7. Prediction

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7. Prediction

Read Wooldridge (2013), Chapter 6.4

7. Prediction . Quantitative Methods of Economic Analysis . 2949605 . Chairat Aemkulwat

Outline: Read Section 6.4

• I. Mean Prediction

• II. Individual Prediction

• III. Predicting y from log(y)

• IV. Choose between y‐Model and log(y)‐Model

I. Mean II. Individual III. y^ from logy IV. R

²

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I. Mean Prediction

• Predictions are useful

– But they are subject to sampling variations because they are obtained from the OLS estimators

• 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.

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7. Prediction . Quantitative Methods of Economic Analysis . 2949605 . Chairat Aemkulwat I. Mean Prediction

Mean Prediction

Example: Find the predicted value of wage:

= ‐2.873 +.599educ + .022exper +.169tenure

s.e. (.729) (.0513) (.012) (.0216) t‐stat [‐3.94] [11.67] [1.85] [7.82]

n=526, R

²

=.306422

• What is the predicted value when educ=16; exper=4; tenure=3?

= -2.873 +.599(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 =

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Dependent Variable: WAGE Method: Least Squares Date: 05/22/03 Time: 11:39 Sample: 1 526

Included observations: 526

Variable Coefficient Std. Error t-Statistic Prob.

C -2.87274 0.728964 -3.940844 0.0001

EDUC 0.598965 0.051284 11.67948 0

EXPER 0.02234 0.012057 1.852849 0.0645

TENURE 0.169269 0.021645 7.820361 0

R-squared 0.306422 Mean dependent var 5.896103

Adjusted R-squared 0.302436 S.D. dependent var 3.693086

S.E. of regression 3.084476 Akaike info criterion 5.098216

Sum squared resid 4966.303 Schwarz criterion 5.130652

Log likelihood -1336.83 F-statistic 76.87317

Durbin-Watson stat 1.791222 Prob(F-statistic) 0

Eviews: wage c educ exper tenure

²

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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.

• Given the estimated wage equation in y and x j form:

= + x 1 + x 2 + x 3

+ + + 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 +  1 c 1 +  2 c 2 +  3 c 3

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Confidence Interval

• To find confidence interval, we need 1)

2) s.e( )

• A 95% confidence interval is

 2*s.e( )

²

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Trick to find the s.e.( )

• Trick : rewrite  0 in terms of  0 , c 1 , c 2 and c 3

 0 =  0 +  1 c 1 +  2 c 2 +  3 c 3

 0 =  0 ‐  1 c 1 ‐  2 c 2 ‐  3 c 3

Substitute  0 = … into the y equation y=  0 +  1 x 1 +  2 x 2 +  3 x 3 + u

y =  0 +  1 (x 1 – c 1 ) +  2 (x 2 – c 2 ) +  3 (x 3 – c 3 ) + u

• 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).

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C 7.30787 0.230378 31.72118 0

EDUC-16 0.598965 0.051284 11.67948 0

EXPER-4 0.02234 0.012057 1.852849 0.0645

TENURE-3 0.169269 0.021645 7.820361 0

Eviews: wage c educ-16 exper-4 tenure-3

²

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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

= 31.72118 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.

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Dependent Variable: WAGE Method: Least Squares Date: 05/22/03 Time: 11:46 Sample: 1 526 Included observations: 526

C 7.30787 0.230378 31.72118 0

EDUC-16 0.598965 0.051284 11.67948 0

EXPER-4 0.02234 0.012057 1.852849 0.0645

TENURE-3 0.169269 0.021645 7.820361 0

Adjusted R-squared 0.302436 S.D. dependent var 3.693086 S.E. of regression 3.084476 Akaike info criterion 5.098216 Sum squared resid 4966.303 Schwarz criterion 5.130652

Interpret:  2 s.e.( )

2. What is the 95% CI for predicting the average value of hourly wage,



₀

?

 2*se( )

= 7.31  2(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.

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Dependent Variable: WAGE Method: Least Squares Date: 05/22/03 Time: 11:46 Sample: 1 526 Included observations: 526

C 7.30787 0.230378 31.72118 0

EDUC-16 0.598965 0.051284 11.67948 0

EXPER-4 0.02234 0.012057 1.852849 0.0645

TENURE-3 0.169269 0.021645 7.820361 0

Adjusted R-squared 0.302436 S.D. dependent var 3.693086 S.E. of regression 3.084476 Akaike info criterion 5.098216 Sum squared resid 4966.303 Schwarz criterion 5.130652

Compare Results from Eviews

Original Equation:

= ‐2.873 +.599educ + .022exper +.169tenure s.e. (.729) (.0513) (.012) (1.85) t‐stat [‐3.94] [11.67] [1.85] [7.82]

n=526, R

²

=.306422 Transformed Equation to Find s.e.( )

= 7.31 +.599(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

=.306422

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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

?

y

⁰

= 

₀

+ 

₁

x

₁⁰

+ 

₂

x

₂⁰

+ 

₃

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

₂⁰

+ x

₃⁰

²

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7. Prediction . Quantitative Methods of Economic Analysis . 2949605 . Chairat Aemkulwat II. Prediction Interval

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

+ 

₁

x

₁⁰

+ 

₂

x

₂⁰

+ 

₃

x

₃⁰

+ u

⁰

‐

⁰

• Mean and Variance of ̂

⁰

E( ̂

⁰

) = 0

Var( ̂

⁰

) = Var(

⁰

) + 

²

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• What is the value of s.e.( ̂

⁰

) found in Eviews?

s.e.( ̂

⁰

) = [s.e.(

⁰

)

^2 + ²

]

^1/2

is S.E. of regression (=3.084475)

s.e.(

⁰

) from the previous section (=0.2304)

 s.e.( ̂

⁰

) = 3.09

Included observations: 526

C -2.87274 0.728964 -3.940844 0.0001

EDUC 0.598965 0.051284 11.67948 0

EXPER 0.02234 0.012057 1.852849 0.0645

TENURE 0.169269 0.021645 7.820361 0

Eviews: wage c educ exper tenure

²

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16 Dependent Variable: WAGE

Method: Least Squares Date: 05/22/03 Time: 11:39 Sample: 1 526 Included observations: 526

C -2.87274 0.728964 -3.940844 0.0001

EDUC 0.598965 0.051284 11.67948 0

EXPER 0.02234 0.012057 1.852849 0.0645

TENURE 0.169269 0.021645 7.820361 0

R‐squared 0.306422 Mean dependent var 5.896103 Adjusted R‐squared 0.302436 S.D. dependent var 3.693086 S.E. of regression 3.084476 Akaike info criterion 5.098216 Sum squared resid 4966.303 Schwarz criterion 5.130652

Log likelihood -1336.83 F‐statistic 76.87317

Durbin‐Watson stat 1.791222 Prob(F‐statistic) 0

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t‐Statistic and CI

• 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.

• A 95% confidence interval is P(‐t

_0.025

< t < t

_0.025

) = .95

Thus , a 95% confidence interval for unknown y

⁰

is

⁰

 2*s.e.(

⁰

).

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Example: Individual Prediction of wage

• 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 confidence interval of y

⁰

is

0

 2*se( ̂

⁰

) = 7.31  2*3.09 (1.24, 13.36) dollars per hour

²

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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)

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III. Predicting y from log(y)

• Suppose we have a model

log(salary) =  0 + 1 log(sales)+ 2 log(mktval)+ 3 ceoten+u

• General form:

log(y) =  0 +  1 x 1 +  2 x 2 +  3 x 3 + u x 1 = log(sales)

x 2 = log(mktval) x 3 = ceoten

• Estimate

log( ) = + x 1 + x 2 + x 3

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7. Prediction . Quantitative Methods of Economic Analysis . 2949605 . Chairat Aemkulwat III. Predicting y from log(y)

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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( )]

 Motivation why we need to find

With assumptions MLR1-MLR6, it can be shown that y = exp(u)*exp(

₀

+ 

₁

x

₁

+ 

₂

x

₂

+ 

₃

x

₃

) E(y|x) = exp(

²

/2)*exp(

₀

+ 

₁

x

₁

+ 

₂

x

₂

+ 

₃

x

₃

)

The sample counterpart is

= exp(

²

/2)*exp(log( ))

²

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Given = exp[log( )]

We can find the coefficient by regressing y i on exp[log( )].

log( ) = 4.504 + .163log(sales) +.109log(mktval) +.0117ceoten log( ) = 7.013

= 1,110.983 (inexact value = exp[log( ))]

 The correct method to find predicted y is

= exp[log( )]

= exp[7.01] How to find ?

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 Example: Prediction of CEO when sales=5,000, mktal=10,000 and ceoten=10

Eviews: Steps to find from log( ) where

= exp[log( )]

(1) Given values of x 1 0 , x 2 0 , x 3 0 , obtain log( ).

(2) Find the fitted values of log(y) from the estimation equation

log(y) =  0 +  1 x 1 +  2 x 2 +  3 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.

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Example

• Step (1): Estimate the model to obtain the predicted value of log(salary) or lsalary when

sales = 5,000 log(sales) = 8.517193 mktval = 10,000 log(mktval) = 9.21034

ceoten = 10 ceoten = 10

= 4.504 +.163lsales +.109lmktval + .0117ceoten n =177 R

²

= 00.318

 log( ) = lsalaryhat = 7.013

= 1,110.983 (inexact value = exp[log( )])

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Dependent Variable: LSALARY Method: Least Squares Date: 07/05/03 Time: 20:42 Sample: 1 177

Included observations: 177

C 4.503795 0.257234 17.50852 0

LSALES 0.162854 0.039242 4.149993 0.0001

LMKTVAL 0.109243 0.049595 2.202717 0.0289

CEOTEN 0.011705 0.005326 2.19776 0.0293

Step 1: lsalary c lsales lmktval ceoten

Proc/Make Residuals Series … use the default name, “resid01”

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obs Actual Fitted Residual Residual Plot

1 7.057037 7.047401 0.009636 | . * . | 2 6.39693 6.305271 0.091659 | . * . | 3 5.937536 6.139374 -0.20184 | .*| . |

.. .. ..

174 5.220356 6.288329 -1.06797 | * . | . | 175 5.958425 6.317993 -0.35957 | * | . | 176 7.705262 6.317698 1.387564 | . | . * | 177 6.098074 6.136342 -0.03827 | . * . |

Step 2: obtained the fitted values in Eviews by generating a series: lsalaryhat=lsalary-resid01

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Step (3): Find by regressing y on exp[log( ))

In Eviews, regress salary on exp(lsalaryhat with no intercept.

 

Dependent Variable: SALARY Method: Least Squares Date: 07/05/03 Time: 20:57 Sample: 1 177

Included observations: 177

EXP(LSALARYHAT) 1.116857 0.047095 23.7148 0

Sum squared resid 46113901 Schwarz criterion 15.3376

Log likelihood -1354.79 Durbin-Watson stat 2.08316

²

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• Step 4: Find the predicted value of y



log( ) = lsalaryhat = 7.013

exp(log( )) = exp(lsalaryhat) = 1,110,983

= exp[log( )]

salaryhat = exp[lsalaryhat]

salaryhat = 1.117[exp(7.013)]

= 1.117[1,110.982]

= 1,240.967

In dollars, the predicted salary is $1,240,967.

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Which model is preferred?

• 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

log( ) = 4.504 + .163log(sales) +.109log(mktval) + .0117ceoten t‐stat [17.51] [4.15] [2.20] [2.20]

n=177, R

²

=0.31815, R

²

‐bar=.306327

= 613.43 +.019sales + .0234mktval + 12.703ceoten t‐stat [9.40] [1.89] [2.47] [2.26]

n=177, R

²

= 0.201274, R

²

‐bar=.187424

²

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IV. Comparing R 2 for dependent variables: y and log(y)

• Comparing two models:

log( ) = 4.504 +.163log(sales) +.109log(mktval)+.117log*(ceoten)

= 613.43 +.019sales + .0234mktval + 12.703ceoten

• log( ) = 4.504 +.163log(sales) +.109log(mktval)+.117log*(ceoten)

Trick : find the R 2 – equivalence from the log (y) equation (find salary variation)

R 2 = [corr(y i , i )] 2 is the square of the sample correlation between y i and

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7. Prediction . Quantitative Methods of Economic Analysis . 2949605 . Chairat Aemkulwat IV. Comparing R2 for dependent variables: y and log(y)

R 2 for dependent variables

• 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)

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Generating new series,

salaryhat =1.117exp(lsalaryhat)

Open two series – salary and salaryhat as “group”

obs SALARY SALARYHAT

1 1161 1284.401

2 600 611.5016

3 379 518.0234

.. ..

174 185 601.2289

175 387 619.3311

176 2220 619.1484

177 445 516.4551

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In “Group” window, Choose View/Correlation

Eviews: find the correlation between salaryhat and salary and R

²

. corr(

_i

, y

_i

) = 0.493

R

²

= [corr(

_i

, y

_i

)]

²

= 0.243

SALARY SALARYHAT

SALARY 1 0.493032

SALARYHAT 0.493032 1

²

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R 2 ‐equivalence = 0.243

Log dependent variable:

log( ) = 4.504 + .163lsales + .109lmktval + .117ceoten R

_log²

=0.31815: variation in log(salary)

R

²

‐equivalence = (0.49302)

²

= 0.243

: variation in salary!

Level dependent variable:

= 613.43 +.019sales + .0234mktval + 12.703ceoten R

_level²

= 0.201274: variation in salary

Which model do you prefer?

²

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Dependent Variable: LSALARY Method: Least Squares Date: 07/05/03 Time: 20:42 Sample: 1 177

C 4.503795 0.257234 17.50852 0

LSALES 0.162854 0.039242 4.149993 0.0001

LMKTVAL 0.109243 0.049595 2.202717 0.0289

CEOTEN 0.011705 0.005326 2.19776 0.0293

Durbin-Watson stat 2.043631 Prob(F-statistic) 0

Eviews: lsalary c lsales lmktval ceoten

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Eviews: salary sales mktval ceoten c

Dependent Variable: SALARY Method: Least Squares Date: 07/05/03 Time: 21:06 Sample: 1 177

Included observations: 177

C 613.4361 65.23685 9.403214 0

SALES 0.019019 0.010056 1.89129 0.0603

MKTVAL 0.0234 0.009483 2.467719 0.0146

CEOTEN 12.70337 5.618052 2.261169 0.025

Adjusted R-squared 0.187424 S.D. dependent var 587.5893 S.E. of regression 529.6707 Akaike info criterion 15.40473

Sum squared resid 48535332 Schwarz criterion 15.47651

I. Mean II. Individual III. y^ from logy IV. R

²

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Recap of Prediction

• Mean Prediction

• Individual Prediction

• Predicting y from log(y)

• Choose between y‐Model and log(y)‐Model

²

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