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2.6. LLNANDCLT
141
Proof TheproofofKolmogorov’sstronglawisnontrivial–see,forexample,theorem8.3.5of
[Dud02]
Ontheotherhand,wecanproveaweakerversionoftheLLNveryeasilyandstillgetmostofthe
intuition
Theversionweproveisasfollows: IfX
1
,...,X
n
isIIDwithEX
2
i
<¥,then,forany>0,we
have
Pfj
¯
X
n
mjeg!0 as n!¥
(2.20)
(Thisversionisweakerbecauseweclaimonlyconvergenceinprobabilityratherthanalmostsure
convergence,andassumeaﬁnitesecondmoment)
Toseethatthisisso,ﬁxe>0,andlets
2
bethevarianceofeachX
i
RecalltheChebyshevinequality,whichtellsusthat
Pfj
¯
X
n
mje
g
E[(
¯
X
n
m)
2
]
e2
(2.21)
Nowobservethat
E[(
¯
X
n
m)
2
]=E
8
<
:
"
1
n
n
å
i=1
(X
i
m)
#
2
9
=
;
=
1
n2
n
å
i=1
n
å
j=1
E(X
i
m)(X
j
m)
=
1
n
2
n
å
i=1
E(X
i
m)
2
=
s
2
n
Herethecrucialstepisatthethirdequality,whichfollowsfromindependence
Independencemeansthatifi6=j,thenthecovariancetermE(X
i
m)(X
j
m)dropsout
Asaresult,n
2
Combiningourlastresultwith(2.21),wecometotheestimate
Pfj
¯
X
n
mjeg
s
2
ne2
(2.22)
Theclaimin(2.20)isnowclear
Ofcourse,ifthesequenceX
1
,...,X
n
iscorrelated,thenthecross-producttermsE(X
i
m)(X
j
m)
arenotnecessarilyzero
Whilethisdoesn’tmeanthatthesamelineofargumentisimpossible,itdoesmeanthatifwewant
asimilarresultthenthecovariancesshouldbe“almostzero”for“most”oftheseterms
Inalongsequence,thiswouldbetrueif,forexample,E(X
i
m)(X
j
m)approachedzerowhen
thedifferencebetweeniandjbecamelarge
T
HOMAS
S
ARGENTAND
J
OHN
S
TACHURSKI
April20,2016
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2.6. LLNANDCLT
142
Inotherwords,theLLNcanstillworkifthesequenceX
1
,...,X
n
hasakindof“asymptoticin-
dependence”,inthesensethatcorrelationfallstozeroasvariablesbecomefurtherapartinthe
sequence
Thisideaisveryimportantintimeseriesanalysis,andwe’llcomeacrossitagainsoonenough
Illustration Let’snowillustratetheclassicalIIDlawoflargenumbersusingsimulation
Inparticular,weaimtogeneratesomesequencesofIIDrandomvariablesandplottheevolution
of
¯
X
n
asnincreases
Belowisaﬁgurethatdoesjustthis(asusual,youcanclickonittoexpandit)
ItshowsIIDobservationsfromthreedifferentdistributionsandplots
¯
X
n
againstnineachcase
ThedotsrepresenttheunderlyingobservationsX
i
fori=1,...,100
Ineachofthethreecases,convergenceof
¯
X
n
tomoccursaspredicted
Theﬁgurewasproducedbyillustrates_lln.jl,whichisshownbelow(andcanbefoundinthe
lln_cltdirectoryoftheapplicationsrepository)
T
HOMAS
S
ARGENTAND
J
OHN
S
TACHURSKI
April20,2016
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2.6. LLNANDCLT
143
The three e distributions s are e chosen n at t random m from a a selection stored d in the e dictionary
distributions
#=
Visual illustration of f the law of large e numbers.
@author : Spencer Lyon <spencer.lyon@nyu.edu>
References
----------
Based off the original python n file illustrates_lln.py
=#
using PyPlot
using Distributions
100
srand(42# reproducible results
# == Arbitrary y collection n of distributions == #
distributions {"student's t with 10 degrees of f freedom" => TDist(10),
"beta(2, 2)" => Beta(2.02.0),
"lognormal LN(0, 1/2)" => LogNormal(0.5),
"gamma(5, 1/2)" => Gamma(5.02.0),
"poisson(4)" => Poisson(4),
"exponential with lambda = 1" => Exponential(1)}
# == Create a a figure e and some axes == #
num_plots 3
fig, axes plt.subplots(num_plots, 1, figsize=(1010))
bbox [0.1.021..102]
legend_args {:ncol => 2,
:bbox_to_anchor => bbox,
:loc => 3,
:mode => "expand"}
for ax in axes
dist_names collect(keys(distributions))
# == Choose e a a randomly selected distribution == #
name dist_names[rand(1:length(dist_names))]
dist pop!(distributions, name)
# == Generate n draws from the distribution == #
data rand(dist, n)
# == Compute sample mean at each n == = #
sample_mean Array(Float64, n)
for i=1:n
sample_mean[i] mean(data[1:i])
end
T
HOMAS
S
ARGENTAND
J
OHN
S
TACHURSKI
April20,2016
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2.6. LLNANDCLT
144
# == Plot == #
ax[:plot](1:n, data, "o", color="grey", alpha=0.5)
axlabel LaTeXString("\$\\bar{X}_n\$ for \$X_i \\sim\$ $name") ax[:plot](1:n, sample_mean, "g-", lw=3, alpha=0.6, label=axlabel) mean(dist) ax[:plot](1:n, ones(n)*m, "k--", lw=1.5, label=L"$\mu$") ax[:vlines](1:n, m, data, lw=0.2) ax[:legend](;legend_args...) end InﬁniteMean WhathappensiftheconditionEjXj < ¥inthestatement oftheLLNisnot satisﬁed? Thismightbethecaseiftheunderlyingdistributionisheavytailed—thebestknownexampleis theCauchydistribution,whichhasdensity f(x)= 1 p(1+x2) (x2R) Thenextﬁgureshows100independentdrawsfromthisdistribution Noticehowextremeobservationsarefarmoreprevalentherethanthepreviousﬁgure Let’snowhavealookatthebehaviorofthesamplemean Herewe’veincreasednto1000,butthesequencestillshowsnosignofconverging Willconvergencebecomevisibleifwetakenevenlarger? Theanswerisno Toseethis,recallthatthecharacteristicfunctionoftheCauchydistributionis f(t)=Ee itX = Z e itx f(x)dx=e jtj (2.23) T HOMAS S ARGENTAND J OHN S TACHURSKI April20,2016 C# PDF Convert to Word SDK: Convert PDF to Word library in C#.net PDF document, keeps the elements (like images, tables and chats) of original PDF file and maintains the original text style (including font, size, color, links .pdf printing in thumbnail size; change paper size pdf Generate Barcodes in Web Image Viewer| Online Tutorials Select "Generate" to process barcode generation; Change Barcode Properties. Select "Font" to choose human-readable text font style, color, size and effects; change font size in pdf form field; pdf reduce file size 2.6. LLNANDCLT 145 Usingindependence,thecharacteristicfunctionofthesamplemeanbecomes Ee it ¯ X n =Eexp ( i t n n å j=1 X j ) =E n Õ j=1 exp i t n X j = n Õ j=1 Eexp i t n X j =[f(t/n)] n Inviewof(2.23),thisisjuste jtj Thus,inthecaseoftheCauchydistribution,thesamplemeanitselfhastheverysameCauchy distribution,regardlessofn Inparticular,thesequence ¯ X n doesnotconvergetoapoint CLT Nextweturntothecentrallimittheorem,whichtellsusaboutthedistributionofthedeviation betweensampleaveragesandpopulationmeans StatementoftheTheorem Thecentrallimittheoremisoneofthemostremarkableresultsinall ofmathematics IntheclassicalIIDsetting,ittellsusthefollowing:IfthesequenceX 1 ,...,X n isIID,withcommon meanmandcommonvariances 2( 0,¥),then p n( ¯ X n m) d !N(0,s 2 as n!¥ (2.24) Here d !N(0,s2)indicatesconvergenceindistributiontoacentered(i.e,zeromean)normalwith standarddeviations T HOMAS S ARGENTAND J OHN S TACHURSKI April20,2016 VB.NET Image: Visual Basic .NET Guide to Draw Text on Image in . Please note that you can change some of the example, you can adjust the text font, font size, font type (regular LoadImage) Dim DrawFont As New Font("Arial", 16 change file size of pdf; advanced pdf compressor online Generate Image in .NET Winforms Imaging Viewer| Online Tutorials Click "Generate" to process barcode generation; Change Barcode Properties. Click "Font" to choose human-readable text font style, color, size and effects. can a pdf file be compressed; pdf optimized format 2.6. LLNANDCLT 146 Intuition ThestrikingimplicationoftheCLTisthatforanydistributionwithﬁnitesecondmo- ment,thesimpleoperationofaddingindependentcopiesalwaysleadstoaGaussiancurve Arelativelysimpleproofofthecentrallimittheoremcanbeobtainedbyworkingwithcharacter- isticfunctions(see,e.g.,theorem9.5.6of[Dud02]) Theproofiselegantbutalmostanticlimactic,anditprovidessurprisinglylittleintuition InfactalloftheproofsoftheCLTthatweknowaresimilarinthisrespect Whydoesaddingindependentcopiesproduceabell-shapeddistribution? PartoftheanswercanbeobtainedbyinvestigatingadditionofindependentBernoullirandom variables Inparticular,letX i bebinary,withPfX i =0g=PfX i =1g=0.5,andletX 1 ,...,X n beindepen- dent ThinkofX i =1asa“success”,sothatY n = å n i=1 X i isthenumberofsuccessesinntrials ThenextﬁgureplotstheprobabilitymassfunctionofY n forn=1,2,4,8 Whenn=1,thedistributionisﬂat—onesuccessornosuccesseshavethesameprobability Whenn=2wecaneitherhave0,1or2successes Noticethepeakinprobabilitymassatthemid-pointk=1 Thereasonisthattherearemorewaystoget1success(“failthensucceed”or“succeedthenfail”) thantogetzeroortwosuccesses Moreover,thetwotrialsareindependent,sotheoutcomes“failthensucceed”and“succeedthen fail”arejustaslikelyastheoutcomes“failthenfail”and“succeedthensucceed” T HOMAS S ARGENTAND J OHN S TACHURSKI April20,2016 2.6. LLNANDCLT 147 (Iftherewaspositivecorrelation,say,then“succeedthenfail”wouldbelesslikelythan“succeed thensucceed”) Here,alreadywehavetheessenceoftheCLT:additionunderindependenceleadsprobability masstopileupinthemiddleandthinoutatthetails Forn=4andn=8weagaingetapeakatthe“middle”value(halfwaybetweentheminimum andthemaximumpossiblevalue) Theintuitionisthesame—therearesimplymorewaystogetthesemiddleoutcomes Ifwecontinue,thebell-shapedcurvebecomesevermorepronounced Wearewitnessingthebinomialapproximationofthenormaldistribution Simulation1 SincetheCLTseemsalmostmagical,runningsimulationsthatverifyitsimplica- tionsisonegoodwaytobuildintuition Tothisend,wenowperformthefollowingsimulation 1. ChooseanarbitrarydistributionFfortheunderlyingobservationsX i 2. GenerateindependentdrawsofY n := p n( ¯ X n m) 3. Usethesedrawstocomputesomemeasureoftheirdistribution—suchasahistogram 4. ComparethelattertoN(0,s 2) Here’ssomecodethatdoesexactlythisfortheexponentialdistributionF(x)=e lx (PleaseexperimentwithotherchoicesofF,butrememberthat,toconformwiththeconditionsof theCLT,thedistributionmusthaveﬁnitesecondmoment) #= Visual illustration of f the central limit theorem @author : Spencer Lyon <spencer.lyon@nyu.edu> References ---------- Based off the original python n file illustrates_clt.py =# using PyPlot using Distributions # == Set parameters == # srand(42# reproducible results 250 # Choice of n 100000 # Number of draws of Y_n dist Exponential(1./2.# Exponential distribution, lambda = 1/2 mu, s mean(dist), std(dist) # == Draw underlying RVs. Each row contains a draw of X_1,..,X_n == = # data rand(dist, (k, n)) T HOMAS S ARGENTAND J OHN S TACHURSKI April20,2016 2.6. LLNANDCLT 148 # == Compute mean of each row, producing g k k draws of \bar X_n == # sample_means mean(data, 2) # == Generate e observations s of Y_n == # sqrt(n) (sample_means .- mu) # == Plot == # fig, ax subplots() xmin, xmax = -s, s ax[:set_xlim](xmin, xmax) ax[:hist](Y, bins=60, alpha=0.5, normed=true) xgrid linspace(xmin, xmax, 200) ax[:plot](xgrid, pdf(Normal(0.0, s), xgrid), "k-", lw=2, label=LaTeXString("\$N(0, \\sigma^2=$(s^2))\$"))
ax[:legend]()
The
ﬁle
is
illustrates_clt.jl,
from
the
applications
repository
https://github.com/QuantEcon/QuantEcon.applications
Theprogramproducesﬁguressuchastheonebelow
YoucanalsoexperimentwithotherspeciﬁcationsofF
Simulation2 Ournextsimulationissomewhat liketheﬁrst, exceptthatweaimtotrackthe
distributionofY
n
:=
p
n(
¯
X
n
m)asnincreases
Inthesimulationwe’llbeworkingwithrandomvariableshavingm=0
Thus,whenn=1,wehaveY
1
=X
1
,sotheﬁrstdistributionisjustthedistributionoftheunder-
lyingrandomvariable
Forn=2,thedistributionofY
2
isthatof(X
1
+X
2
)/
p
2,andsoon
T
HOMAS
S
ARGENTAND
J
OHN
S
TACHURSKI
April20,2016