7. Prediction. Outline: Read Section 6.4. Mean Prediction

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1 Outline: Read Section 6.4 II. Individual Prediction IV. Choose between y Model and log(y) Model 7. Prediction Read Wooldridge (2013), Chapter Mean Prediction Predictions are useful But they are subject to sampling variations because they are obtained from the OLS estimators Example: Find the predicted value of wage: = educ +.022exper +.169tenure s.e. (.729) (.0513) (.012) (.0216) t stat [ 3.94] [11.67] [1.85] [7.82] n=526, R 2 = Goal: to find confidence intervals for a prediction. 1) Mean prediction study a confidence interval around the OLS estimate of E(y x 1,, x k ), for instance, for the average hourly wage. 2) Individual prediction study a confidence interval of the hourly wage for a particular person. What is the predicted value when educ=16; exper=4; tenure=3? = (16) +.022(4) +.169(3) = 7.3 Claim: is the estimate of the expected value of wage given the particular value of explanatory variables. Thus, this value is called mean prediction. Question: What is the s.e. of? Define = 3 4

2 Eviews: wage c educ exper tenure Date: 05/22/03 Time: 11:39 The use of se( ) or se( ) 1) to test the statistical significance of the mean prediction. 2) to find the confidence interval of mean prediction. C EDUC EXPER TENURE R-squared Mean dependent var Adjusted R-squared S.D. dependent var Log likelihood F-statistic Durbin-Watson stat Prob(F-statistic) 0 Given the estimated wage equation in y and x j form: = + x 1 + x 2 + x tenure Let educ = 16 x 1 = c 1 exper = 4 x 2 = c 2 tenure = 3 x 3 = c 3 The parameter we would like to estimate is 0 = 0 c 1 c 2 c Confidence Interval Trick to find the s.e.( ) To find confidence interval, we need 1) 2) s.e( ) Trick : rewrite 0 in terms of 0,c 1, c 2 and c 3 0 = 0 c 1 c 2 c 3 0 = 0 1 c 1 2 c 2 3 c 3 Substitute 0 = into the y equation y= 0 x 1 x 2 x 3 + u y = 0 (x 1 c 1 ) (x 2 c 2 ) (x 3 c 3 ) + u A 95% confidence interval is 2*s.e( ) Estimate to obtain s.e.( ) = + x 1 c 1 ) + (x 2 c 2 ) + (x 3 c 3 ) In Eviews, we find the estimate of s.e.( ) by running the regression of wage on (educ 16) (exper 4) (tenure 3). 7 8

3 Eviews: wage c educ-16 exper-4 tenure-3 Date: 05/22/03 Time: 11:46 C EDUC EXPER TENURE R-squared Mean dependent var Adjusted R-squared S.D. dependent var Log likelihood F-statistic Durbin-Watson stat Prob(F-statistic) 0 t Stat and Confidence Interval for predicting the average value of hourly wage 1. Is the mean prediction statistically significant at the 1% significance level? H 0 : 0 = 0 t = ( 0 )/se( ) = 7.308/0.230 = p value = 0 Note that the predicted value, its standard error, t statistic, and p value are obtained from the intercept of the transformed equation. Date: 05/22/03 Time: 11:46 C EDUC EXPER TENURE R-squared Mean dependent var Adjusted R-squared S.D. dependent var Log likelihood F-statistic Durbin-Watson stat Prob(F-statistic) Interpret: 2 s.e.( ) Compare Results from Eviews 2. What is the 95% CI for predicting the average value of hourly wage, 0? 2*se( ) = (0.2304) = (6.85, 7.77) dollars per hour INTERPRETATION: The predicted mean wage ranges from 6.85 to 7.7 dollars per hour with the 95% confidence. Date: 05/22/03 Time: 11:46 C EDUC EXPER TENURE R-squared Mean dependent var Adjusted R-squared S.D. dependent var Log likelihood F-statistic Durbin-Watson stat Prob(F-statistic) 0 Original Equation: = educ +.022exper +.169tenure s.e. (.729) (.0513) (.012) (1.85) t stat [ 3.94] [11.67] [1.85] [7.82] n=526, R 2 = Transformed Equation to Find s.e.( ) = (educ 16) +.022(exper 4) +.169(tenure 3) s.e. (.230) (.0513) (.012) (1.85) t stat [31.72] [11.67] [1.85] [7.82] n=526, R 2 =

4 II. Prediction Interval Mean Prediction confidence interval for the average hourly wage. Individual Prediction confidence interval of the hourly wage for a particular person, This is called prediction interval. What is the unknown value of y (wage) when x 1 =x 10, x 2 =x 20 and x 3 =x 30? Find a 95% CI for an unknown y The prediction error in using y 0^ to predict y 0 is 0 = y 0 0 = 0 x 10 x 20 x 30 + u 0 0 Mean and Variance of 0 E( 0 ) = 0 Var( 0 ) = Var( 0 ) y 0 = 0 x 10 x 20 x 30 + u 0 What is the predicted value of y when x 1 =x 10, x 2 =x 20 and x 3 =x 30? 0 = + x 10 + x x 3 0 What is the value of s.e.( 0 ) found in Eviews? s.e.( 0 ) = [s.e.( 0 ) ] 1/2 is S.E. of regression (= ) s.e.( 0 ) from the previous section (=0.2304) s.e.( 0 )= 3.09 II. Prediction Interval 13 II. Prediction Interval 14 Eviews: wage c educ exper tenure Date: 05/22/03 Time: 11:39 Date: 05/22/03 Time: 11:39 C C EDUC EXPER TENURE EDUC EXPER TENURE R-squared Mean dependent var R squared Mean dependent var Adjusted R squared S.D. dependent var Log likelihood F statistic Adjusted R-squared S.D. dependent var Durbin Watson stat Prob(F statistic) 0 Log likelihood F-statistic Durbin-Watson stat Prob(F-statistic) 0 II. Prediction Interval 15 16

5 t Statistic and CI Example: Individual Prediction of wage Note that 0 is normally distributed with mean 0 and variance, Var( 0 ): 0 ~ Normal[0,Var( 0 )] The t statistic is t= 0 /s.e.( 0 ), which has a t distribution with n k 1 DFs. Find a 95% Confidence interval for an unknown hourly wage when educ 0 =x 10 =16, exper 0 =x 20 =4 and tenure 0 =x 30 =3 The prediction of wage 0 is 0 = y 0 = 7.31 se( 0 ) = 3.09 A 95% confidence interval is A confidence interval of y 0 is 0 2*se( 0 ) = *3.09 P( t < t < t ) =.95 Thus, a 95% confidence interval for unknown y 0 is 0 2*s.e.( 0 ). (1.24, 13.36) dollars per hour II. Prediction Interval 17 II. Prediction Interval 18 Mean Prediction versus Individual Prediction Compared Mean Prediction Find a confidence interval around the OLS estimate of E(y x 1,,x k ) for any values of regressors. One Source of variation: the variance of sampling error. Individual Prediction Find a confidence interval for a particular unit (individual, family, firm, and so on). y 0 could be a person or firm not in our original sample. Two sources of variations: (1) the variance of sampling error and (2) the variance of unobserved error (in the population) Suppose we have a model log(salary) = log(sales)+ 2 log(mktval)+ 3 ceoten+u General form: log(y) = 0 x 1 x 2 x 3 + u x 1 = log(sales) x 2 = log(mktval) x 3 = ceoten Estimate log( )= + x 1 + x 2 + x 3 II. Prediction Interval 19 20

6 Easy way but inexact! for predicting y = exp[log( )] It will underestimate the expected value of y. The correct method to find predicted y is = exp[log( )] Example: Prediction of CEO when sales=5,000, mktal=10,000 and ceoten=10 log( ) = log(sales) +.109log(mktval) ceoten log( ) = = 1, (inexact value = exp[log( ))] Motivation why we need to find With assumptions MLR1-MLR6, it can be shown that y = exp(u)*exp( 0 x 1 x 2 x 3 ) E(y x) = exp( 2 /2)*exp( 0 x 1 x 2 x 3 ) The correct method to find predicted y is = exp[log( )] = exp[7.01] How to find? The sample counterpart is = exp( 2 /2)*exp(log( )) Given = exp[log( )] We can find the coefficient by regressing y i on exp[log( )] Eviews: Steps to find = exp[log( )] from log( ) where Example (1) Given values of x 10, x 20, x 30, obtain log( ). (2) Find the fitted values of log(y) from the estimation equation log(y) = 0 x 1 x 2 x 3 + u to obtain exp[log( )]. (3) Find from regression with no intercept. regress y on exp[log( )] (4) After is known, we can find the predicted value of y. Step (1): Estimate the model to obtain the predicted value of log(salary) or lsalary when sales = 5,000 log(sales) = mktval = 10,000 log(mktval) = ceoten = 10 ceoten = 10 = lsales +.109lmktval ceoten n =177 R 2 = log( ) = lsalaryhat = = 1, (inexact value = exp[log( )]) 23 24

7 Step 1: lsalary c lsales lmktval ceoten Dependent Variable: LSALARY Date: 07/05/03 Time: 20:42 Sample: Included observations: 177 C LSALES LMKTVAL CEOTEN R-squared Mean dependent var Adjusted R-squared S.D. dependent var S.E. of regression Akaike info criterion Sum squared resid Schwarz criterion Log likelihood F-statistic Durbin-Watson stat Prob(F-statistic) 0 Proc/Make Residuals Series use the default name, resid01 25 Step 2: obtained the fitted values in Eviews by generating a series: lsalaryhat=lsalary-resid01 obs Actual Fitted Residual Residual Plot * * * * * * *. 26 Step (3): Find by regressing y on exp[log( )) In Eviews, regress salary on exp(lsalaryhat with no intercept. Step 4: Find the predicted value of y Dependent Variable: SALARY Date: 07/05/03 Time: 20:57 Sample: Included observations: 177 EXP(LSALARYHAT) R-squared Mean dependent var Adjusted R-squared S.D. dependent var S.E. of regression Akaike info criterion Sum squared resid Schwarz criterion Log likelihood Durbin-Watson stat log( ) = lsalaryhat = exp(log( )) = exp(lsalaryhat) = 1,110,983 = exp[log( )] salaryhat = exp[lsalaryhat] salaryhat = 1.117[exp(7.013)] = 1.117[1, ] = 1, In dollars, the predicted salary is $1,240,

8 Which model is preferred? IV. Comparing R 2 for dependent variables: y and log(y) Example: CEO salary of 177 firms salary 1990 compensation, $1000s sales 1990 firm sales, millions mktval market value, end 1990, millions. ceoten years as CEO with company Comparing two models: log( ) = log(sales) +.109log(mktval)+.117log*(ceoten) = sales mktval ceoten log( ) = log(sales) +.109log(mktval) ceoten t stat [17.51] [4.15] [2.20] [2.20] n=177, R 2 = , R 2 bar= log( ) = log(sales) +.109log(mktval)+.117log*(ceoten) Trick : find the R 2 equivalence from the log (y) equation (find salary variation) = sales mktval ceoten t stat [9.40] [1.89] [2.47] [2.26] n=177, R 2 = , R 2 bar= R 2 = [corr(y i, i )] 2 is the square of the sample correlation between y i and 29 IV. Comparing R2 for dependent variables: y and log(y) 30 R 2 for dependent variables Generating new series, salaryhat =1.117exp(lsalaryhat) Open two series salary and salaryhat as group R 2 = [corr(y i, i )] 2 y i = salary i = fitted value of salary from log( ) equation (called salaryhat). In log(salary) equation, we can find = exp[log( )] In Eviews, generating a new series, salaryhat = 1.117exp(lsalaryhat) obs SALARY SALARYHAT IV. Comparing R2 for dependent variables: y and log(y) 31 IV. Comparing R2 for dependent variables: y and log(y) 32

9 In Group window, Choose View/Correlation R 2 equivalence = SALARY SALARYHAT SALARY SALARYHAT Eviews: find the correlation between salaryhat and salary and R 2. corr( i, y i ) = R 2 = [corr( i, y i )] 2 = Log dependent variable: log( ) = lsales +.109lmktval +.117ceoten R log2 = : variation in log(salary) R 2 equivalence = ( ) 2 = : variation in salary! Level dependent variable: = sales mktval ceoten R level2 = : variation in salary Which model do you prefer? IV. Comparing R2 for dependent variables: y and log(y) 33 IV. Comparing R2 for dependent variables: y and log(y) 34 Eviews: lsalary c lsales lmktval ceoten Eviews: salary sales mktval ceoten c Dependent Variable: LSALARY Date: 07/05/03 Time: 20:42 Sample: Included observations: 177 C LSALES LMKTVAL CEOTEN R-squared Mean dependent var Adjusted R-squared S.D. dependent var S.E. of regression Akaike info criterion Sum squared resid Schwarz criterion Log likelihood F-statistic Durbin-Watson stat Prob(F-statistic) 0 Dependent Variable: SALARY Date: 07/05/03 Time: 21:06 Sample: Included observations: 177 C SALES MKTVAL CEOTEN R-squared Mean dependent var Adjusted R-squared S.D. dependent var S.E. of regression Akaike info criterion Sum squared resid Schwarz criterion Log likelihood F-statistic Durbin-Watson stat Prob(F-statistic)

10 Recap of Prediction Mean Prediction Individual Prediction Predicting y from log(y) Choose between y Model and log(y) Model 37