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Use the data in NYSE to answer these questions.$\begin{array}{l}{\text { (i) Estimate the model in equation }(12.47) \text { and obtain the squared OLS residuals. Find the average, }} \\ {\text { minimum, and maximum values of } \hat{u}_{t}^{2} \text { over the sample. }}\end{array}$(ii) Use the squared OLS residuals to estimate the following model of heteroskedasticity:$$\operatorname{Var}\left(u_{t} | \operatorname{return}_{t-1}, \text { return}_{t-2}, \ldots\right)=\operatorname{Var}\left(u_{t} | r e t u r n_{t-1}\right)=\delta_{0}+\delta_{1} r e t u r n_{t-1}+\delta_{2} r e t u r n_{t-1}^{2}$$Report the estimated coefficients, the reported standard errors, the $R$ -squared, and the adjusted$R$ -squared.(iii) Sketch the conditional variance as a function of the lagged return $_{-1} .$ For what value of return $_{-1}$is the variance the smallest, and what is the variance?$\begin{array}{l}{\text { (iv) For predicting the dynamic variance, does the model in part (ii) produce any negative variance }} \\ {\text { estimates? }}\end{array}$$\begin{array}{l}{\text { (iv) For predicting the dynamic variance, does the model in part (ii) produce any negative variance }} \\ {\text { estimates? }}\end{array}$$\begin{array}{l}{\text { (v) Does the model in part (ii) seem to fit better or worse than the ARCH(1) model in }} \\ {\text { Example } 12.9 ? \text { Explain. }}\end{array}$$\begin{array}{l}{\text { (vi) To the ARCH(1) regression in equation }(12.51), \text { add the second lag, } \hat{u}_{t-2}^{2} . \text { Does this lag seem }} \\ {\text { important? Does the ARCH(2) model fit better than the model in part (ii)? }}\end{array}$

(i) (ii) (iii) see video (iv) No (v) Compared by adjusted R squared, model in part (ii) is better (vi) No

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Chapter 12

Serial Correlation and Heteroskedasticity in Time Series Regressions

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part one. The average of you I had square over. The simple is four 0.44 with the smallest value being 0.0 00 They're five zeroes 74 and the largest being 232.89 Part two. What we obtain is the same as in part two of problem 12.4, and recall from that problem, we can sketch the estimated variance function. As follow. The function takes a U shape that bottoms out at the return T minus one value of one point 33. The variance is then about 2.74 The graph in part three makes it clear that there is no negative estimated variance. Part five, the are square for their arch. One model is 10.114 and the adjusted are square is 0.112 the quadratic model in the first part. The second part is it is having a R square of one point 0.130 and the adjusted our square is 0.1 to 8. Yeah, we need to compare the models by adjusted our square because they have different number of explanatory variables. Okay, or they have different degrees of freedom. Yeah, right. Comparing by adjusted our square. We see that the quadratic model. It's better heart. Sixth, we will estimate on our sh to model The estimate on the coefficient of you had t minus two square is 0.42 And the T statistic is barely above one. The it's just our square of this model is 0.113 So we can conclude that this model arch to fits the data worse than the quadratic model in part two. Yes. Yeah.

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