Temple Universtity

Department of Economics

Homework 4: Workouts in OLS

1. It is pretty straightforward to show that the alternative estimator is unbiased.
2. By expanding the multiplication and taking expectations it is obvious that is unbiased.

Now show that has a larger variance.

Numerically this is

The variance for the OLS estimator is

Although both estimators are unbiased, OLS has the smaller variance in the positive definite sense. This result is not surprising, given our knowledge of the Gauss-Markov Theorem.

3. A. You can show this several different ways:
1. Compute the inner product of x1 and x2 to show that they are not orthogonal
2. Orthogonality is sufficient but not necessary for linear independence.

3. Compute the coefficients in the simple regression of x2 on x1 and show that the residual sum of squares is not zero. LS essentially tries to find the linear relationship between two variables.
4. Compute the determinant of the augmented matrix made from the column vectors x1 and x2.

We already knew that the off-diagonals wouldn't be zero, but it is still possible for the determinant to be zero, which would indicate linear dependence in the columns of x.

2.b. The OLS estimates, using the original data are

2.c. After adding 5 to each obs on y and 10 to each obs on x the OLS estimates are now

Notice that the intercept changed, but neither of the slopes is changed.

2.d. After multiplying each observation on the dependent variable by 10 the OLS coefficients change to:

2.e. If x1 is multiplied by 10 and the data is otherwise unchanged then the OLS results change to:

2.f. The effect of adding a constant to both sides of the equation is to change the intercept. The effect of multiplying the dependent variable by 10 is to rescale all of the coefficients by a power of 10. Multiplying just one of the independent variables by 10 has the effect of rescaling only the corresponding coefficient.  Notice also, in your regression output, that R2,  t-statistics and the F-statistic are unaffected by rescaling the data.

3.b. The estimated coefficients are

 Ordinary least squares regression.Dep. Variable = LNRENT Observations = 82 Mean of LHS = 0.2020316E+01 StdDev of residuals= 0.3841586E+00 R-squared = 0.9084267E+00 F[ 8, 73] = 0.9052197E+02 Log-likelihood = -0.3313692E+02 Amemiya Pr. Criter.= 0.1027730E+01 Durbin-Watson stat.= 2.2420910 Ordinary least squares regression. Weights = ONEStd.Dev of LHS = .1205161E+01 Sum of squares = 0.1077318E+02 Adjusted R-squared= .89839E+00 Restr.(á=0) Log-l = -0.13115E+03 Akaike Info.Crit. = 0.1637754E+00 Autocorrelation = -0.1210455 ANOVA SourceRegression Residual Total Variation0.1068723E+03 0.1077318E+02 0.1176455E+03 D of Fr8 73 81 Mean Square0.1335903E+02 0.1475778E+00 0.1452413E+01 Variable Coefficients Std Error t-ratio Pr|t|>x Mean of X Std Dev of x ConstantD61 D62 D63 D64 D65 LNMULT LNMEM LNACCES -0.10446 -0.13980 -0.48911 -0.59385 -0.92482 -1.1632 -.6536E-01 0.57933 -0.14060 0.3149 0.1665 0.1738 0.1661 0.1663 0.1661 .284E-01 .353E-01 .293E-01 -0.332 -0.840 -2.815 -3.575 -5.561 -7.003 -2.301 16.369 -4.794 0.74100.4037 0.0062 0.0006 0.0000 0.0000 0.0242 0.0000 0.0000 0.14634 0.13415 0.18293 0.21951 0.19512 4.7598 5.7263 1.8266 0.35562 0.34291 0.38899 0.41646 0.39873 2.6499 1.5140 2.3422

3.b. The quality adjusted price index series is:

The quality adjusted price index is declining throughout.

3.c. The results for the new model are:

 Ordinary least squares regression.Dep. Variable = LNRENT Observations = 82 Mean of LHS = 0.2020316E+01 StdDev of residuals= 0.4907885E+00 R-squared = 0.8423462E+00 F[ 4, 77] = 0.1028530E+03 Log-likelihood = -0.5541066E+02 Amemiya Pr. Criter.= 0.1473431E+01 Durbin-Watson stat.= 1.5506978 Ordinary least squares regression. Weights = ONEStd.Dev of LHS = .1205161E+01 Sum of squares = 0.1854725E+02 Adjusted R-squared= 0.8341564E+00 Restr.(á=0) Log-l = -0.1311522E+03 Akaike Info.Crit. = 0.2555608E+00 Autocorrelation = 0.2246511 ANOVA SourceRegression Residual Total Variation0.9909821E+02 0.1854725E+02 0.1176455E+03 Deg of Freedom4 77 81 Mean Square0.2477455E+02 0.2408734E+00 0.1452413E+01 Variable Coefficients Std Error t-ratio Pr|t|>x Mean of X Std Dev of x Constant -DUMM1 LNMULT LNMEM LNACCES -0.57377 -0.60048 -.3732E-01 0.61148 -0.11087 0.38560.1762 0.353E-01 0.439E-01 0.369E-01 -1.488-3.407 -1.055 13.926 -2.999 0.1408.00105 0.2947 0.0000 0.0036 0.8780 4.7598 5.7263 1.8266 0.32924 2.6499 1.5140 2.3422

The result of using a single dummy for the period 61 to 65 is to increase the magnitude of each of the slope coefficients, to reduce both the R2 and the adjusted R2 , and the coefficient on LNMULT is no longer different from zero.