Psych156A/Ling150:
AcquisitionofLanguageII
Announcements
Reviewquestionsavailableforwordmeaning
HW1returned
Lecture8
WordmeaningI
BeworkingonHW2(due5/5/16)
Midtermreviewinclasson4/28/16
Midtermexamduringclasson5/3/16
Whatdoes“gavagai”mean?
Gavagai!
Whatdoes“gavagai”mean?
Rabbit?
Mammal?
grayrabbit?
Animal?
Carroteater?
vegetarian?
Ears?
Longears?
Isitgray?
Fluffy?
Whatacutie!
Thumping
Hopping
Scurrying
Whatdoes“gavagai”mean?
hNp://www.thelingspace.com/episode-35
https://www.youtube.com/watch?v=Ci-5dVVvf0U
~2:03-2:32
Stay!
Look!
Meal!
Rabbitonlyuntileaten!
Cheeksandleftear!
That’snotadog!
Sameproblemthechildfaces
Alittlemorecontext…
“Look!There’sagoblin!”
Goblin=????
Themappingproblem
Evenifsomethingisexplicitlylabeledintheinput(“Look!There’sa
goblin!”),howdoesthechildknowwhatspecificallythatwordrefers
to?(Isitthehead?Thefeet?Thestaff?Thecombinationofeyesand
hands?Attachedgoblinparts?…)
Quine(1960):Aninfinitenumberofhypothesesaboutwordmeaningare
possiblegiventheinputthechildhas.Thatis,theinput
underspecifiestheword’smeaning.
Sohowdochildrenfigureitout?Obviously,theydo….
Evenby6to9months,infantsrecognizemanyfamiliarwordsin
theirlanguage,likebodypartsandfooditems—thatis,
concreteobjects(Bergelson&Swingley2012,2015).
eyes,mouth,hands,…
milk,spoon,juice,cookie,…
Sohowdochildrenfigureitout?Obviously,theydo….
By10to13monthsold,infantsunderstandwordslike“all
gone”,“hug”,“bye”,and“wet”(Bergelson&Swingley2013)
gone,hug,bye…
wet
Computationalproblem
“Ilovemydaxandmykleeg.”
dax=??
kleeg=??
Onesolution:Fastmapping
Onesolution:Fastmapping
Childrenbeginbymakinganinitialfastmappingbetweenanewword
theyhearanditslikelymeaning.Theyguess,andthenmodifythe
guessasmoreinputcomesin.
Childrenbeginbymakinganinitialfastmappingbetweenanewword
theyhearanditslikelymeaning.Theyguess,andthenmodifythe
guessasmoreinputcomesin.
Experimentalevidenceoffastmapping
(Carey&Bartlett1978,Dollaghan1985,Mervis&Bertrand1994,
Medina,Snedecker,Trueswell,&Gleitman2011)
Experimentalevidenceoffastmapping
(Carey&Bartlett1978,Dollaghan1985,Mervis&Bertrand1994,
Medina,Snedecker,Trueswell,&Gleitman2011)
ball
bear
ball
kitty
bear
[unknown]
[unknown]
Onesolution:Fastmapping
Aslightproblem…
Childrenbeginbymakinganinitialfastmappingbetweenanewword
theyhearanditslikelymeaning.Theyguess,andthenmodifythe
guessasmoreinputcomesin.
Experimentalevidenceoffastmapping
(Carey&Bartlett1978,Dollaghan1985,Mervis&Bertrand1994,
Medina,Snedecker,Trueswell,&Gleitman2011)
ball
bear
“CanIhavethezib?”
kitty
20months
[unknown]
“CanIhavetheball?”
kitty
“…notallopportunitiesforwordlearningareasunclutteredasthe
experimentalsettingsinwhichfast-mappinghasbeendemonstrated.In
everydaycontexts,therearetypicallymanywords,manypotential
referents,limitedcuesastowhichwordsgowithwhichreferents,and
rapidattentionalshiftsamongthemanyentitiesinthescene.”-Smith&
Yu(2008)
Aslightproblem…
“…manystudiesfindthatchildrenevenasoldas18monthshavedifficulty
inmakingtherightinferencesabouttheintendedreferentsofnovel
words…infantsasyoungas13or14months…canlinkanametoanobject
givenrepeatedunambiguouspairingsinasinglesession.Overall,
however,theseeffectsarefragilewithsmallexperimentalvariationsoften
leadingtonolearning.”-Smith&Yu(2008)
Cross-situationallearning
Differentapproach:infantsaccruestatisticalevidenceacrossmultipletrialsthatare
individuallyambiguousbutcanbedisambiguatedwhentheinformationfromthe
trialsisaggregated.
Howdoeslearningwork?
Howdoeslearningwork?
Bayesianinferenceisoneway.
Bayesianinferenceisoneway.
InBayesianinference,thebeliefinaparticularhypothesis(H)(ortheprobabilityof
thathypothesis),giventhedataobserved(D)canbecalculatedthefollowing
way:
InBayesianinference,thebeliefinaparticularhypothesis(H)(ortheprobabilityof
thathypothesis),giventhedataobserved(D)canbecalculatedthefollowing
way:
P(H|D)=
P(D|H)*P(H)
P(H|D)=
P(D|H)*P(H)
P(D)
P(D)
PosteriorprobabilityofhypothesisH,giventhatdataDhavebeenobserved
Howdoeslearningwork?
Howdoeslearningwork?
Bayesianinferenceisoneway.
Bayesianinferenceisoneway.
InBayesianinference,thebeliefinaparticularhypothesis(H)(ortheprobabilityof
thathypothesis),giventhedataobserved(D)canbecalculatedthefollowing
way:
InBayesianinference,thebeliefinaparticularhypothesis(H)(ortheprobabilityof
thathypothesis),giventhedataobserved(D)canbecalculatedthefollowing
way:
P(H|D)=
P(H|D)=
P(D|H)*P(H)
P(D)
Posteriorprobability
P(D|H)*P(H)
P(D)
Posteriorprobability
LikelihoodofseeingdataD,giventhatHistrue
Likelihood
Howdoeslearningwork?
PriorprobabilityofhypothesisH
Howdoeslearningwork?
Bayesianinferenceisoneway.
Bayesianinferenceisoneway.
InBayesianinference,thebeliefinaparticularhypothesis(H)(ortheprobabilityof
thathypothesis),giventhedataobserved(D)canbecalculatedthefollowing
way:
InBayesianinference,thebeliefinaparticularhypothesis(H)(ortheprobabilityof
thathypothesis),giventhedataobserved(D)canbecalculatedthefollowing
way:
P(H|D)=
P(D|H)*P(H)
P(H|D)=
P(D|H)*P(H)
Σh P(D|h)*P(h)
P(D)
Posteriorprobability
Likelihood
Prior
Probabilityofobservingthedata,no
matterwhathypothesisistrue
Posteriorprobability
Likelihood
Prior
Probabilityofobservingthedata,no
matterwhathypothesisistrue:
Calculatebysummingoverallhypotheses
Howdoeslearningwork?
Bayesianinferenceisoneway.
Cross-situationallearning
Let’sapplyBayesianinferencetothisscenario.
InBayesianinference,thebeliefinaparticularhypothesis(H)(ortheprobabilityof
thathypothesis),giventhedataobserved(D)canbecalculatedthefollowing
way:
P(H|D)=
P(D|H)*P(H)
Σh P(D|h)*P(h)
data
Posteriorprobability
Likelihood
Prior
Cross-situationallearning
Let’sapplyBayesianinferencetothisscenario.
Cross-situationallearning
Let’sapplyBayesianinferencetothisscenario.
Posteriorprobabilitythat“ball”refersto
Observabledata
Hypothesis1(H1):“ball”=
Hypothesis2(H2):“ball”=
Sincetherearetwohypothesesinthehypothesis
spaceatthispoint
P(H1)=1/2=0.5
P(H2)=1/2=0.5
Cross-situationallearning
Let’sapplyBayesianinferencetothisscenario.
Cross-situationallearning
Let’sapplyBayesianinferencetothisscenario.
Observabledata
Observabledata
Hypothesis1(H1):“ball”=
Ifthisistheonlydataavailable,
Hypothesis1(H1):“ball”=
Hypothesis2(H2):“ball”=
P(D|H1)=wouldthisbeobservedifH1were
true?Yes.Thereforep(D|H1)=1.0.
Hypothesis2(H2):“ball”=
Ifthisistheonlydataavailable,
P(D)=Σ P(D|h)P(h)=
h
P(D|H1)*P(H1)=1.0*0.5=0.5
P(D|H2)*P(H2)=1.0*0.5=0.5
P(D|H2)=wouldthisbeobservedifH2were
true?Yes.Thereforep(D|H2)=1.0.
so
Σ P(D|h)P(h)=0.5+0.5=1.0
h
Cross-situationallearning
Let’sapplyBayesianinferencetothisscenario.
Cross-situationallearning
Let’sapplyBayesianinferencetothisscenario.
Observabledata
Observabledata
Hypothesis1(H1):“ball”=
Ifthisistheonlydataavailable,
Hypothesis1(H1):“ball”=
Hypothesis2(H2):“ball”=
=P(D|H1)*P(H1)
P(D)
Hypothesis2(H2):“ball”=
=1.0*0.5=0.5
1.0
Thisfeelsintuitivelyright,since“ball”couldrefertoeitherobject,giventhisdata
point.
Hypothesis3(H3):“ball”=
Sincetherearethreehypothesesinthehypothesis
spaceatthispoint
P(H1)=1/3=0.33
P(H2)=1/3=0.33
P(H3)=1/3=0.33
Cross-situationallearning
Let’sapplyBayesianinferencetothisscenario.
Observabledata
Cross-situationallearning
Let’sapplyBayesianinferencetothisscenario.
Observabledata
Hypothesis1(H1):“ball”=
Hypothesis2(H2):“ball”=
Ifthisistheonlydataavailable,
P(D|H1)=wouldthisbeobservedifH1were
true?Yes.Thereforep(D|H1)=1.0.
Hypothesis3(H3):“ball”=
Hypothesis1(H1):“ball”=
Hypothesis2(H2):“ball”=
Hypothesis3(H3):“ball”=
Cross-situationallearning
Let’sapplyBayesianinferencetothisscenario.
Observabledata
Hypothesis1(H1):“ball”=
Hypothesis2(H2):“ball”=
Hypothesis3(H3):“ball”=
Ifthisistheonlydataavailable,
P(D|H2)=wouldthisbeobservedifH2were
true?No.(Whywould“ball”besaidinthesecond
scene?)Thereforep(D|H2)=0.0.
P(D|H3)=wouldthisbeobservedifH3were
true?No.(Whywould“ball”besaidinthefirst
scene?)Thereforep(D|H3)=0.0.
Cross-situationallearning
Let’sapplyBayesianinferencetothisscenario.
Observabledata
Ifthisistheonlydataavailable,
P(D)=Σ P(D|h)P(h)=
h
P(D|H1)*P(H1)=1.0*0.33=0.33
P(D|H2)*P(H2)=0.0*0.33=0.0
P(D|H3)*P(H3)=0.0*0.33=0.0
so
Σ P(D|h)P(h)=0.33+0.0+0.0=0.33
h
Hypothesis1(H1):“ball”=
Ifthisistheonlydataavailable,
Hypothesis2(H2):“ball”=
=P(D|H1)*P(H1)
P(D)
=1.0*0.33=1.0
0.33
Hypothesis3(H3):“ball”=
Thisfeelsintuitivelyright,since“ball”couldonlyrefertotheball,whenthesetwo
scenesarereconciledwitheachother.
Smith&Yu(2008)
Yu&Smith(2007):Adultsseemabletodocross-situationallearning
(inexperimentalsetups).
Smith&Yu(2008)ask:Can12-and14-month-oldinfantsdothis?
(Relevantageforbeginningword-learning.)
Smith&Yu(2008):Experiment
InfantsweretrainedonsixnovelwordsobeyingphonotacticprobabilitiesofEnglish:
bosa,gasser,manu,colat,kaki,regli
Thesewordswereassociatedwithsixbrightlycoloredshapes
(sadlygreyscaleinthepaper)
Figurefrompaper
Smith&Yu(2008):Experiment
Whattheshapesareprobablymorelike
Smith&Yu(2008):Experiment
Training:30slideswith2objectsnamedwithtwowords(totaltime:4min)
manu
colat
Testing:12trialswithonewordrepeated4timesand2objects(correctoneand
distracter)present
manu
manu
Whichonedoestheinfant
thinkismanu?Thatshouldbe manu
theonetheinfantprefersto manu
lookat.
Exampletrainingslides
bosa
manu
Smith&Yu(2008):Experiment
Results:Infantspreferentiallylookattargetoverdistracter,and14-month-olds
lookedlongerthan12-month-olds.Thismeanstheywereabletotabulate
distributionalinformationacrosssituations.
Somethingtothinkabout…
Therealworldisn’tnecessarilyas
simpleastheseexperimental
setups-oftentimes,therewill
bemanypotentialreferents.
(Asimilarissuetotheonefastmappinghas.)
Implication:12and14-month-oldinfantscandocross-situationallearning
Somethingtothinkabout…
Astrategywherelearnershangon
toonehypothesisatatimeuntil
it’sprovenincorrectandonlythen
switchtoadifferentonemaywork
betterbecauseofthis.There’s
someevidencethatitmatches
infantbehavioralresultsquitewell
(Stevens,Trueswell,Yang,&
Gleitman2013)andmaybemore
effectivefornavigatingthe
hypothesisspace(Romberg&Yu
2014).
Somemorediscussionaboutthis:http://facultyoflanguage.blogspot.com/2013/03/learningfast-and-slow-i-how-children.html
Somethingelsetothinkabout…
Havingmorereferentsmaynotbeabadthing.
Whynot?
It’seasierforthecorrectassociationstoemergefromspurious
associationswhentherearemoreobject-referentpairing
opportunities.Let’sseeanexampleofthis.
Whymoremaynotalwaysbeharder…
Supposetherearesixobjectstotal,the
amountusedintheSmith&Yu
(2008)experiment.
First,let’sconsidertheircondition,
wheretwoobjectsareshownata
time.Let’ssaywegetthreeslides/
scenesofdata.
“manu”
“colat”
Supposetherearesixobjectstotal,the
amountusedintheSmith&Yu
(2008)experiment.
“manu”
“colat”
“bosa”
“gasser”
Canwetellwhether“manu”refersto
or
?
“bosa”
“gasser”
“kaki”
“regli”
No-bothhypothesesareequally
compatiblewiththesedata.
“kaki”
“regli”
Whymoremaynotalwaysbeharder…
Supposetherearesixobjectstotal,the
amountusedintheSmith&Yu
(2008)experiment.
Now,let’sconsideramorecompex
condition,wherefourobjectsare
shownatatime.Let’ssayweget
threeslides/scenesofdata.
Whymoremaynotalwaysbeharder…
“manu”
“colat”
“bosa”
“regli”
“bosa”
“gasser”
“manu”
“colat”
“manu”
“gasser”
“kaki”
“regli”
Whymoremaynotalwaysbeharder…
Supposetherearesixobjectstotal,the
amountusedintheSmith&Yu
(2008)experiment.
Canwetellwhether“manu”refersto
or oror?
Well,thefirstslideisn’thelpfulin
distinguishingbetweenthesefour
hypotheses…
“manu”
“colat”
“bosa”
“regli”
“bosa”
“gasser”
“manu”
“colat”
“manu”
“gasser”
“kaki”
“regli”
Whymoremaynotalwaysbeharder…
Supposetherearesixobjectstotal,the
amountusedintheSmith&Yu
(2008)experiment.
Canwetellwhether“manu”refersto
or oror?
Thesecondslidesuggests“manu”
can’tbe-otherwise,that
objectwouldappearinthesecond
slide.
“manu”
“colat”
“bosa”
“regli”
“bosa”
“gasser”
“manu”
“colat”
“manu”
“gasser”
“kaki”
“regli”
Whymoremaynotalwaysbeharder…
Supposetherearesixobjectstotal,the
amountusedintheSmith&Yu
(2008)experiment.
Therefore,“manu”is.
Thisshowsusthathavingmorethings
appear(andbenamed)atonce
actuallyoffersmoreopportunities
forthecorrectassociationsto
emerge.
“manu”
“colat”
“bosa”
“regli”
“bosa”
“gasser”
“manu”
“colat”
“manu”
“gasser”
“kaki”
“regli”
Whymoremaynotalwaysbeharder…
Supposetherearesixobjectstotal,the
amountusedintheSmith&Yu
(2008)experiment.
Canwetellwhether“manu”refersto
or or?
Thethirdslidesuggests“manu”can’t
beor-otherwise,those
objectswouldwouldappearinthe
thirdslide.
“manu”
“colat”
“bosa”
“regli”
“bosa”
“gasser”
“manu”
“colat”
“manu”
“gasser”
“kaki”
“regli”
Whymoremaynotalwaysbeharder…
Let’swalkthroughthisscenario
usingBayesianinference.
P(H|D)=
P(D|H)*P(H)
Σ P(D|h)*P(h)
We’llseeanexampleofhow
sequentialupdatingwouldwork
(insteadofcalculatingthe
posteriorjustonce,basedonall
ofthedata).
“manu”
“colat”
“bosa”
“regli”
“bosa”
“gasser”
“manu”
“colat”
“manu”
“gasser”
“kaki”
“regli”
Sequentialupdating
datapoint1
Hypothesis1(H1):“manu”=
Hypothesis2(H2):“manu”=
Hypothesis3(H3):“manu”=
Sequentialupdating
“manu”
“colat”
“bosa”
“regli”
Hypothesis1(H1):“manu”=
Hypothesis2(H2):“manu”=
Hypothesis3(H3):“manu”=
Hypothesis4(H4):“manu”=
Hypothesis4(H4):“manu”=
Sincetherearefourhypothesesinthehypothesis
spaceatthispoint,thepriorsare:
Wecancalculatethelikelihoods,giventhisdatapoint:
P(H1)=1/4=0.25
P(H2)=1/4=0.25
P(H3)=1/4=0.25
P(H4)=1/4=0.25
P(D|H1)=1
P(D|H2)=1
P(D|H3)=1
P(D|H4)=1
Sequentialupdating
Hypothesis1(H1):“manu”=
datapoint1
Hypothesis2(H2):“manu”=
Hypothesis3(H3):“manu”=
“manu”
“colat”
“bosa”
“regli”
Hypothesis1(H1):“manu”=
Hypothesis2(H2):“manu”=
Hypothesis3(H3):“manu”=
Hypothesis4(H4):“manu”=
Wecancalculatethelikelihood*priorforeachhypothesis:
Wecannowcalculatetheposteriorforeachhypothesis:
P(D|H1)*P(H1)=1*0.25=0.25
P(D|H2)*P(H2)=1*0.25=0.25
P(D|H3)*P(H3)=1*0.25=0.25
P(D|H4)*P(H4)=1*0.25=0.25
P(H1|D)=0.25/1=0.25
P(H2|D)=0.25/1=0.25
P(H3|D)=0.25/1=0.25
P(H4|D)=0.25/1=0.25
Σ P(D|h)*P(h)
“manu”
“colat”
“bosa”
“regli”
Sequentialupdating
Hypothesis4(H4):“manu”=
Thesum(whichwe’llneedforthedenominatoroftheposterior)=1
datapoint1
datapoint1
“manu”
“colat”
“bosa”
“regli”
Sequentialupdating
Sequentialupdating
Hypothesis1(H1):“manu”=
Hypothesis1(H1):“manu”=
Hypothesis2(H2):“manu”=
Hypothesis2(H2):“manu”=
Hypothesis3(H3):“manu”=
datapoint2
Hypothesis3(H3):“manu”=
“bosa”
“gasser”
“manu”
“colat”
Hypothesis4(H4):“manu”=
Thesebecomethepriorsforthenextdatapoint.
P(H1)=0.25
P(H2)=0.25
P(H3)=0.25
P(H4)=0.25
datapoint2
“bosa”
“gasser”
“manu”
“colat”
Hypothesis4(H4):“manu”=
Wecancalculatethelikelihoods,giventhisdatapoint:
P(D|H1)=1
P(D|H2)=1
P(D|H3)=1
P(D|H4)=0(doesn’tappear)
Sequentialupdating
Sequentialupdating
Hypothesis1(H1):“manu”=
Hypothesis1(H1):“manu”=
Hypothesis2(H2):“manu”=
Hypothesis2(H2):“manu”=
Hypothesis3(H3):“manu”=
Hypothesis4(H4):“manu”=
Wecancalculatethelikelihood*priorforeach
hypothesis:
P(D|H1)*P(H1)=1*0.25=0.25
P(D|H2)*P(H2)=1*0.25=0.25
P(D|H3)*P(H3)=1*0.25=0.25
P(D|H4)*P(H4)=0*0.25=0
Thesum(whichwe’llneedforthedenominatorofthe
posterior)=0.75
Σ P(D|h)*P(h)
Hypothesis3(H3):“manu”=
datapoint2
“bosa”
“gasser”
“manu”
“colat”
Hypothesis4(H4):“manu”=
Wecannowcalculatetheposteriorforeachhypothesis:
P(H1|D)=0.25/0.75=0.33
P(H2|D)=0.25/0.75=0.33
P(H3|D)=0.25/0.75=0.33
P(H4|D)=0/0.75 =0
datapoint2
“bosa”
“gasser”
“manu”
“colat”
Sequentialupdating
Sequentialupdating
Hypothesis1(H1):“manu”=
Hypothesis1(H1):“manu”=
Hypothesis2(H2):“manu”=
Hypothesis2(H2):“manu”=
Hypothesis3(H3):“manu”=
Hypothesis3(H3):“manu”=
Hypothesis4(H4):“manu”=
Hypothesis4(H4):“manu”=
Thesebecomethepriorsforthenextdatapoint.
Wecancalculatethelikelihoods,giventhisdatapoint:
datapoint3
“manu”
“gasser”
“kaki”
“regli”
P(H1)=0.33
P(H2)=0.33
P(H3)=0.33
P(H4)=0
Sequentialupdating
Hypothesis1(H1):“manu”=
Hypothesis2(H2):“manu”=
Hypothesis2(H2):“manu”=
Hypothesis3(H3):“manu”=
Hypothesis3(H3):“manu”=
Hypothesis4(H4):“manu”=
Hypothesis4(H4):“manu”=
P(D|H1)*P(H1)=0*0.33=0
P(D|H2)*P(H2)=1*0.33=0.33
P(D|H3)*P(H3)=0*0.33=0
P(D|H4)*P(H4)=1*0=0
Thesum(whichwe’llneedforthedenominatorofthe
posterior)=0.33
Σ P(D|h)*P(h)
“manu”
“gasser”
“kaki”
“regli”
Sequentialupdating
Hypothesis1(H1):“manu”=
Wecancalculatethelikelihood*priorforeach
hypothesis:
datapoint3
P(D|H1)=0(doesn’tappear)
P(D|H2)=1
P(D|H3)=0(doesn’tappear)
P(D|H4)=1
Wecannowcalculatetheposteriorforeachhypothesis:
datapoint3
“manu”
“gasser”
“kaki”
“regli”
P(H1|D)=0/0.33 =0
P(H2|D)=0.33/0.33=1
P(H3|D)=0/0.33 =0
P(H4|D)=0/0.33 =0
datapoint3
“manu”
“gasser”
“kaki”
“regli”
Sequentialupdating
Theutilityofprobabilities
Hypothesis1(H1):“manu”=
Hypothesis2(H2):“manu”=
Hypothesis3(H3):“manu”=
Hypothesis4(H4):“manu”=
Wecannowcalculatetheposteriorforeachhypothesis:
P(H1|D)=0/0.33 =0
P(H2|D)=0.33/0.33=1
P(H3|D)=0/0.33 =0
P(H4|D)=0/0.33 =0
datapoint3
“manu”
“gasser”
“kaki”
“regli”
Someotherfactorsincross-situationallearning
Eveniftherearemore
referents,cross-situational
learningismoresuccessful
whensomereferentsare
immediatelyrepeatedfrom
situationtosituation
(Kachergis,Yu,&Shiffrin
2012).
“manu”
“colat”
“bosa”
“regli”
“bosa”
“gasser”
“manu”
“colat”
“manu”
“gasser”
“kaki”
“regli”
Partialknowledgeofsome
wordsappearstobevery
helpfulforhelpinglearners
figureoutthemeaningof
wordstheydon’tknowyet
(Yurovsky,Fricker,&Yu
2013).
“bosa”
“gasser”
“manu”
“colat”
P(H1|D)=0.25/0.75=0.33
P(H2|D)=0.25/0.75=0.33
P(H3|D)=0.25/0.75=0.33
P(H4|D)=0/0.75 =0
Someotherfactorsincross-situationallearning
Thechild’sperspectiveofreal
worldeventsmaymakecrosssituationallearningmore
feasible,ascomparedtoa
neutralthirdparty(thewaya
photographrepresentsthe
world).Thisislikelybecause
certainthingsaremoresalient
fromachild’sperspectivedueto
objectforegroundingand
degreeofclutterinlineofsight
(Yurovsky,Smith,&Yu2013).
Recap:Word-meaningmapping
Questions?
Cross-situationallearning,whichreliesondistributionalinformation
acrosssituations,canhelpchildrenlearnwhichwordsreferto
whichthingsintheworld.
OnewaytoimplementthereasoningprocessbehindcrosssituationallearningisBayesianinference.Itcanbedoneinabatch
overallthedataobserved,orsequentiallyasthedataareobserved
onebyone.
Experimentalevidencesuggeststhatinfantsarecapableofthiskind
ofreasoningincontrolledexperimentalsetups.
Youshouldbeabletodoupthroughquestion7onHW2andupthrough
question5onthewordmeaningreviewquestions.