\[ \begin{aligned} \text{Corr}(X_t,X_{t+h}) &= \frac{\text{Cov}(X_t,X_{t+h})}{\sqrt{\text{Var}(X_t)\text{Var}(X_{t+h})}} \\ &= \frac{\gamma_h}{\sqrt{\gamma_0^2}} = \frac{\gamma_h}{\gamma_0} \end{aligned} \]
\[\text{E}(X_t) = \text{E}(e_t - 0.5e_{t-1}+ 0.5e_{t-2}) = 0\] \[\text{Var}(X_t) = \text{Var}(e_t - 0.5e_{t-1}+ 0.5e_{t-2}) = 1 + 0.25 +0.25 = 1.5\] because the variance of \(e_t = 1\) for all \(t\) and independence of \(e_t\).
\[ \begin{aligned} \text{Cov}(X_t, X_{t+1}) &= \text{E}(X_tX_{t-1}) - \text{E}(X_t)\text{E}(X_{t-1}) \\ &= \text{E}[(e_t-0.5e_{t-1} + 0.5e_{t-2})(e_{t+1}-0.5e_{t} + 0.5e_{t-1})] - 0 \\ &= -0.5\text{E}(e_t^2) - 0.25\text{E}(e_{t-1}^2) = -0.75 \end{aligned} \]
And,
\[ \begin{aligned} \text{Var}(X_t) &= \text{Var}(X_{t+1}) = 1.5 \\ \implies \text{Corr}(X_t, X_{t+1}) &= \frac{\text{Cov}(X_t, X_{t+1})}{\sqrt{\text{Var}(X_t)\text{Var}(X_{t+1})}} \\ &= -0.75/1.5 = -\frac{1}{2}. \end{aligned} \]
Similarly,
\[ \begin{aligned} \text{Cov}(X_t, X_{t+2}) &= \text{E}[(e_t-0.5e_{t-1} + 0.5e_{t-2})(e_{t+2}-0.5e_{t+1} + 0.5e_{t})] - 0 \\ &= \text{E}[0.5e_t^2] = 0.5 \\ \implies \text{Cor}(X_t, X_{t+2}) &= \frac{\text{Cov}(X_t, X_{t+2})}{\sqrt{\text{Var}(X_t)\text{Var}(X_{t+2})}} \\ &= 0.5/1.5 = \frac{1}{3} \end{aligned} \]
0, since zero covariance among \(e_t\) and \(e_{t+h}\) as \(h\) > 2.
Yes, \(\text{Corr}(X_{t}, X_{t+h}) = 0\) as \(h \to n\) is obvious as \(h > 2\), \(\text{Corr}(X_{t}, X_{t+h}) = 0.\)
\[ \begin{aligned} \text{E}(y_t) &= \text{E}(z_t + e_t) = 0 \\ \text{Var}(y_t) &= \text{Var}(z_t + e_t) = \sigma_z^2 + \sigma_e^2 \end{aligned} \]
They are independent on t.
\[ \begin{aligned} \text{Cov}(y_t,y_{t+h}) &= \text{E}(y_ty_{t+h}) - \text{E}(y_t)\text{E}(y_{t+h}) \\ &= \text{E}(y_ty_{t+h}) = \text{E}[(e_t + z)(e_{t+h} + z)] \\ &= \text{E}(z^2) = \text{Var}(z^2) = \sigma_z^2 \end{aligned} \] Yes, it is covariance stationary as \(\text{Cov}(y_t,y_{t+h})\) depends on neither \(t\) nor \(h\).
\[ \begin{aligned} \text{Corr}(y_t,y_{t+h}) &= \frac{\text{Cov}(y_t,y_{t+h})}{\sqrt{\text{Var}(y_t)\text{Var}(y_{t+h})}} \\ &= \frac{\sigma_z^2}{\sigma_z^2 + \sigma_e^2} \end{aligned} \]
No, as \(\text{Corr}(y_t,y_{t+h}) = \frac{\sigma_z^2}{\sigma_z^2 + \sigma_e^2} > 0\) for all \(h\).
library(dplyr)
library(tidyverse)
library(broom)
library(wooldridge)
nyse$return_1_sqr <- (nyse$return_1)^2
reg1=lm(return~return_1+return_1_sqr,data=nyse)
tidy(reg1)The null hypo is \(H_0: \beta_1 = \beta_2 = 0\)
summary(reg1)##
## Call:
## lm(formula = return ~ return_1 + return_1_sqr, data = nyse)
##
## Residuals:
## Min 1Q Median 3Q Max
## -15.1867 -1.3051 0.1005 1.3229 8.1718
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.225549 0.087234 2.586 0.00993 **
## return_1 0.048572 0.038722 1.254 0.21013
## return_1_sqr -0.009735 0.007030 -1.385 0.16654
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.109 on 686 degrees of freedom
## (2 observations deleted due to missingness)
## Multiple R-squared: 0.006259, Adjusted R-squared: 0.003362
## F-statistic: 2.16 on 2 and 686 DF, p-value: 0.1161The F-statistics = 2.16 with p-value = 0.1161, implying that we fail to reject the null hypothesis at 10% level.
The code is shown below:
# construct the return_{t-2}
return_2 <- numeric(nrow(nyse))
for (t in 3:nrow(nyse)){
return_2[1] = NA
return_2[2] = (nyse$return_1)[1]
return_2[t] = (nyse$return_1)[t-1]
}
nyse$return_2 <- return_2
reg2 <- lm(nyse$return ~ nyse$return_1 + I(nyse$return_1 * nyse$return_2))
summary(reg2)##
## Call:
## lm(formula = nyse$return ~ nyse$return_1 + I(nyse$return_1 *
## nyse$return_2))
##
## Residuals:
## Min 1Q Median 3Q Max
## -15.2578 -1.2820 0.0991 1.3374 8.1103
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.17316 0.08096 2.139 0.0328 *
## nyse$return_1 0.06871 0.03925 1.751 0.0804 .
## I(nyse$return_1 * nyse$return_2) 0.01134 0.01001 1.132 0.2579
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.111 on 685 degrees of freedom
## (3 observations deleted due to missingness)
## Multiple R-squared: 0.005234, Adjusted R-squared: 0.002329
## F-statistic: 1.802 on 2 and 685 DF, p-value: 0.1658We see that F-statistic = 1.802 with p-value = 0.1658, implying that we fail to reject the null hypothesis.
It is not reasonable to predict \(return_t\) by past returns as the null hypotheses in (ii) and (iii) cannot be rejected.