Econometrics I
Professor William Greene
Stern School of Business
Department of Economics
10-1/21
Part 10: Prediction
Econometrics I
Part 10 - Prediction
10-2/21
Part 10: Prediction
Forecasting
Objective: Forecast
Distinction: Ex post vs. Ex ante forecasting




Prediction vs. model validation.



10-3/21
Ex post: RHS data are observed
Ex ante: RHS data must be forecasted
Within sample prediction
“Hold out sample”
Part 10: Prediction
Prediction Intervals
Given x0 predict y0.
Two cases:
Estimate E[y|x0] = x0;
Predict y0 = x0 + 0
Obvious predictor, b’x0 + estimate of 0. Forecast 0 as 0, but allow for
variance.
Alternative: When we predict y0 with bx0, what is the 'forecast error?'
Est.y0 - y0 = bx0 - x0 - 0, so the variance of the forecast error is
x0Var[b - ]x0 + 2
How do we estimate this? Form a confidence interval. Two cases:
If x0 is a vector of constants, the variance is just x0 Var[b] x0. Form
confidence interval as usual.
If x0 had to be estimated, then we use a random variable. What is the
variance of the product? (Ouch!) One possibility: Use bootstrapping.
10-4/21
Part 10: Prediction
Forecast Variance
Variance of the forecast error is
2 + x0’ Var[b]x0 = 2 + 2[x0’(X’X)-1x0]
If the model contains a constant term, this is

1 K 1 K 1 0
0
0
jk 
Var[e ]   1    ( x j  x j )( xk  xk )( ZM Z) 
 N j 1 k 1

0
2
In terms squares and cross products of deviations from
means. Interpretation: Forecast variance is smallest
in the middle of our “experience” and increases as we
move outside it.
10-5/21
Part 10: Prediction
Butterfly Effect
10-6/21
Part 10: Prediction
Internet Buzz Data
10-7/21
Part 10: Prediction
A Prediction Interval
Prediction includes a range of uncertainty
Point estimate: yˆ  a  bx*
The range of uncertainty around the prediction:
2


1
(x
*

x)
2
a  bx*  1.96 S e  1+  N
2 
 N i1(xi  x) 
The usual 95%
10-8/21
Due to ε
Due to estimating α and β with a and b
Part 10: Prediction
Slightly Simpler Formula for Prediction
Prediction includes a range of uncertainty
Point estimate: yˆ  a  bx*
The range of uncertainty around the prediction:
2
 1
2
a  bx*  1.96 S  1+   (x *  x)  SE(b) 
 N
2
e
10-9/21
Part 10: Prediction
Prediction from Internet Buzz Regression
Buzz
= 0.48242
Max(Buzz)= 0.79
10-10/21
Part 10: Prediction
Prediction Interval for Buzz = .8
Predict Box Office for Buzz = .8
a+bx = -14.36 + 72.72(.8) = 43.82
1

se2 1    (.8  Buzz)2 SE(b)2
N

1 

2
2
 13.38632 1 

(.8

.48242)
10.94
62 

 13.93
Interval = 43.82  1.96(13.93)
= 16.52 to 71.12
10-11/21
Part 10: Prediction
Dummy Variable for One Observation
A dummy variable that isolates a single
observation. What does this do?
Define d to be the dummy variable in question.
Z = all other regressors. X = [Z,d]
Multiple regression of y on X. We know that
X'e = 0 where e = the column vector of
residuals. That means d'e = 0, which says that
ej = 0 for that particular residual. The
observation will be predicted perfectly.
Fairly important result. Important to know.
10-12/21
Part 10: Prediction
Oaxaca Decomposition
Two groups, two regression models: (Two time periods,
men vs. women, two countries, etc.)
y1 = X11 + 1 and y2 = X22 + 2
Consider mean values,
y1* = E[y1|mean x1] = x1* 1
y2* = E[y2|mean x2] = x2* 2
Now, explain why y1* is different from y2*. (I.e., departing
from y2, why is y1 different?) (Could reverse the roles
of 1 and 2.)
y1* - y2* = x1* 1 - x2* 2
= x1*(1 - 2)
+ (x1* - x2*) 2
(change in model)
(change in conditions)
10-13/21
Part 10: Prediction
The Oaxaca Decomposition
Two groups (e.g., men=1, women=2)
Regression predictions:
ˆ
y1  x1 b1 , ˆ
y 2  x 2b2 (e.g., wage equations)
Explain ˆ
y1 - ˆ
y2.
ˆ
y1 - ˆ
y 2  x1 (b1 - b2 )
+ (x1 - x 2 )b2
discrimination + qualifications
Var[x1 (b1 - b2 )]=x1 {12 ( X1 X1 ) 1  22 ( X 2 X 2 ) 1 } x 1
Wald : W=(x1 (b1 - b2 ))2 / [x1 {12 ( X1 X1 ) 1  22 ( X 2 X 2 ) 1 } x 1 ]
What is the hypothesis?
10-14/21
Part 10: Prediction
Application - Income
German Health Care Usage Data, 7,293 Individuals, Varying Numbers of
Periods
Variables in the file are
Data downloaded from Journal of Applied Econometrics Archive. This is an
unbalanced panel with 7,293 individuals. They can be used for regression, count
models, binary choice, ordered choice, and bivariate binary choice. This is a
large data set. There are altogether 27,326 observations. The number of
observations ranges from 1 to 7. (Frequencies are: 1=1525, 2=2158, 3=825,
4=926, 5=1051, 6=1000, 7=987).
HHNINC = household nominal monthly net income in German marks / 10000.
(4 observations with income=0 were dropped)
HHKIDS = children under age 16 in the household = 1; otherwise = 0
EDUC = years of schooling
AGE = age in years
MARRIED = 1 if married, 0 if not
FEMALE = 1 if female, 0 if male
10-15/21
Part 10: Prediction
Regression: Female=0 (Men)
10-16/21
Part 10: Prediction
Regression Female=1 (Women)
10-17/21
Part 10: Prediction
Pooled Regression
10-18/21
Part 10: Prediction
Application
namelist ; X = one,age,educ,married,hhkids$
? Get results for females
include ; new ; female=1$
Subsample females
regr
; lhs=hhninc;rhs=x$
Regression
matrix
; bf=b ; vf = varb ; xbarf = mean(x) $ Coefficients, variance, mean X
calc
; meanincf = bf'xbarf $
Mean prediction for females
? Get results for males
include ; new ; female=0$
Subsample males
regr
; lhs=hhninc;rhs=x$
Regression
matrix
; bm=b ; vm = varb ; xbarm = mean(x) $ Coefficients, etc.
calc
; meanincm = bm'xbarm $
Mean prediction for males
? Examine difference in mean predicted income
calc
; list
; meanincm ; meanincf
Display means
; diff = xbarm'bm - xbarf'bf $
Difference in means
matrix
; vdiff = xbarm'[vm]xbarm + xbarf'[vf]xbarf $ Variance of difference
calc
; list ; diffwald = diff^2 / vdiff $
Wald test of difference = 0
? “Discrimination” component of difference
matrix
; db = bm-bf ; discrim = xbarm'db
Difference in coeffs., discrimination
; vdb = vm+vf ; vdiscrim = xbarm'[vdb]xbarm $ Variance of discrimination
calc
; list ; discrim ; dwald = discrim^2 / vdiscrim $ Walt test that D = 0.
? “Difference due difference in X”
matrix
; dx = xbarm - xbarf $
Difference in characteristics
matrix
; qual = dx'bf ; vqual = dx'[vf]dx $
Contribution to total difference
calc
; list ; qual ; qualwald = qual^2/vqual $ Wald test.
10-19/21
Part 10: Prediction
Results
+------------------------------------+
| Listed Calculator Results
|
+------------------------------------+
MEANINCM =
.359054
MEANINCF =
.344495
DIFF
=
.014559
DIFFWALD =
52.006502
DISCRIM
DWALD
10-20/21
=
=
-.005693
7.268757
QUAL
=
QUALWALD =
.020252
1071.053640
Part 10: Prediction
Decompositions
10-21/21
Part 10: Prediction