exercicio 03 reposta

Prévia do material em texto

EXERCICIO 03
ECONOMETRIA DE SERIES TEMPORAIS
Econometria de series temporais
Amanda Steffany da silva Araujo
Ciências econômicas
Noite
 EXERCÍCIO 03
 Seja o banco de dados “quebra”, utilize a variável B1 para: 
i) fazer o gráfico dos dados; 
ii) realizar teste ADF; 
iii) realizar o teste de Perron para quebra estrutural discutido em sala. 
Realizando o teste:
Baixei o arquivo e rodei
	y <- quebra_dados_exercico_03$B1;y
 [1] 0.320200 -0.272600 -0.351300 -0.245550 -0.064580 -0.494390 -0.787190 -0.231300
 [9] 0.246252 0.473626 0.441513 0.051656 0.424328 0.193964 0.547082 0.292541
 [17] 0.345371 0.330885 0.112343 0.016671 0.099936 -0.211130 -0.069070 0.161267
 [25] 0.149933 0.054967 0.254983 -0.038110 0.202546 0.622673 0.008936 -0.320930
 [33] 0.139434 0.501717 -0.063940 -0.138670 -0.240040 -0.412120 -0.410160 0.064821
 [41] -0.104090 0.205455 0.381328 0.572064 0.190532 -0.105330 0.313233 -0.060580
 [49] -0.463890 -0.077050 1.063677 1.866639 1.711019 0.179441 1.746205 2.019402
 [57] 2.496501 2.219551 2.229775 2.383988 2.357794 2.654697 2.309248 1.974724
 [65] 1.857962 1.741781 1.756091 1.999045 2.181823 2.535211 2.392806 1.948903
 [73] 1.833851 1.522326 1.616363 2.030281 2.044741 0.176877 2.257085 2.010143
 [81] 2.002171 1.619686 1.668243 1.545621 1.810511 1.785255 1.392528 1.815964
 [89] 1.996382 1.897291 1.509845 1.681823 1.687011 2.140506 2.413353 1.885376
 [97] 1.950988 1.711594 2.058297 1.347049
> plot.ts(y)
d1 <- ifelse(y[1:50],0); d2 <- ifelse(y[51:100],1)
> DL <- ts(c(d1,d2));# create level dummy
> d3 <- ifelse(y[1:49],0); d4 <- ifelse(y[50:50],1); d5 <- ifelse(y[51:100],0)
> Dp <- ts(c(d3,d4,d5)); # create pulse dummy
> T <- ts(1:100, frequency=1); T; #create trend
Time Series:
Start = 1 
End = 100 
Frequency = 1 
 [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
 [22] 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42
 [43] 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63
 [64] 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84
 [85] 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
> library(dyn)
> z <- zoo(y)
> perr1 <- dyn$lm(z ~ lag(z, -1) + DL + Dp + T)
> summary(perr1)
Call:
lm(formula = dyn(z ~ lag(z, -1) + DL + Dp + T))
Residuals:
 Min 1Q Median 3Q Max 
-1.7483 -0.1588 0.0028 0.2241 0.9273 
Coefficients:
 Estimate Std. Error t value Pr(>|t|) 
(Intercept) 0.060153 0.089191 0.674 0.501699 
lag(z, -1) 0.318050 0.088499 3.594 0.000521 ***
DL 1.313941 0.212453 6.185 1.6e-08 ***
Dp 0.073979 0.397259 0.186 0.852671 
T -0.001273 0.002754 -0.462 0.645055 
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.3839 on 94 degrees of freedom
 (1 observation deleted due to missingness)
Multiple R-squared: 0.8576,	Adjusted R-squared: 0.8516 
F-statistic: 141.6 on 4 and 94 DF, p-value: < 2.2e-16
> Dy <- diff(y)
> Dz <- zoo(Dy)
> Dz <- lag(Dz, c(0,1))
> perr2 <- dyn$lm(z ~ lag(z, -1) + DL + Dp + T + Dz)
> summary(perr2)
Call:
lm(formula = dyn(z ~ lag(z, -1) + DL + Dp + T + Dz))
Residuals:
 Min 1Q Median 3Q Max 
-0.67716 -0.15655 0.03202 0.17526 0.54300 
Coefficients:
 Estimate Std. Error t value Pr(>|t|) 
(Intercept) 0.081158 0.064629 1.256 0.21246 
lag(z, -1) 0.184498 0.064828 2.846 0.00548 ** 
DL 1.550786 0.154694 10.025 2.54e-16 ***
Dp 0.946824 0.297305 3.185 0.00199 ** 
T -0.001259 0.002017 -0.624 0.53416 
Dz1 -0.591875 0.064919 -9.117 1.96e-14 ***
Dz2 -0.350373 0.061433 -5.703 1.48e-07 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.2724 on 90 degrees of freedom
 (3 observations deleted due to missingness)
Multiple R-squared: 0.9304,	Adjusted R-squared: 0.9257 
F-statistic: 200.5 on 6 and 90 DF, p-value: < 2.2e-16
> AIC(perr1);AIC(perr2)
[1] 98.25387
[1] 31.71127
	
		
o resultado mostra que AIC do modelo perr2 é menor e portanto o modelo é melhor.