TAREA 5

1. FUNCION AUTOCORRELACION.

\[ Y_t= 5+e_t- \frac{1}{2}e_{t-1}+{\frac{1}{4}}e_{t-2}\\ \epsilon(Y_t)=0\\ var(Y_t)= var(5+e_t- \frac{1}{2}e_{t-1}+{\frac{1}{4}}e_{t-2})\\ var(Y_t)= \sigma^2+\frac{1}{4}\sigma^2+\frac{1}{16}\sigma^2\\ Var(Y_t)= \frac{21}{16}\sigma^2\\ Cov(Y_t,Y_{t-1})= Cov(e_t-\frac{1}2e_{t-1}+\frac{1}4e_{t-2},e_{t-1}-\frac{1}2e_{t-2}+\frac{1}4e_{t-3})\\ Cov(Y_t,Y_{t-1})=Cov(-\frac{1}2e_{t-1},e_{t-1})-Cov(\frac{1}{4}e_{t-2},\frac{1}{2}e_{t-2})\\ Cov(Y_t,Y_{t-1})=-\frac{1}{2}Var(e_{t-1})- \frac{1}{8}Var(e_{t-2})\\ Cov(Y_t,Y_{t-1})=-\frac{1}{2}\sigma^2-\frac{1}{8}\sigma^2\\ Cov(Y_t,Y_{t-1})=-\frac{5}{8}\sigma^2\\ Corr(Y_t,Y_{t-1})=-\frac{\frac{5}{8}{\sigma^2}}{\frac{21}{16}\sigma^2}\\ Corr(Y_t,Y_{t-1})=-\frac{10}{21}\sigma^2 \]

2. Grafica funcion de autocorrelacion MA(2)

\[ \theta 1= -1.2 \\ \theta 2= 0.7 \]

\[ \theta 1= 1 \\ \theta 2= 0.6 \]

3. Demostar si cambia la funcion de correlacion.

\[ \frac{-1}{\theta}/1+(1\theta) = \frac{-\theta}{1+\theta^2} \]

4. Calcular y graficar las funciones de autocorrelacion.

\[ a) \Phi= 0.6 \]

ARMAacf(ar=0.6, lag.max = 12)

          0           1           2           3           4           5           6           7           8 
1.000000000 0.600000000 0.360000000 0.216000000 0.129600000 0.077760000 0.046656000 0.027993600 0.016796160 
          9          10          11          12 
0.010077696 0.006046618 0.003627971 0.002176782 

\[ b) \Phi= 0.6 \]

\[ c) \Phi= 0.95 \]

\[ d) \Phi= 0.3 \]

5. Caracteristicas generales de la funcion de autocorrelacion.

1. MA(1) Es correlación no nula solo en el retraso 1. Debe estar entre -0.5 y +0.5, puede ser positivo o negativo.

2. MA(2) En los retrasos 1 y 2 tiene correlacion distinta de cero.

3. AR(1) Son autocorrelaciones en descomposición exponencial a partir del retraso 0. Si > 0, entonces todas las autocorrelaciones son positivas.

6. Ecuacion Yule- Walke.

\[ \Phi1= 0.6 \\ \Phi2=0.3 \]

rho 

 [1] 0.8571429 0.8142857 0.7457143 0.6917143 0.6387429 0.5907600 0.5460789 0.5048753 0.4667488 0.4315119 0.3989318
[12] 0.3688126 0.3409671 0.3152241 0.2914246 0.2694220 0.2490806 0.2302749 0.2128891 0.1968159

\[ \Phi1= -0.4 \\ \Phi2=0.5 \]

rho 

 [1] -0.8000000  0.8200000 -0.7280000  0.7012000 -0.6444800  0.6083920 -0.5655968  0.5304347 -0.4949723  0.4632063
[11] -0.4327687  0.4047106 -0.3782686  0.3536627 -0.3305994  0.3090711 -0.2889281  0.2701068 -0.2525068  0.2360561

\[ \Phi1= 1.2 \\ \Phi2= -0.7 \]

rho

 [1]  0.70588235  0.14705882 -0.31764706 -0.48411765 -0.35858824 -0.09142353  0.14130353  0.23356071  0.18136038
[10]  0.05413996 -0.06198431 -0.11227915 -0.09134596 -0.03101975  0.02671848  0.05377599  0.04582826  0.01735071
[19] -0.01125892 -0.02565621

\[ \Phi1= -1 \\ \Phi2= -0.6 \]

rho 

 [1] -0.6250000000  0.0250000000  0.3500000000 -0.3650000000  0.1550000000  0.0640000000 -0.1570000000  0.1186000000
 [9] -0.0244000000 -0.0467600000  0.0614000000 -0.0333440000 -0.0034960000  0.0235024000 -0.0214048000  0.0073033600
[17]  0.0055395200 -0.0099215360  0.0065978240 -0.0006449024

\[ \Phi1= 0.5 \\ \Phi2= -0.9 \]

rho

 [1]  0.26315789 -0.76842105 -0.62105263  0.38105263  0.74947368  0.03178947 -0.65863158 -0.35792632  0.41380526
[10]  0.52903632 -0.10790658 -0.53008597 -0.16792707  0.39311384  0.34769128 -0.17995682 -0.40290056 -0.03948914
[19]  0.34286593  0.20697320

\[ \Phi1= -0.5 \\ \Phi2= -0.6 \]

rho

 [1] -0.3125000000 -0.4437500000  0.4093750000  0.0615625000 -0.2764062500  0.1012656250  0.1152109375 -0.1183648437
 [9] -0.0099441406  0.0759909766 -0.0320290039 -0.0295800840  0.0340074443  0.0007443282 -0.0207766307  0.0099417184
[17]  0.0074951192 -0.0097125907  0.0003592238  0.0056479425

7. Graficas de las funciones de autocorrelacion

\[ a) ARMA(1,1)= \phi=0.7 \\ \theta=-0.4 \]

\[ b)ARMA(1,1)= \phi=0.7 \\ \theta=0.4 \]

8. Considerando los procesos MA(2)

\[ a1) \phi1.1= \phi2=-1/6 \]

\[ \phi=1 \\ \phi2=-6 \]

ARMAacf(ma=c(1,-6))

         0          1          2          3 
 1.0000000 -0.1315789 -0.1578947  0.0000000 

ARMAacf(ma=c(-1/6,-1/6))

         0          1          2          3 
 1.0000000 -0.1315789 -0.1578947  0.0000000 

9 Considere un proceso AR(1) Yt=Yt???1+et. Mostrar que si Phi= 1 el proceso no puede ser estacionario. (Pista: Tomar varianzas de ambos lados).

\[ Var(Y_t)=\phi^2Var(Y_{t-1})+\sigma^2_e \]

10. MA(6) con

ARMAacf(ma=c(-0.5, 0.25, -0.125, 0.0625, -0.0325, 0.015625))

          0           1           2           3           4           5           6           7 
 1.00000000 -0.49995181  0.24987336 -0.12474625  0.06199227 -0.03023441  0.01171876  0.00000000 

ARMAacf(ar= -0.5, lag.max = 7)

         0          1          2          3          4          5          6          7 
 1.0000000 -0.5000000  0.2500000 -0.1250000  0.0625000 -0.0312500  0.0156250 -0.0078125