Time Series Analysis and Its Applications (Instructor's Solution Manual) (Solutions)

81 Pages·2017·0.842 MB·English

by  Robert H. Shumway, David S. Stoffer

Time Series Analysis and Its Applications
(cid:81) 4th Edition (cid:83)
Instructor’s Manual
© 2017,R.H.ShumwayandD.S.Stoffer
Weassumethatthepackageforthetext,astsa,hasbeenloaded.SeethePackageNotesattheweb
pageforthetext:http://www.stat.pitt.edu/stoffer/tsa4/
(cid:33) (cid:33)
Please Do Not Reproduce
Chapter 1
Solutions
1.1 Themajordifferencesarehowquicklythesignaldiesoutintheexplosionversustheearthquakeandthe
largeramplitudeofthesignalsintheexplosion.
plot(EQ5, ylab="Earthquate/Explosion")
lines(EXP6, col=2)
legend("topright", c("EQ","EXP"), col=1:2, lty=1)
1.2 Figure 1.1 shows contrived data simulated according to this model. The modulating functions are also
plotted.Thecodeisgivenforpart(b);part(a)isgiveninthetext.Forpart(c),basicallyremovethecos()
part.
s = c(rep(0,100), 10*exp(-(1:100)/200)*cos(2*pi*1:100/4)) # part (b)
x = ts(s + rnorm(200, 0, 1))
plot(x)
lines(c(rep(0,100), 10*exp(-(1:100)/200))) # modulator on same plot as the series
The first signal bears a striking resemblance to the two arrival phases in the explosion. The second signal
decays more slowly and looks more like the earthquake. The periodic behavior is emulated by the cosine
functionwhichwillmakeonecycleeveryfourpoints.Ifweassumethatthedataaresampledat4pointsper
second,thedatawillmake1cycleinasecond,whichisaboutthesamerateastheseismicseries.
1.3 BelowisRcodeforparts(a)-(c).Inallcasesthemovingaveragenearlyannihilates(completelyinthe2nd
case)thesignal.Thesignalsinpart(a)and(c)aresimilar.
w = rnorm(150,0,1) # 50 extra to avoid startup problems
x = filter(w, filter=c(0,-.9), method="recursive")[-(1:50)] # AR
x2 = 2*cos(2*pi*(1:100)/4) # sinusoid
x3 = x2 + rnorm(100,0,1) # sinusoid + noise
v = filter(x, rep(1,4)/4) # moving average
v2 = filter(x2, rep(1,4)/4) # moving average
v3 = filter(x3, rep(1,4)/4) # moving average
par(mfrow=c(3,1))
plot.ts(x, main="autoregression")
lines(v,lty="dashed")
plot.ts(x2, main="sinusoid")
lines(v2,lty="dashed")
plot.ts(x3, main="sinusoid + noise")
lines(v3,lty="dashed")
1.4 Simplyexpandthebinomialproductinsidetheexpectationandusethefactthatµ isanonrandomconstant,
t
i.e.,
γ(s,t)=E[(x x −µ x −x µ +µ µ ]
s t s t s t s t
=E(x x )−µ E(x )−E(x )µ +µ µ
s t s t s t s t
2 1 Chapter1Solutions
Series (a) Modulator (a)
10 1
0.8
5
0.6
0
0.4
−5
0.2
−10 0
0 50 100 150 200 0 50 100 150 200
Series (b) Modulator (b)
15 1
10 0.9
0.8
5
0.7
0
0.6
−5
0.5
−10 0.4
−15 0.3
0 50 100 150 200 0 50 100 150 200
Fig.1.1.Simulatedserieswithexponentialmodulations
=E(x x )−µ µ −µ µ +µ µ
s t s t s t s t
1.5 (a) In each case the signals are fixed, so E(x ) = E(s +w ) = s +E(w ) = s . To plot the means,
t t t t t t
repeatProblem1.2andjustplotthesignals(s)withoutthenoise.
(b) Theautocovariancefunctionγ(t,u)=E[(x −s )(x −s )=E(w w ),whichisone(1)whent=u
t t u u t u
andzero(0)otherwise.
1.6 (a) SinceEx =β +β t,themeanisnotconstant,sotheprocessisnotstationary.Notethat
t 1 2
x −x =β +β t+w −β −β (t−1)−w
t t−1 1 2 t 1 2 t−1
=β +w −w ,
2 t t−1
whichisclearlystationary.Verifythatthemeanisβ andtheautocovarianceis2fors = 2and−1for
2
|s−t|=1andiszerofor|s−t|>1.
(b) First,write
q
1 (cid:88)
E(y )= [(β +β (t−j)]
t 2q+1 1 2
j=−q
(cid:20) q (cid:21)
1 (cid:88)
= (2q+1)(β +β t)−β j
2q+1 1 2 2
j=−q
=β +β t
1 2
becausethepositiveandnegativetermsinthelastsumcancelout.Togetthecovariancewritetheprocess
as
∞
(cid:88)
y = a w ,
t j t−j
j=−∞
wherea =1,j =−q,...,0,...,qandiszerootherwise.Togetthecovariance,notethatweneed
j
1 Chapter1Solutions 3
γ (h)=E[(y −Ey )(y −Ey )]
y t+h t+h t t
(cid:88)(cid:88)
=(2q+1)−2 a a Ew w
j k t+h−j t−k
j k
σ2 (cid:88)
= a a δ ,
(2q+1)2 j k h+k−j
j,k
∞
(cid:88)
= a a ,
j+h j
j=−∞
whereδ =1,j =k+handiszerootherwise.Writingoutthetermsinγ (h),forh=0,±1,±2,...,
h+k−j y
weobtain
σ2(2q+1−|h|)
γ (h)=
y (2q+1)2
forh=0,±1,±2,...,±2qandzerofor|h|>q.
1.7 ByacomputationanalogoustothatappearinginExample1.17,wemayobtain

6σ2 h=0
4σw2 h=±1
γ(h)= w
σ2 h=±2
0w |h|>2.
Theautocorrelationisobtainedbydividingtheautocovariancesbyγ(0)=6σ2.
w
1.8 (a) Simplysubstituteδs+(cid:80)s w forx toseethat
k=1 k s
t t−1
(cid:88) (cid:16) (cid:88) (cid:17)
δt+ w =δ+ δ(t−1)+ w +w .
k k t
k=1 k=1
(cid:124) (cid:123)(cid:122) (cid:125) (cid:124) (cid:123)(cid:122) (cid:125)
xt xt−1
Alternately,theresultcanbeshownbyinduction.
(b) Forthemean,
(cid:32) t (cid:33) t
(cid:88) (cid:88)
Ex =E δt+ w =δt+ Ew =δt.
t k k
k=1 k=1
Forthecovariance,withoutlossofgenerality,considerthecases≤t.
γ(s,t)=cov(x ,x )=E{(x −δs)(x −δt)}
s t s t
(cid:26) s t (cid:27)
(cid:88) (cid:88)
=E w w
j k
j=1 k=1
(cid:26) (cid:27)
=E (w +···+w )(w +···+w +w +...+w )
1 s 1 s s+1 t
s
(cid:88)
= E(w2)=sσ2. [or min(s,t)σ2]
j w w
j=1
(c) Theseriesisnonstationarybecauseboththemeanfunctionandtheautocovariancefunctiondependon
time,t.
(d) From(b),ρ (t−1,t) = (t−1)σ2/(cid:112)(t−1)σ2(cid:112)tσ2,whichyieldstheresult.Theimplicationisthat
x w w w
theseriestendstochangeslowly.
4 1 Chapter1Solutions
(e) One possibility is to note that ∇x = x −x = δ+w , which is stationary because µ = δ and
t t t−1 t x,t
γ (t+h,t)=σ2δ (h)arebothindependentoftimet,whereδ (h)isthedeltameasure.
x w 0 0
1.9 BecauseE(U )=E(U )=0,wehaveE(x )=0.Then,
1 2 t
γ(h)=E(x x )
t+h t
(cid:26) (cid:27)
(cid:0) (cid:1)(cid:0) (cid:1)
=E U sin[2πω (t+h)]+U cos[2πω (t+h)] U sin[2πω t]+U cos[2πω t]
1 0 2 0 1 0 2 0
(cid:18) (cid:19)
=σ2 sin[2πω (t+h)]sin[2πω t]+cos[2πω (t+h)]cos[2πω t]
w 0 0 0 0
=σ2 cos[2πω (t+h)−2πω t]
w 0 0
=σ2 cos[2πω h]
w 0
bythestandardtrigonometricidentity,cos(A−B)=sinAsinB+cosAcosB.
1.10 (a)
(cid:26) (cid:27)
MSE(A)=E x2 −2AE(x x )+A2E(x2) =γ(0)−2Aγ((cid:96))+A2γ(0).
t+(cid:96) t+(cid:96) t t
SettingthederivativewithrespecttoAtozeroyields
−2γ((cid:96))+2Aγ(0)=0
andsolvinggivestherequiredvalue.
(b)
(cid:20) (cid:21) (cid:20) (cid:21) (cid:20) (cid:21)
ρ((cid:96))γ((cid:96))
MSE(A)=γ(0) 1−2 +ρ2((cid:96)) =γ(0) 1−2ρ2((cid:96))+ρ2((cid:96)) =γ(0) 1−ρ2((cid:96)) .
γ(0)
(c) Ifx =Ax withprobabilityone,then
t+(cid:96) t
(cid:20) (cid:21)
E(x −Ax )2 =γ(0) 1−ρ2((cid:96)) =0
t+(cid:96) t
implyingthatρ((cid:96))=±1.SinceA=ρ((cid:96)),theconclusionfollows.
1.11 (a) Sincex =(cid:80)∞ ψ w ,
t j=−∞ j t−j
∞ ∞ ∞
(cid:88) (cid:88) (cid:88) (cid:88)
γ(h)=E ψ w w ψ =σ2 ψ ψ δ =σ2 ψ ψ ,
j t+h−j t−k k w j k h−j+k w k+h k
j=−∞k=−∞ j,k k=−∞
whereδ =1fort=0andiszerootherwise.
t
(b) TheproofisidenticaltotheonegiveninAppendixA,ExampleA.2.
1.12
γ (h)=E[(x −µ )(y −µ )]=E[(y −µ )(x −µ )]=γ (−h)
xy t+h x t y t y t+h x yx
1.13 (a)

σ2(1+θ2)+σ2 h=0
 w u
γ (h)= −θσ2 h=±1
y w
0 |h|>1.
1 Chapter1Solutions 5
(b)
(cid:40)σ2 h=0
w
γxy(h)= −θσw2 h=−1
0 otherwise.
(cid:26)
σ2 h=0
γ (h)= w
x 0 otherwise.
γ (h)
ρ (h)= xy .
xy (cid:112)
γ (0)γ (0)
x y
(c) The processes are jointly stationary because the autocovariance and cross-covariance functions depend
onlyonlagh.
1.14 (a) Forthemean,write
(cid:26) (cid:27)
1
E(y )=E(exp{x })=exp µ + γ (0) ,
t t x 2 x
usingthegivenequationatλ=1.
(b) Fortheautocovariancefunction,notethat
(cid:0) (cid:1)
E(y y )=E exp{x }exp{x }
t+h t t+h t
(cid:0) (cid:1)
=E exp{x +x }
t+h t
=exp{2µ +γ (0)+γ (h)},
x x x
sincex +x isthesumoftwocorrelatednormalrandomvariablesandwillbenormallydistributed
t t+h
withmean2µ andvariance
x
(cid:0) (cid:1)
γ (0)+γ (0)+2γ (h)=2 γ (0)+γ(h) .
x x x x
Fortheautocovarianceofy ,
t
γ (h)=E(y y )−E(y )E(y )
y t+h t t+h t
(cid:0) (cid:8) 1 (cid:9)(cid:1)2
=exp{2µ +γ (0)+γ (h)}− exp µ + γ (0)
x x x x 2 x
(cid:0) (cid:1)
=exp{2µ +γ (0)} exp{γ (h)}−1 .
x x x
1.15 Theprocessisstationarybecause
µ =E(x )=E(w w )=E(w )E(w )=0;
x,t t t t−1 t t−1
γ (0)=E(w w w w )=E(w2)E(w2 )=σ2σ2 =σ4,
x t t−1 t t−1 t t−1 w w w
γ (1)=E(w w w w )=E(w )E(w2)E(w )=0=γ(−1),
x t+1 t t t−1 t+1 t t−1
andsimilarcomputationsestablishthatγ (h)=0,for|h|≥1.Theseriesiswhitenoise.
x
1.16 (a) Fort=1,2,...,
(cid:90) 1 1 (cid:12)(cid:12)1 1 (cid:2) (cid:3)
E(xt)= sin(2πut)du=−2πtcos(2πut)(cid:12)(cid:12) =−2πt cos(2πt)−1 =0;
0 0
(cid:90) 1
γ(h)= sin[2πu(t+h)]sin[2πut]du.
0
Usingtheidentity2sin(α)sin(β)=cos(α−β)−cos(α+β)givesγ(0)=1/2andγ(h)=0,forh(cid:54)=0.
6 1 Chapter1Solutions
(b) This part of the problem is harder than it seems at first and it might be a good idea to omit it in
moreelementarypresentations.Theeasiestwaytotackletheproblemistocalculatesomeprobabilities
(whichcanbegivenasahint),e.g.,becauseU isuniformon[0,1],
Pr{x ≤0, x ≤0}=Pr{sin(2πU)≤0, sin(2π3U)≤0}
1 3
(cid:110) (cid:16) (cid:17)(cid:111) 1
=Pr U ∈[1/2,1]∩ [1/6,1/3]∪[1/2,2/3]∪[5/6,1] =0+1/6+1/6=
3
but,similarly,
1
Pr[x ≤0,x ≤0]= .
2 4 4
1.17 (a) Theessentialpartoftheexponentofthecharacteristic[ormomentgenerating]functionis
n n
(cid:88) (cid:88)
λ x = λ (w −θw )
j j j j j−1
j=1 j=1
n−1
(cid:88)
=−λ θw + (λ −θλ )w +λ w .
1 0 j j+1 j n n
j=1
Becausethew areindependentandidenticallydistributed,thecharacteristicfunctioncanbewrittenas
t
n−1
(cid:89)
φ(λ ,...,λ )=φ (−λ θ) φ (λ −θλ )φ (λ )
1 n w 1 w j j+1 w n
j=1
(b) Becausethejointdistributionofthew willnotchangesimplybyshiftingx ,...,x tox ,...,x ,
j 1 n 1+h n+h
thecharacteristicfunction[orMGF]remainsthesame.
1.18 Lettingk =j+h,holdingj fixedaftersubstitutingfrom(1.29)yields
(cid:88)∞ (cid:88)∞ (cid:12) (cid:88)∞ (cid:12) (cid:88)∞ (cid:88)∞
|γ(h)|=σ2 (cid:12) ψ ψ (cid:12)≤σ2 |ψ ||ψ |
w (cid:12) j+h j(cid:12) w j+h j
h=−∞ h=−∞ j=−∞ h=−∞j=−∞
∞ ∞
(cid:88) (cid:88)
=σ2 |ψ | |ψ |<∞.
w k j
k=−∞ j=−∞
1.19 (a) E(x )=E(µ+w +θw )=µ.
t t t−1
(b) γ(h)=cov(w +θw ,w +θw ,soγ (0)=(1+θ2)σ2,γ (±1)=θσ2,and0otherwise.
t+h t+h−1 t t−1 x w x w
(c) From(a)and(b)weseethat,foranyθ,boththemeanfunctionandautocovariancefunctionareindependent
oftime.
(d) Fromthegivenformula,andbecauseγ (h)=0for|h|>1,wehave
x
(cid:20) (cid:21)
1 2(n−1)
var(x¯)= γ (0)+ γ (1) .
n x n x
• Whenθ =0,γ (±1)=0andit’stheclassicalcase,var(x¯)=σ2/n.
x w
• When θ = 1, γx(0) = 2σw2 and γx(±1) = σw2, so var(x¯) = σnw2 [2+ 2(nn−1)] = σnw2 [4− n2]. In this
case,thevarianceisabout4timesaslargeastheuncorrelatedcase.
• When θ = −1, γx(0) = 2σw2 and γx(±1) = −σw2, so var(x¯) = σnw2 [2− 2(nn−1)] = σnw2 [n2]. In this
case, the variance is smaller (n > 2) than the uncorrelated case and the variance is nearly zero for
moderaten.
1 Chapter1Solutions 7
(e) It’seasiertoestimatethemeanwhenθisnegative.Negativelycorrelateddatavaryaroundthemeanmore
tightlythanpositivelycorrelateddata.Inessence,youneedfewerdatatoidentifythemeanifthedataare
negativelycorrelated.
1.20 Codeforparts(a)and(b)isbelow.Studentsshouldhaveabout1in20ACFvalueswithinthebounds,
butthevaluesforpart(b)willbelargeringeneralthanforpart(a).
wa = rnorm(500,0,1)
wb = rnorm(50,0,1)
par(mfrow=c(2,1))
(acf(wa, 20)) # plot and print results
(acf(wb, 20)) # plot and print results
1.21 This is similar to the previous problem. Generate 2 extra observations due to loss of the end points in
makingtheMA.
wa = rnorm(502,0,1)
wb = rnorm(52,0,1)
va = filter(wa, sides=2, rep(1,3)/3)
vb = filter(wb, sides=2, rep(1,3)/3)
par(mfrow=c(2,1))
(acf(va, 20, na.action = na.pass)) # plot and print results
(acf(vb, 20, na.action = na.pass)) # plot and print results
1.22 Generate the data as in Problem 1.2 and then type acf(x). The sample ACF will exhibit significant
correlations at one cycle every four lags, which is the same frequency as the signal. (The process is not
stationarybecausethemeanfunctionisthesignal,whichdependsontimet.)
1.23 ThesampleACFshouldlooksinusoidal,makingonecycleevery50lags.
x = 2*cos(2*pi*(1:500)/50 + .6*pi)+ rnorm(500,0,1)
acf(x,100)
1.24 γ (h) = cov(y ,y ) = cov(x − .7x ,x − .7x ) = 0 if |h| > 1 because the x s are
y t+h t t+h t+h−1 t t−1 t
independent.Whenh=0,γ (0)=σ2(1+.72),whereσ2isthevarianceofx .Whenh=1,γ (1)=−.7σ2.
y x x t y x
Thus,ρ (1)=−.7/(1+.72)=−.47
y
1.25 (a) Thevarianceisalwaysnon-negative,soforx astationaryseries
t
(cid:26) n (cid:27) (cid:26) (cid:27) n n
(cid:88) (cid:88) (cid:88) (cid:88)(cid:88)
var a x =cov a x , a x = a γ(s−t)a =aaa(cid:48)Γaaa≥0,
s s s s t t s t
s=1 s t s=1t=1
thusΓ ={γ(s−t),s,t=1,...,n}isanon-negativedefinitematrix.
(b) LetY =x −x¯fort=1...nandconstructthen×2nmatrix
t t
0 0 ··· 0 Y Y ··· Y 
1 2 n
0 ··· 0 Y Y ··· Y 0
 1 2 n 
D =.. .. .. 
. . . 
0 Y Y ··· Y 0 ··· 0
1 2 n
WithΓˆ ={γˆ(s−t)}n ,itiseasytoshowthat
n s,t=1
1
Γˆ = DD(cid:48).
n n
Then,foranyvectoraaa∈IRn,
n
aaa(cid:48)Γˆ aaa= 1aaa(cid:48)DD(cid:48)aaa= 1ccc(cid:48)ccc=(cid:88)c2 ≥0
n n n i
i=1
8 1 Chapter1Solutions
forccc=D(cid:48)aaa.NotingthatΓˆ issymmetric,itwillbepositivedefinite(p.d.)ifitseigenvaluesarepositive.
n
SincethediagonalelementsofΓˆ areγˆ(0),thesumoftheeigenvaluesofΓˆ isnγˆ(0).Consequently,Γˆ
n n n
isp.d.aslongasγˆ(0)>0.Inotherterms,ifthesamplevarianceofthedataisnotzero,Γˆ isp.d.
n
1.26 (a)
N n
Ex¯ = 1 (cid:88)x = 1 (cid:88)µ = Nµt =µ
t N jt N t N t
j=1 j=1
(b)
N N N
1 (cid:88)(cid:88) 1 (cid:88) 1
E[(x¯ −µ )2]= E(x −µ )(x −µ )= e2 = γ (t,t)
t t N2 jt t kt t N2 jt N e
j=1k=1 j=1
(c) Aslongastheseparateseriesareobservingthesamesignal,wemayassumethatthevariancegoesdown
proportionallytothenumberseriesasintheiidcase.Ifnormalityisreasonable,pointwise100(1−α)%
intervalscanbecomputedas
√
x¯ ±z γ (t,t)/ N
t α/2 e
1.27
1 1
V (h)= E[(x −µ)−(x −µ)]2 = [γ(0)−γ(h)−γ(−h)+γ(0)]=γ(0)−γ(h).
x 2 s+h s 2
1.28 Thenumeratoranddenominatorofρˆ(h)are
1 n(cid:88)−h β2(cid:20)n(cid:88)−h n(cid:88)−h (cid:21)
γˆ(h)= [β (t−t¯)+β h][β (t−t¯)]= 1 (t−t¯)2+h (t−t¯)
n 1 1 1 n
t=1 t=1 t=1
and
β2 (cid:88)n
γˆ(0)= 1 (t−t¯)2.
n
t=1
Now,writethenumeratoras
β2(cid:20) (cid:88)n (cid:88)n (cid:21)
γˆ(h)=γˆ(0)+ 1 − (t−t¯)2−h (t−t¯)
n
t=n−h+1 t=n−h+1
Hence,wecanwrite
ρˆ(h)=1+R
where
β2 (cid:20) (cid:88)n (cid:88)n (cid:21)
R= 1 − (t−t¯)2−h (t−t¯)
nγˆ(0)
t=n−h+1 t=n−h+1
isaremaindertermthatneedstoconvergetozero.Wecanevaluatethetermsintheremainderusing
m
(cid:88) m(m+1)
t=
2
t=1
and
m
(cid:88) m(2m+1)(m+1)
t2 =
6
t=1
Thedenominatorreducesto
1 Chapter1Solutions 9
(cid:20) n (cid:21)
(cid:88)
nγˆ(0)=β2 t2−nt¯2
1
t=1
(cid:20)n(n+1)(2n+1) n(n+1)2(cid:21)
=β2 −
1 6 4
n(n+1)(n−1)
=β2 ,
1 12
whereasthenumeratorcanbesimplifiedbylettings=t−n+hsothat
β2 (cid:20) (cid:88)h (cid:88)h (cid:21)
R= 1 − (s+n−h−t¯)2−h (s+n−h−t¯)
nγˆ(0)
s=1 s=1
ThetermsinthenumeratorofR areO(n2),whereasthedenominatorisO(n3)sothattheremainderterm
convergestozero.
1.29 (a)
√ E[x¯2]
Pr{ n|x¯|>(cid:15)}≤n
(cid:15)2
Notethat,
∞
(cid:88)
nE[x¯2]→ γ(h)=0,
u=−∞
wherethelaststepemploysthesummabilitycondition.Thevarianceofx¯isderivedin(1.33).
(b) Anexampleofsuchaprocessisx =∇w =w −w ,wherew iswhitenoise.Thissituationarises
t t t t−1 t
whenastationaryprocessisover-differenced(i.e.,w isalreadystationary,so∇w wouldbeconsidered
t t
over-differencing).
1.30 Lety =x −µ andwritethedifferenceas
t t x
n n−h
n1/2(cid:0)γ˜(h)−γˆ(h)(cid:1)=n−1/2(cid:88)y y −n−1/2(cid:88)(y −y¯)(y −y¯)
t+h t t+h t
t=1 t=1
(cid:20) n n−h n−h (cid:21)
(cid:88) (cid:88) (cid:88)
=n−1/2 y y +y¯ y +y¯ y −(n−h)y¯2
t+h t t t+h
t=n−h+1 t=1 t=1
Forthefirstterm
(cid:20) n (cid:21) n
E n−1/2(cid:12)(cid:12) (cid:88) yt+hyt(cid:12)(cid:12) ≤ n−1/2E (cid:88) |yt+hyt|
t=n−h+1 t=n−h+1
n
(cid:88)
≤ n−1/2 E1/2[y2 ]E1/2[y2]
t+h t
t=n−h+1
= n−1/2hγ (0)
x
→0,
as n → ∞. Applying the Markov inequality in the hint then shows that the first term is o (1). In order to
p
handletheotherterms,whichdiffertriviallyfromn−1/2ny¯2,notethat,fromTheoremA.5,n1/2y¯converging
indistributiontoastandardnormalimpliesthatny¯2convergesindistributiontoachi-squarerandomvariable
with1degreeoffreedomandhenceny¯2 =O (1).Hence,n−1/2ny¯2 =n−1/2O (1)=o (1)andtheresult
p p p
isproved.