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Lecture #6

Lecture #6. Studenmund(2006) Chapter 7. Objectives:. 1. Suppressing the intercept 2. Alternative Functional forms 3. Scaling and units of measurement. Estimated relationship Suppressing the intercept. ^. Such an effect potentially biases the β s and inflates their t-values. Y.

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Lecture #6

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  1. Lecture #6 Studenmund(2006) Chapter 7 Objectives: 1. Suppressing the intercept 2. Alternative Functional forms 3. Scaling and units of measurement

  2. Estimated relationship Suppressing the intercept ^ Such an effect potentially biases the βs and inflates their t-values Y True relation X

  3. The intercept term is absent or zero. = b + Y X  i.e., i 1 i i Y ^ ^ i = SRF : Y  X i 1 i ^ b 1 1 X 0 i Regression through the origin

  4. The estimated model: ~ ~ = b Y X 1 i ~ ~ or = b +  Y X 1 i i Applied OLS method: ^ ( ) 2 s ~ å X Y ~ = b Var and i i å b = 1 X 2 1 i 2 å X i ~ å 2  and ^ 2 s = N -1 Regression through the origin

  5. ~ å 1. need not be zero  i can be occasions turn out to be negative. may not be appropriate for the summary of statistics. R2 2. df 3. does not include the constant term, i.e., (n-k) Some feature of no-intercept model • In practice: • 1. A very strong priori or theoretical expectation, otherwise stick to the conventional intercept-present model. • 2. If intercept is included in the regression model but it turns out to be statistically insignificant, then we have to drop the intercept to re-run the regression.

  6. Regression through origin = b + Y X ’ = b + b + Y X  1 i 0 1 i å ^ xy å XY b = ~ b = 1 x2 å å X2 1 s ^ 2 ^ ( ) s 2 ^ ) b = ( ~ Var b = Var å å 1 1 2 x X2 ~ å 2 ’ s ^ ^ å = 2 2 N-k-1  N-k ^ - s = 2 n 1 [ ] ( ) ( ) - n 2 2 - - å X X Y Y = 2 R å 2 ( ) ( ) ( XY ) 2 2 - - å å X raw X Y Y = 2 R å å X2 Y2 ( ) 2 å xy or = 2 R å å 2 2 x y

  7. ( ) ( ) - = b - ER r ER r 1 i f m f risk free of return expected rate of return on security i expected rate of return on market portfolio b as a measure of systematic risk. 1 b >1 ==> implies a volatile or aggressive security. 1 b <1 ==> implies a defensive security. 1 Example 1: Capital Asset Pricing Model (CAPM) security i’s expect risk premium=expected market risk premium

  8. Example 1:(cont.) Security market line - ER r i f b 1 1 - ER f m

  9. - F e - = = * ( i i ) f N N e International interest rate differentials equal exchange rate forward premium. - F e ( ) - = b * i i ( ) i.e., 1 e Covered interest parity line b 1 1 Example 2:Covered Interest Parity

  10. in regression: a = E ( ) 0 0 - F e - = a + b + * ( i i ) ( ) u 0 1 i e a If covered interest parity holds, is expected to be zero. 0 Example 2:(Cont.) Use the t-test to test the intercept to be zero

  11. y: Return on A Future Fund, % X: Return on Fisher Index, % ^ Formal report: = R2=0.714 Y 1 . 0899 X N=10 SEE=19.54 (5.689)

  12. H0: 0 = 0 1.279 - 0 7.668 ^ = + Y 1 . 2797 1 . 0691 X R2=0.715 N=10 (0.166) (4.486) SEE=20.69 The t-value shows that b0 is statistically insignificant different from zero

  13. Functional Forms of Regression The term linear in a simple regression model means that there are linear in the parameters; variables in the regression model may or may not be linear.

  14. True model is nonlinear Y Income SRF PRF X Age 15 60 But run the wrong linear regression model and makes a wrong prediction

  15. Linear vs. Nonlinear Examples of Linear Statistical Models Yi = 0 + 1Xi + i Yi = 0 + 1 ln(Xi)+ i 2 ln(Yi) = 0 + 1Xi + i Yi = 0 + 1Xi + i Examples of Non-linear Statistical Models 2 2 Yi = 0 + 1Xi + i Yi = 0 + 1Xi + i Yi = 0 + 1Xi + exp(2Xi)+ i

  16. Different Functional Forms 1. Linear Attention to each form’s slope and elasticity 2. Log-Log • 3. Semilog • Linear-Log or Log-Linear 4. Polynomial 5. Reciprocal (or inverse)

  17. 2. Log-log model: This is a non- linear model = b - b1 i Y X e 0 Transform into linear log-form: X = b0 - b + ln ln Y ln  1 i X = b - b + * ln Y ln  ==> 0 1 i = b + b + * * ==> * * Y X  b = - b * where 0 1 i 1 1 dY * elasticity coefficient dY d ln Y Y * = = = b1 dX * dX d ln X X Functional Forms of Regression models

  18. Functional Forms of Regression models Quantity Demand lnY Y = b - b ln Y ln ln X = b - b Y X 0 1 1 0 lnY Quantity Demand Y lnX X = b + b ln Y ln ln X = b b Y X price 0 1 1 0 lnX X price

  19. 3.Semi log model: Log-lin model or lin-log model: = a + a + ln Y X  i 0 1 i i or = b + b + Y ln X  i 0 1 i i dY relative change in Y d ln Y dY 1 Y a = = = = absolute change in X dX dX dX Y 1 and b = absolute change in Y 1 dY dY X = = relative change in X dX d ln X 1 Functional Forms of Regression models

  20. Functional Forms of Regression models(Cont.) 4. Polynomial: Quadratic term to capture the nonlinear pattern Yi= 0 + 1 Xi +2X2i + i 1<0, 2>0 1 Yi Yi = b + b + Y ( )  i 0 1 i X 1>0, 2<0 i 1 Xi Xi = b + b + * = Where * Y ( X )  X ==> i i 0 1 i i 5. Reciprocal (or inverse) transformations X i

  21. Some features of reciprocal model Y Y 1 = b0 + b1 Y X and and and b b b < < > b0 b b > > < 0 0 0 0 0 0 0 0 1 1 1 -b b -b b 0 0 0 0 X X 0 0 - b b / 1 0 Y 1 = b0 + b1 Y Y X + and X b > b < 0 0 0 0 1 0 X - - b b / 1 0

  22. Example 1: = b + b + b + b 2 Suppose Y X X X X 0 1 1 2 1 3 1 2 X2* = X12 transforming X3* = X1X2 = b + b + b + b * * rewrite Y X X X 0 1 1 2 2 3 3 Two conditions for nonlinear, non-additive equation transformation. 1. Exist a transformation of the variable. 2. Sample must provide sufficient information.

  23. Example 2: b = b + Y 1 + b 0 X 2 1 transforming = * X + b 1 X 2 = b + b * rewrite Y X 0 1 1 However, X1* cannot be computed, because b is unknown. 2

  24. Application of functional form regression 1. Cobb-Douglas Production function: = b b2 b1  Y L K e 0 Transforming: = b + b + b + ln Y ln ln L ln K  0 1 2 = b + b + b + ln Y ln L ln K  ==> 0 1 2 d ln Y : elasticity of output w.r.t. labor input = b 1 d ln L d ln Y : elasticity of output w.r.t. capital input. = b 2 d ln K > Information about the scale of returns. b + b = 1 < 1 2

  25. 2. Polynomial regression model: Marginal cost function or total cost function costs costs (MC) = b + b + b + i.e. Y X X 2  0 1 2 MC y y or (TC) = b + b + b + b + costs Y X X 2 X 3  0 1 2 3 TC y

  26. Slope Elasticity dY dY Model Equation Y = ( ) = ( ) dX dX X dY X = b + b Y X = b b linear ( ) 0 1 1 1 dX Y dY Log-log = b + b ln Y ln X d ln Y Y b = = b 0 1 dX 1 1 d ln X X dY Y = b ==> ( ) 1 dX X Summary

  27. Slope Elasticity dY Log-lin = b + b ln Y X d ln Y Y b X = = b 0 1 1 1 dX dX dY = b ==> Y 1 dX 1 dY dY = b + b b = = b Y ln X Lin-log dX 1 1 Y d ln X 0 1 X dY 1 = b ==> 1 dX X 1 dY dY -1 = = b = b + b Y b ( ) Reciprocal - 1 1 1 0 1 1 XY X d ( ) dX 2 X X dY -1 = b ==> 1 dX X2 Summary(Count.)

  28. Linear model ^ = + G N P 100 . 304 1 . 5325 M 2 (1.368) (39.20)

  29. ^ GNP = -1.6329.21 + 2584.78 lnM2 (-23.44) (27.48) Lin-log model

  30. ^ lnGNP = 6.8612 + 0.00057 M2 (100.38) (15.65) Log-lin model

  31. ^ = + ln G NP 0 . 5529 0 . 9882 ln M 2 (3.194) (42.29) Log-log model

  32. Wage(y) 10.43 ^ wage=10.343-3.808(unemploy) SRF (4.862) (-2.66) unemp.(x)

  33. Reciprocal Model (1/unemploy) uN: natural rate of unemployment y uN 1 1 ( ) ^ ( ) Wage = -1.4282+8.7243 x x The 0 is statistically insignificant Therefore, -1.428 is not reliable (-.0690) (3.063) -1.428 SRF

  34. ^ lnwage = 1.9038 - 1.175ln(unemploy) (10.375) (-2.618)

  35. ^ 1 Lnwage = 1.9038 + 1.175 ln ( ) X (10.37) (2.618) Antilog(1.9038) = 6.7113, therefore it is a more meaningful and statistically significant bottom line for min. wage Antilog(1.175) = 3.238, therefore it means that one unit X increase will have 3.238 unit decrease in wage

  36. Procedures: 1. Run OLS on the linear model, obtain Y ^ Y = 0 + 1 X1 + 2 X2 ^ ^ ^ ^ 2. Run OLS on the log-log model and obtain lnY ^ ^ lnY = 0 + 1 lnX1 + 2 lnX2 ^ ^ ^ ^ 3. Compute Z1 = ln(Y) - lnY ^ 4. Run OLS on the linear model by adding z1 ^ Y = 0’ + 1’X1 + 2’X2 + 3’ Z1 ^ ^ ^ ^ and check t-statistic of 3’ If t*3 > tc ==> reject H0 : linear model ^ If t*3 < tc ==> not reject H0 : linear model ^ (MacKinnon, White, Davidson) MWDTest for the functional form (Wooldridge, pp.203)

  37. 5. Compute Z2 = antilog (lnY) - Y ^ ^ 6. Run OLS on the log-log model by adding Z2 ^ lnY = 0’ + 1’ ln X1 + 2’ ln X2 + 3’ Z2 ^ ^ ^ ^ ^ and check t-statistic of b’3 If t*3 > tc ==> reject H0 : log-log model ^ If t*3 < tc ==> not reject H0 : log-log model ^ MWD test for the functional form (Cont.)

  38. Example:(Table 7.3) Step 1: Run the linear model and obtain Y ^ C X1 X2  1583.279 ^ CV1 = = _ Y 24735.33 = 0.064 MWD TEST: TESTING the Functional form of regression

  39. Step 2: Run the log-log model and obtain lnY ^ C LNX1 LNX2 fitted or estimated  _ Y ^ 0.07481 CV2 = = = 0.0074 10.09653

  40. Step 4: H0 : true model is linear C X1 X2 Z1 MWD TEST tc0.05, 11 = 1.796 tc0.10, 11 = 1.363 t* < tc at 5% => not reject H0 t* > tc at 10% => reject H0

  41. Step 6: H0 : true model is log-log model C LNX1 LNX2 Z2 C.V.1 Comparing the C.V. = 0.064 = 0.0074 C.V.2 MWD Test tc0.025, 11 = 2.201 tc0.05, 11 = 1.796 tc0.10, 11 = 1.363 Since t* < tc => not reject H0

  42. Criterion for comparing two different functional models: The coefficient of variation: C.V. = It measures the average error of the sample regression function relative to the mean of Y. Linear, log-linear, and log-log equations can be meaningfully compared. ^  Y The smaller C.V. of the model, the more preferredequation (functional model).

  43. Compare two different functional form models: Model 1 linear model Model 2 log-log model  / Yof model 1 ^ 0.0236 2.1225/89.612 Coefficient Variation (C.V.) = = 0.0048 0.0217/4.4891  / Yof model 2 ^ = 4.916 means that model 2 is better

  44. = b + b Y X + i 0 1 b : the slope of the regression line. 1 D Y dY Units of change of y b = = or 1 D X dX Units of change of x if Y* = 1000Y X* = 1000X ^ ^ ^ then b = b + X + 1000 Y 1000 1000  1000 1 0 i ^ ^ * ^ = * + b + * * Y X  ==> 0 1 Scaling and units of measurement

  45. Yi = 0 + 1Xi + i Yi/k = (0/k)+(1)Xi/k+ i/k * * * Yi = 0 + 1Xi+ i * * i = i/k * Yi = Yi/k where * * 0 = 0/k Xi = Xi/k and Changing the scale of X and Y R2 and the t-statistics are no change in regression results for 1 but all other statistics are change.

  46. Y 25 ^ 0* 5 ^ 0 X 10 50

  47. Yi = 0 + 1Xi + i Yi = 0 + (k1)(Xi/k)+ i * Yi = 0 + 1Xi+ i * where and * * 1 = k1 Xi = Xi/k Changing the scale of x The estimated coefficient and standard error change but the other statistics are unchanged.

  48. Y 5 ^ 0 X 10 50 50

  49. Yi = 0 + 1Xi + i Yi/k = (1/k)+ (1/k)Xi + i/k * * * * Yi = 0 + 1Xi + i * i = i/k * Yi = Yi/k where * * 0 = 0/k 1 = 1/k and Changing the scale of Y All statistics are changed except for the t-statistics and R2.

  50. Y 25 5 ^ 0 X 10

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