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$\begin{array}{l}{\text { (i) Use NYSE to estimate equation }(12.48) . \text { Let } \hat{h}_{t} \text { be the fitted values from this equation (the esti-i- }} \\ {\text { mates of the conditional variance). How many } \hat{h}_{t} \text { are negative? }}\end{array}$$\quad \text {(ii) Add } \text {return}_{t-1}^{2} \text { to }(12.48) \text { and again compute the fitted values, } \hat{h}_{t} \text { . Are any } \hat{h}_{t}$$\begin{array}{l}{\text { (iii) Use the } \hat{h}_{t} \text { from part (ii) to estimate }(12.47) \text { by weighted least squares (as in Section } 8-4 ) \text { . }} \\ {\text { Compare your estimate of } \beta_{1} \text { with that in equation }(11.16) . \text { Test } \mathrm{H}_{0} : \beta_{1}=0 \text { and compare the }} \\ {\text { outcome when OLS is used. }}\end{array}$$\begin{array}{l}{\text { (iv) Now, estimate }(12.47) \text { by WLS, using the estimated ARCH model in }(12.51) \text { to obtain the } \hat{h}_{t}} \\ {\text { Does this change your findings from part (iii)? }}\end{array}$

(i) $\hat{\mathrm{h}}_{\mathrm{t}}=4.66-1.104$ return $_{\mathrm{t}}$ for each $\mathrm{t}$(ii) $\frac{.789}{[2(.297)]} \approx 1.33$(iii) $\hat{\beta}_{0} \approx .155(\mathrm{se} \approx .078)$ and $\hat{\beta}_{1} \approx .039(\mathrm{se} \approx .046)$(iv) $\hat{\beta}_{0} \approx .159(\mathrm{se} \approx .078)$ and $\hat{\beta}_{1} \approx .024(\mathrm{se} \approx .047)$

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Chapter 12

Serial Correlation and Heteroskedasticity in Time Series Regressions

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part one. After obtaining the residuals, UT head from Equation 11.16 and then estimating equation 12.48. We can compute the fitted values H T V Hat as a function of return T for each T. H T hat is equal to 4.66 minus 1.104 times return sub d. When we examine h T had, it turns out that 12 out of almost 700 fitted values are negative. This means we cannot directly apply weighted least square using the hetero Scholastic City function in 12.48. In Part two, we add the first leg of return Square to the equation we get based on this equation. The conditional variance is a quadratic. In the first leg of return, additional variants takes a U shape that bottoms out at the value of first leg of return that satisfy this equation. It's a value that makes this partial derivative equal to zero. Mhm solving. For the value we get 1.33 This is the value of return T minus one that bottoms out the value of conditional mean, sorry conditional variance from Patou. We find out that the smallest conditional variance is greater than zero. So we can use weighted least squares with the fitted values H T hat obtained from the quadratic hetero Scholastic City function. When we apply weighted least squares to equation 12.47 we obtain later zero hat of 00.155 It's centered. Errol is going 078 The estimate on beta one is 0.39 and the standard error is 0.46 So the coefficient on return T minus one once waited lee squares has been used is even less significant. The T statistic is about 0.85 This is compared to you when we use old ls part four We obtained the Weighted Lee Square using an outer regressive, conditional hetero status titty variance function. We win first estimate the equation in 12 point 51 and then obtained the fitted values. The weighted least square results are now fine. 159 for beta not had 0.24 for beta one hat and the centre's heroines are 0.76 and 0.47 respectively. We find that the coefficients and T statistics are even smaller, okay? And we agree Thio are finding in part three. Yeah,

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