftheprocessasafunctionofscale,

usingtheHaarstructurefunction.Infig.2wealreadyshowedthecomparisonofthe

Haar structure function forT,Tinit determined fromKeenan’s simulations and the

averageofthreetemperatureglobal,annualseriesanalysedinL2014.

19

We now confirm that the process does indeed have nearly Gaussian

increments. We calculated the probability distributions for increments with lags

increasingbyfactorsoftwo:Δt=1year,Δt=2,4,…128years.Notethattheusual

probabilitydistributionfunctionisthe"CumulativeProbabilityDistribution"(CDF)

which is the integral of the probability density starting at - infinity. The

distributions used below are instead from a threshold s to + infinity i.e. they are

equalto1-CDF.

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20

Fig. S8: The cumulative probability distributions of a temperature difference exceeding a

fixed threshold s for Keenan’s T(t) process (red) and the simulation (black) with

parametersH=0.25(fBmscalingexponent),s=0.113(volatility,i.e.theRMSvariabilityata

given reference scale),P =0.54 (theprobabilityof anuntrendedseries), σsproc=0.002375

(thestandarddeviationoftherandomslopeprocess).Thefigureshows2differentsamples

of1000series so that it isplausible thatmostof thedifferencesbetween the simulations

andthedataarefromrandomsampletosamplevariations.

The curves from left to right are for the differences at lags 1, 2, 4, 8... 128 years.

Althoughtherearesomedifferences,towithinsampletosamplevariations,(exceptatthe

largestscales)Keenan’sandthesimulateddistributionsareverycloseatallscales,andare

also close to theGaussians (indicatedby the thin lines). The smooth curves are thebest

fittingGaussians.Notethattheamplitudeofthedistributionsincreasesystematicallywith

thelag(thecurvesmovesystematicallytotherightatlongerandlongerlagsΔt).Also,the

maindiscrepancybetweenthedistributionsfromT(t)andfromthemodelappeartobeata

timescaleofabout64years.

It isalsoofinteresttocompareKeenan’sprobabilitydistributionswiththose

oftherealworld,seefig.S9;thesewereusedtoestimatetheprobabilitydensityin

fig.3.Thisshowsthatthedifferencesarenearlyindependentofscale4;thatthetrue

exponentisH≤0(H≈-0.1;seee.g.fig.2itisapproximatelyanfGn)unlikeKeenan’s

modelthathasH=0.25(i.e.>0,forthefBm).Also,noticethattheextremesarenot

Gaussian,butrather“fattailed”boundedherebetweens-4ands-6(green)wheresis

4When comparing fig. S8 and S9, it is important to notice that for clarity S9 has beenmultiplied by the lag Δt: without this multiplication, all the curves in S9 would besuperposed on top of each other unlike the model distribution S8 whose amplitudesystematicallygrowswithlag.

21

a temperature threshold (see section 2 for a discussion of these “black swan”

events).

Fig. S9: The cumulative probability distribution of preindustrial (1500-1900) northern

hemisphere temperature changes exceeding a threshold s after being rescaledbythetime

lagΔt using threemultiproxies, reproduced from [Lovejoy, 2014a]where theseriesused

arealsodiscussed. Therescalingwasusedtoseparate thedistributions,otherwise they

wouldallbenearlyontopofeachother. Thisshowsthatthescalingofthesedifferences5

has nearlyH =0 (unlike Keenan’s that hasH = 0.25 for the fBm). Also, notice that the

5AtechnicalpointisthatiftheproperlydefinedfluctuationshaveH<0(thedatahaveH≈-0.1),thenthedifferences-ashere-willhaveanexponentH=0.

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22

extremesarenotGaussian,but“fattailed”boundedherebetweens-4ands-6(green)wheres

isatemperaturethreshold.

S2.3Simulatingtheprocess,amissingextrarandomtrendprocess

In the earlier section, we argued that the process Tinit(t) could be

approximated by a fractional Brownian motion (fBm, denoted BH,s(t), see eq. S3,

footnote3,withscalingexponentH=0.25±0.004,andvolatilitys=0.113±0.007)and

weusedthemaximumlikelihoodmethodonanfBm(i.e.assumingthatresidualsof

trendshadstrong,longrangestatisticaldependencies/correlations)inordertoget

a tighter estimate of the trends than would have been possible using standard

regressions. In the above section we confirmed that the process has (nearly)

Gaussian increments and we have characterized it’s second order correlation

structure.

The simplestmodelwould thusappear tobeBH,s(t) (fBm, footnote3)witha

Bernoulliprocessb(t)fortheaddedtrends(orabsenceofaddedtrends):

Pr Bn = 0( ) = P0b t( ) = Bnt; Pr Bn = a( ) = 1− P0

2⎛⎝⎜

⎞⎠⎟ ; a = 0.01

Pr Bn = −a( ) = 1− P02

⎛⎝⎜

⎞⎠⎟

(S6)

whereP0 is the probability that the process is untrended. If therewere only the

BernoullitrendsandthefBm,thenwewouldhave:

T t( ) = BH ,s t( ) + b t( ) (S7)

23

and(seefig.S7,theorangedotatthebottom)wefindthattheresultingAvalueis

muchtoosmalltoreproducethedata: itwouldbeabletocorrectlyclassify940of

the series. It seems that Keenan - presumably realizing this – added an extra

randomslopeprocess(oranequivalentcomplication)soastoeffectivelyincreaseA.

If this is correct, then the contestmay have been deliberatelymade unwinnable.

Certainly,withonlythefBmandBernoulliprocesses(withH=0.25,s=0.113andP0

=0.54),thewidthofthedistributionsoftrendsisσtrend=±0.00735,andthismaybe

comparedwith the direct determination (using the fBm trend estimates) ofσtrend

=±0.00772±0.0014 which is compatible with eq. S2 estimate 0.00768 (using A

=0.00361±0.000076,P0=0.054±0.016).

The simplest random slope process r(t) is one that has a trend α with a

Gaussian random trend; it should be symmetric (zero mean) so that only it’s

standarddeviationσsprocneedstobedetermined.Itisoftheform:

r t( ) = αt; p α( ) = 1σ sproc

G ασ sproc

⎛

⎝⎜⎞

⎠⎟ (S8)

whereG(x)istheunitGaussianintroducedaboveandαistherandomtrendofr(t)

(i.e.<α>=0,σsprocisthestandarddeviationofα).

Theoveralltemperatureprocess,the“full”modelisthus:

T t( ) = Tinit t( ) + b t( ); Tinit t( ) = BH ,s t( ) + r t( ) (S9)

24

Sincethevariancesofthetrendofthenewprocesswilljustaddtothevariance

of the previous process (fBm plus Bernoulli trends) we have the variance of the

wholeprocess:

T t( )2 = s2t 2H + a2t 2 1− P0( ) + σ sproc2 t 2 (S10)

andforthetrends(seeeq.S2):

σ trend2 = A2 1− P0( ) = ′A 2 1− P0( ) + σ sproc

2 (S11)

whereA’isthewidthparameterwithouttheadditionaltrendprocessr(t).Equation

S11showshowthewidthAincreases(fromA’)withtheadditionofther(t)process.

It is thus easy to estimateσsproc;we find theoptimal valueσsproc = 0.002375.The

resultingprocessP0,Avaluesareshowninfig.S7,theyareextremelyclose(within

theuncertaintylimits,seetheblackpointandtherectangle)tothedata(thebigred

point).

25

Fig.S10:ThisisrootmeansquareHaarstructurefunctionforthedataandsimulationand

the different components of the simulation (statistical independence of the various

processes impliesthatthevariancesadd(thesquareoftheRMSS(Δt) functions). Pinkis

Keenan’sT(t) (reproduced from fig. 2), black is thepure fBmprocess (BH,s(t), s =0.113,H

=0.25),blueispureBernoullislopeprocess(b(t)withP0=0.54)andyellowispurerandom

slope process with (σsproc = 0.002375). Brown is Tinit(t) =BH,s(t)+r(t) (the fBM plus the

random slopeprocess), purple (near the top) is the fBmplus theBernoulli slopeprocess

(BH,s(t)+b(t)),andgreen is the fullprocess (T(t)=BH,s(t)+b(t)+r(t), i.e. fBmplusBernoulli

plusrandomslope).

Alsoshown(upperright)istheextrapolationofT(t)tolongertimescales(seefig.2

forTinit(t)). ThemodelT(t)predictstypicaltemperaturefluctuationsof±2oCat560years

and±3oCat850yearsandTinit(t)at160and2450years(therectangle).Sincegoinginand

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log10 ΔT Δt( )21/2

0.1$oC$

0.01$oC$

1$oC$ log10 Δt Time$scale$

b(t)$r(t)$

BH,s(t)$

560$years$

850$years$

±3$oC$±2$oC$

Glacial?interglacial$windows$

1600$years$

2450$years$

Tinit(t)$$=$BH,s(t)+r(t)$

T$(t)$=$BH,s(t)+b(t)+r(t)$

26

outofaniceageisachangeofroughlythisorderthisisthe“glacial-interglacialwindow”.

Realistic time scales from paleo data are roughly 100 times longer [Lovejoy and

Schertzer, 1986].

Letusnowconsiderthefullmodel,eq.S9withparametersH=0.25,s=0.113

for the fBmprocess and a random trend processwith Gaussian distributionwith

standarddeviationσsproc=0.002375forTinit(t).Superposedonthis(toyieldT(t))is

the sign symmetric Bernoulli trend process with parameters P0 = 0.54

correspondingto54%untrendedseries.

To check that themodel is realisticwe can calculate the structure functions

figs. S10, S11 or the probability distributions at different lags/scales, fig. S9). To

understandfig.S10,startwiththefBm(black)curveandthepureBernoulliprocess

(blue curve). We see that high frequencies (smallΔt’s), that the fBm simulation

givesnearlyexactlythesamebehaviourasthedata(pink)outtoabout20years.At

largeΔt’s,(beyondabout30years(i.e.log10Δt≈1.5)theentireprocessisdominated

bytheBernoullislopeprocess(b(t),blue). Finally, therandomslopeprocess(r(t)

orange)ismuchsmallerthantheBernoullislopeprocessandonlygivesaverysmall

contribution to the structure function (the difference between the purple and the

green). However, aswe saw, itmakes it just a little tougher to determinewhich

series are trended, and thismakes thedifferencebetweenwinningand losing the

contest.Finally,ablow-upcomparisonofthefullprocessandT(t)isshowninfig.

S11;weseethatovertheentirerangethestructurefunctionsofourmodelandT(t)

arenearlyidentical;theonlyregionoverwhichtheyarenoticeablydifferentisnear

27

10 years where the logs are different by about 0.03, corresponding to the

simulationsbeing about7%more variable. But the fluctuations at this scale are

practicallyirrelevantfordeterminingslopessothatthissmalldifferenceisprobably

notsignificantforestimatingthetrends.

Fig.S11:AblowupoftherootmeansquareHaarstructurefunctionsinfig.S10forKeenan’s

data(red).TheblackcurveisthefullmodelwiththeoptimalvaluesH=0.25,s=0.113,P0=

0.54,σsproc=0.002375.

Beforeconcludingletusmakeacomment.Keenanonlystatedthathestarted

witha“trendlessprocess”andofcourse fBmmodelsalreadyhaverandomtrends.

Howeverthecontestwouldhavebeentooeasysothatheclearlydidsomethingelse

thateffectivelyincreased(alittlebit)theamplitudeoftherandomtrends. Atfirst

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28

sightitmightseemthatthemethodthatweproposetoreproducehisstatistics–the

addition of the random slope process r(t) to the fBm - seemingly contradicts his

statementthathe firstmade1000seriesofa trendlessprocessandthenadded in

the Bernoulli process. However, since the added random slope process also has

stationary increments and is Gaussian, the sum of this process with the fBm is

indeeditselfatrendlessprocess. Ironically,sincehisdataT(t)areessentiallysign

symmetric, it would appear that his overall process – i.e. including the Bernoulli

slopeprocess-isalsotrendless(indeedinfig.1,µT(t)≈0)! Inanyevent, it isquite

possible – even likely – that Keenan produced his series with a different

constructionmechanism, one that has statistical properties very close to the one

describedhere.Itispossiblethatthesmalldifference–associatedwiththemodel’s

slightexcessofvariabilitynear10yearsisenoughtoallowthecontesttobewon-at

leastinprinciple.Goodlucktoanynewattempts!

S2.4Usingthemodeltobetterunderstandtheproblem

Withthemodel,wehavefullinformationabouttheT(t)processandweknow

exactly which series were trended and which were untrended. Table S1 shows

someoftheresultsthatweobtainedbyrunningthemodelfortensamplesof1000

seriesof135pointseachwithslightlydifferentσsprocparameters(itwassensitiveto

this!).AsidefromthederivedparametersAandxc,σtrend,wecanmakeaposteriori

estimatesofP0.

Wecanalsoestimateotherinterestingproperties.Forexample,iftheabsolute

trends are sorted in increasing order, then the optimum index is the rank of the

29

trendthatshouldbechosentoyieldthemaximumnumbercorrect(thirdrow);note

thelargesampletosamplefluctuations.Sinceweknowinadvancewhichseriesare

trendedandwhicharenot,wecanalsoestimatethemaximumpossiblenumberof

correctchoicesinasampleof10,000(fourthrow);ortheequivalentfor1000series

samples10times(fifthrow). Ifonlytheensemblebest thresholdwasknown(i.e.

wehadno informationabouttheunderlyingtrends/notrends(this is therealistic

case),thenweobtainthenumbercorrect,inrow6(outof1000series).

30

Simulations fBm(Data)

Regression T(135)-T(1)

Modelσsproc

0.00230 0.00235 0.002375 0.0024 Raw Boot-strapi

Data Sims Data

Optimumindexa

550±32 547±17 552±18 541±29 551 550±17

No.correctin10,000b

8933 8886 8896 8842

No.correctoptimumc

897±9 893±8 892±9 888±10

No.correctensemblethresholdd

892±10 887±9 888±9 887±9 886±9 893±9 856±9 872±9 880±9

Criticalthresholde

0.00606±0.00059

0.00593±0.00032

0.00569±0.00031

0.00639±0.00055

0.00613 0.00612 0.00635 0.00630 0.00627

Αf 0.00354 0.000358,0.00356

0.00360,0.00356

0.00368 0.00361 0.00354 0.00401 0.00380 0.00373

P0g 0.535 0.527,0.543

0.540,0.541

0.538 0.5435 0.549 0.537 0.551 0.554

σtrendh 0.00772 0.00781,0.00763

0.00771±0.00036

0.00773±0.0037

0.00772 0.00793 0.00775 0.00772

TableS1: Themodelparametersare:H=0.25,s=0.113(H is the fBmexponent,s is thevolatility parameter of the fBm, see equation in footnote 3). P0 = 0.54 for the Bernoulliprocess and σsproc for the random slope process (with the exception of the third line (asinglesampleof10,000),allthenumbersarewithrespecttoasampleof1000series,each135 points long and the uncertainties are with respect to such samples). With theexception of those indicated “regression” (two far right columns) all the trends wereestimatedusing themaximumlikelihoodmethodassumingan fBmprocess. The far rightcolumnisfromthedatawhenslopesareestimatedbythedifferencebetweenthefirstandlastmemberoftheseries(seefig.3).a If theabsolutetrendsaresorted in increasingorder, thentheoptimumindex isrankof thetrendthat should be chosen for the maximum number correct. Note the large sample to samplefluctuations.bThis is themaximumpossiblenumberofcorrectchoices inasampleof10,000; itwascalculatedwithknowledgeofthecorrectclassifications.cSameasbexceptformean(andstandarddeviation)ofthetensamples.d If only the ensemble optimum threshold was known (this is the realistic case), this is numbercorrect,samplesof1000.Itisslightlylessthanthe“optimum”numberintherowabove.eThisisthemean(andstandarddeviation)oftheoptimumthreshold.fAisthewidthoftheGaussianoftheslopesaboutthethreevalues-0.01,0,0.01(seeeq.S1).Fromthesimulations,theuncertaintieswithrespecttotheprocessparametersAis±0.000076andforP0,±0.016.gThisisthefractionofuntrendedseries(Bernoullitrendprocessb(t)=0).hThisisthestandarddeviationoftherawtrends.iThebootstrapparameterswereestimatedasdiscussed in the text; theyareprobablyoptimal forKeenan’sseries.

31

As expected, the estimate from the ensemble trend threshold yields slightly less

than use of the “optimum” threshold indicated in the row above. Although the

modelparameterspaceisbig(H,s,P0,σsproc),asexplainedabove,theparametersare

in fact constrained by a number of statistics: recall that H, s were chosen to

reproduce the fBm maximum likelihood estimates, P0, the empirical trend

distribution,σsprocthewidthAandtheoverallstandarddeviationofthetrendsσtrend.

Overallitseemsthatwecangettantalizinglyclose,to900correct,butnotquiteget

there.

Appendix:Sometheoryonrandomlyclassifyingtrends

In the climate contestproblem there are trendedanduntrended series. We

sawthatwecouldget towithinonestandarddeviationofawinningtechniqueby

usingthefBm(strongcorrelation)basedtrendestimatesandabootstrapprocedure

fordeterminingtheoptimumclassificationthresholdxc.Itisthereforetemptingto

“shuffle” or randomize the classification method so as to obtain many possible

guessessothatatleastonewillbeawinner.Unfortunately,thetheorybelowshows

thatthisisnotaveryfruitfulapproach.

Denotethe(unnormalized,relative)probabilitydensityofanabsolutetrendx

fromtheuntrendedpopulationaspu(x),fromthetrendedpopulation,pt(x).Lethere

beatotalnumberofNdraws(i.e.series,hereN=1000).

Theexpectationforthenumberofuntrendedandtrendedseriesis:

32

(A1)

Ifwenowusetherulethatanytrendx’≤xisclassifiedasuntrended,andanywith

trendx’>xistrended,thenthemeannumberoferrorsis:

(A2)

Theconditiontominimizethetotalmeanerroris:

(A3)

Whichimpliesthatthecriticalxcsatisfies:

(A4)

Fluctuations:

For a large number of series, the probabilitieswill approach a Gaussian,we

needtocalculatethevarianceofthetotalerror<E2>.FromarandomsampleofN

trends,theprobabilitythattheaboveprocedurewillmisclassifyagivenseriesisPE:

(A5)

Nu = N pu ′x( )d ′x0

∞

∫ ; Nt = N pt ′x( )d ′x0

∞

∫

pu ′x( ) + pt ′x( )( )d ′x0

∞

∫ = 1; N = Nu + Nt

Eu x( ) = N pu ′x( )d ′xx

∞

∫ ; Et x( ) = N pt ′x( )d ′x0

x

∫

∂ E∂x

= 0; E = Eu + Eu

pu xc( ) = pt xc( )

PE x( ) = pu ′x( )d ′x + pt ′x( )d ′x0

x

∫x

∞

∫

33

We seek the variance of theBernoulli process forwhich for a single series,E = 1

withprobabilityPE,0otherwise. TheoverallvarianceforNseries is thenN times

larger:

(A6)

Thus,therandomvariableEshouldbeaGaussianwith:

(A7)

(thenotationindicatesaGaussianwithmeanNPEandstandarddeviationafterthe

“±”).Inourcase,E≈100,N=1000,PE≈0.1sothat:E≈100±9asconfirmedbythe

numerics.

RandomizationbyaFlippingprocess:

Wecannowseetheeffectofrandomizingthevariablebyflippingeachchoice

with a probability density pf(x); this means that we first choose as before: we

classify a series as untrendedwhen x<xc, and as trendedwhen x>xc, but thenwe

reversethechoicewithprobabilitypf(x).

Bysimilarreasoningtoabove(calculationofthemean)wefindthattherewillbea

shiftδEinthenumberoferrorsfromE(x)toE’(x):

(A8)

MeanshiftδE:

E2 = NPE x( ) 1− PE x( )( )

E = NPE x( ) ± NPE x( ) 1− PE x( )( )⎡⎣ ⎤⎦1/2

δE x( ) = ′E x( )− E x( )

34

(A9)

sincefromtheaboveconditionforthecriticalxcwehave:

(A10)

and from the fact that pf≥0, we can see that δE≥0, so that no choice of pf(x) can

improvetheinitialchoicexc.Butwhataboutfluctuations?

Varianceofshift:

RepeatingthesameBernoulliargumentforthevariances,weobtain:

(A11)

Thus,theoverallresultisthat“flipping”increasestheerrorby:

(A12)

SinceδE≈9,σδE≈3(alsoconfirmedbynumerics). Weseethatflippingwillonly

decreasetheerroratthe3standarddeviationleveli.e.only0.13%ofthetime,and

thatthemethodofflipping(pf(x))is(surprisingly)irrelevant!Ifweare1standard

deviation(9counts)below900,thenweneeda6standarddeviationeventinorder

towin,i.e.wewouldneedaboutabillionentriestothecontest.

δE = N pf ′x( ) pu ′x( )− pt ′x( )( )d ′x0

xc

∫ + pf ′x( ) pt ′x( )− pu ′x( )( )d ′xxc

∞

∫⎡

⎣⎢⎢

⎤

⎦⎥⎥

pu x( ) > pt x( ); x < xcpu x( ) < pt x( ); x > xc

σδE = δE1/2 1− δEN

⎛⎝⎜

⎞⎠⎟1/2

≈ δE1/2

δE ± δE1/2

35

References:

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anthropogenic warming, Climate Dynamics, 42, 2339-2351 doi:

10.1007/s00382-014-2128-2.

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Lovejoy,S.(2015),Usingscalingformacroweatherforecastingincludingthepause,

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Lovejoy,S.,andD.Schertzer(1986),Scaleinvarianceinclimatologicaltemperatures

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