0bA Final: Open book Total points: 10 Duration: 120 minutes ate : -2018/1/1. lec

Question

0bA Final: Open book Total points: 10 Duration: 120 minutes ate : -2018/1/1. lecturer: Yucan Liu Authorizer: Edit d Section 1. (30 points) Subsection 1(8 points) 040008 040008 040008 040008 040008 040008 040008 040008 040008 040008 040008 040008 040008 040008 040008 040008 040008 040008 date 2007/9/30 0.0190 0.0470 2007/12/31 0.0282 0.1468 2008/3/31 0.0515 0.1953 2008/6/30 0.0490 0.2310 2008/9/30 0.0176 0.0694 2008/12/31 0.0054 0.0131 2009/3/31 0.0309 0.1859 2009/6/30 0.075 0.1151 2009/9/30 0.02390.0506 2009/12/31 0.0043 0.0290 2010/3/31 0.00540.0241 2010/6/30 0.0197 -0.0773 2010/9/30 0.0091 0.0096 2010/12/31 0.0084 -0.0102 2011/3/31 0.0214 0.0004 2011/6/30 0.0109 0.0214 2011/9/30 -0.0256 -0.0917 2011/12/31 -0.0184 -0.0759 The sample data are given in the above table. Suppose that the model of the relationship between y(quarterly fund return ) and x(quarterly stock market return) is given as follows: y bi+ bxte. Some values are given as follow: x1-0.193 Question: ccording to the formulas which you learned in Econometrics, what are values of the estimators by and b2? 1

Explanation / Answer

Subsection-1).

Consider the given problem here “Y=Quarterly fund return” and “X=Quarterly stock market return” and our regression model is given below.

=> Yi = b1 + b2*Xi, there are “n=18” observations.

We have also given that Y = -0.0370, X = -0.109, XY = 0.045 and X^2 = 0.193.

Now, here “mean(X) = X/n = (-0.109)/18 = (-0.00606)” and “mean(Y) = Y/n = (-0.0370)/18 = (-0.00206).

=> The estimated value of “b2” is “[XY – n*mean(X)*mean(Y) ] / [ X^2 – n*mean(X)^2]”.

Now, “[XY – n*mean(X)*mean(Y)]”, => [0.045 – 18*(-0.00606)* (-0.00206)] = [0.045 – 0.00022] = 0.04478.

Similarly, “[X^2 – n*mean(X)^2]”, => [0.193 – 18*(-0.00606)^2] = [0.193 – 0.00066] = 0.19234.

=> The estimated value of “b2” is “[XY – n*mean(X)*mean(Y)] / [X^2 – n*mean(X)^2]”,

=> 0.04478/0.19234 = 0.2328.

So, now the estimated value of “b1” is “mean(Y) – b2*mean(X)”, => “(-0.00206) - 0.2328*(-0.00606) = (-0.00065).

So, the estimated value of the “b1” is “-0.00065” and the estimated value of the “b2” is “0.2328”.

So, the estimated regression model is given below.

=> Y = (-0.00065) + 0.2328*X.

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