Economics 130
Lecture 8
Final Comments on Dummy Variables
Heteroskedasticity
Serial Correlation
Teams and Topics
Midterm Preparation
Multiple Regression
• We continue with addressing our second issue
+ add in how we evaluate these relationships:
– Where do we get data to do this analysis?
– How do we create the model relating the data?
– How do we relate data to on another?
– How do we evaluate these relationships?
Multiple Regression
• Tonight we will work on:
– Finish Dummy Variables (Indicator Variables)
– Heteroskedascity
– Serial Correlation (a little)
– Talk about Teams and Topics
Regressions with Dummy Variables
• Simple Introduction:
• Dummy variable is either 0 or 1.
• Use to turn qualitative (Yes/No) data into 1/0.
Multiple Regression
• Simple Regression with a Dummy Variable
Y = b1 + b2D + e
• OLS estimation, confidence intervals, testing, etc. carried out in standard way
• Interpretation a little different.
Dummy Variables
• Simple Regression with a Dummy
Variable
• Fitted value for ith observation (point on
regression line):
• Since Di = 0 or 1 either
• or6
Ŷi = b1 + b2Di
Ŷi = b1
Ŷi = b1 + b2Di
Dummy Variables• Example: Explaining house prices (continued)
• Regress Y = house price on D = dummy for air
conditioning (=1 if house has air conditioning, = 0
otherwise).
• Result:
• Average price of house with air conditioning is $85,881
• Average price of house without air conditioning is
$59,8857
b1 = 59,885
b2 = 25,996
b1 + b2 = 85,881
Dummy Variables• Multiple Regression with Dummy Variables
• Example: Explaining house prices (continued)
• Y = b1 + b2D1 . . . bkDk + e
• Regress Y = house price on D1 = driveway dummy and
D2 = rec room dummy
•
• Four types of houses:
• Houses with a driveway and a rec room (D1=1, D2=1)
• Houses with a driveway but no rec room (D1=1, D2=0)
• Houses with a rec room but no driveway (D1=0, D2=1)
• Houses with no driveway and no rec room (D1=0, D2=0)
Dummy Variables
Example: Explaining house prices (continued)
• If D1=1 and D2=1, then
Ŷi = b1 + b2 + b3 = 47,099 + 21,160 + 16,024 = 84,283
• “The average price of houses with a driveway and rec room is
$84,283”.
Coeff. St.
Error
t Stat P-
value
Lower
95%
Upper
95%
Inter. 47099.1 2837.6 16.60 2.E-50 41525 52673
D1 21159.9 3062.4 6.91 1.E-11 15144 27176
D2 16023.7 2788.6 5.75 1.E-08 10546 21502
Dummy Variables• If D1 = 1 and D2=0, then
• “The average price of houses with a driveway but no rec room is $68,259”.
• If D1=0 and D2=1, then
•
• “The average price of houses with a rec room but no driveway is $63,123”.
•
• If D1=0 and D2=0, then
•
• “The average price of houses with no driveway and no rec room is $47,099”.
Ŷi = b1 + b2 = 47,099 + 21,160 = 68,259
Ŷi = b1 + b3 = 47,099 + 16,024 = 63,123
Ŷi = b1 = 47,099
Dummy Variables
• Multiple Regression with Dummy and non-Dummy
Explanatory Variables
• Regress Y = house price on D = air conditioning
dummy and X = lot size.
• OLS estimates: b1= 32,693
b2 = 20,175
b3 = 5.64
Dummy Variables
• For houses with an air conditioner D = 1 and
• For houses without an air conditioner D=0 and
• Two different regression lines depending on whether the house has an air conditioner or not.
• Two lines have different intercepts but same slope (i.e. same marginal effect)
Ŷi = 52,868 + 5.64X
Ŷi = 32,693 + 5.64X
Dummy Variables
• Verbal ways of expressing OLS results:
• “An extra square foot of lot size will tend to add $5.64 onto the price of a house” (Note: no ceteris paribusqualifications to statement since marginal effect is same for houses with and without air conditioners)
• “Houses with air conditioners tend to be worth $20,175 more than houses with no air conditioners, ceteris paribus” (Note: Here we do have ceteris paribusqualification)
• “If we consider houses with similar lot sizes, those with air conditioners tend to be worth an extra $20,175”
Dummy Variables
• Another House Price Regression
• Regress Y = house price on D1 = dummy variable for driveway, D2 = dummy variable for rec room, X1 = lot size and X2 = number of bedrooms
• OLS estimates:
Y = b1 + b2D1 + b3D2 + b4X1 + b5X2 + e
b1= -2,736b2 = 12,598b3 = 10,969b4 = 5.197b5 = 10,564
Dummy Variables
1. If D1=1 and D2=1, then
This is the regression line for houses with a driveway and rec room.
2. If D1=1 and D2=0, then
• This is the regression line for houses with a driveway but no rec room.
3. If D1=0 and D2=1, then
This is the regression line for houses
with a rec room but no driveway.
4. If D1=0 and D2=0, then
This is the regression line for houses
with no driveway and no rec room.
Dummy Variables
1. If D1=1 and D2=1, then
This is the regression line for houses with a driveway and rec room.
2. If D1=1 and D2=0, then
• This is the regression line for houses with a driveway but no rec room.
Dummy Variables
3. If D1=0 and D2=1, then
This is the regression line for houses with a rec
room but no driveway
4. If D1=0 and D2=0, then
This is the regression line for houses with no
driveway and no rec room.
Dummy Variables
• “Houses with driveways tend to be worth
$12,598 more than similar houses with no
driveway.”
• “If we consider houses with the same number
of bedrooms, then adding an extra square foot
of lot size will tend to increase the price of a
house by $5.197.”
• “An extra bedroom will tend to add $10,562 to
the value of a house, ceteris paribus”
Dummy Variables
• Interacting Dummy and Non-Dummy Variables
• Where Z=DX.
• Z is either 0 (for observations with D=0) or X (for observations with D=1)
• If D=1 then
• If D=0, then
Two different regression lines corresponding to D=0 and D=1 exist and have different intercepts and slopes. • The marginal effect of X on Y is different for D=0 and
D=1
Y = b1 + b2D + bX + b4Z + e
Ŷi = (b1 + b2) + (b3 + b4)X
Ŷi = b1 + b3X
Dummy Variables
• Regress Y = house price on D = air conditioner dummy, X = lot size and Z = DX
• OLS estimates:
• The marginal effect of lot size on housing is 7.27 for houses with air conditioners and only $5.02 for houses without.
• Increasing lot size will tend to add more to the value of a house if it has an air conditioner than if it does not.
20
b1= 35,684
b2 = 7,613
b3 = 5.02
b4 = 2.25
Dummy Variables
• Issue here: Using dummy variables on the right side of regression, i.e., as INDEPENDENT variables
• Dummy variables are (0,1) variables
• The illustrative example:
• House prices depend upon house characteristics: size (square feet), location (may be dummy), # of bedrooms, bathrooms, age, whether has a pool (dummy), whether has a tile roof (dummy), etc.
• Note: These sorts of models are called “hedonic price” models.
21
Dummy Variables
Here again is our basic simple regression model:
1 2PRICE SQFT e b b
1 if characteristic is present
0 if characteristic is not presentD
1 if property is in the desirable neighborhood
0 if property is not in the desirable neighborhoodD
Dummy Variables
1 2PRICE D SQFT e b b
1 2
1 2
( ) when 1( )
when 0
SQFT DE PRICE
SQFT D
b b
b b
Dummy Variables
Dummy Variables
1 2 ( )PRICE SQFT SQFT D e b b
1 2
1 2
1 2
( ) when 1( )
when 0
SQFT DE PRICE SQFT SQFT D
SQFT D
b b b b
b b
2
2
when 1 ( )
when 0
DE PRICE
DSQFT
b
b
1 2
1 2
1 2
( ) when 1( )
when 0
SQFT DE PRICE SQFT SQFT D
SQFT D
b b b b
b b
2
2
when 1 ( )
when 0
DE PRICE
DSQFT
b
b
Dummy variables can also be used to determine whether the
marginal impact of one variable depends on the presence (or
absence) of the dummy characteristic. Does price per square
foot depend upon whether the house is in a good
neighborhood?
Dummy Variables
Dummy Variables
• Here’s a model with both intercept and slope
dummy effects. We will estimate it now.
Dummy Variables
• 1000 house sales from two similar neighborhoods,
one bordering a university, the other 3 miles away.
• Variables:
• price house price, in $1000
• sqft square feet of living area, in 100's
• age house age, in years
• utown =1 if close to university
• pool =1 if house has pool
• fplace =1 if house has fireplace
Dummy Variables
• The Model:
• Sample Data:
Dummy Variables
• Model 2: OLS, using observations 1-1000• Dependent variable: p
• coefficient std. error t-ratio p-value • ---------------------------------------------------------• const 24.5000 6.19172 3.957 8.13e-05 ***• sqft 7.61218 0.245176 31.05 1.87e-148 ***• utown 27.4530 8.42258 3.259 0.0012 ***• usqft 0.0129940 0.00332048 3.913 9.72e-05 ***• pool 4.37716 1.19669 3.658 0.0003 ***• fplace 1.64918 0.971957 1.697 0.0901 *• age -0.190086 0.0512046 -3.712 0.0002 ***
• Mean dependent var 247.6557 S.D. dependent var 42.19273• Sum squared resid 230184.4 S.E. of regression 15.22521• R-squared 0.870570 Adjusted R-squared 0.869788• F(6, 993) 1113.183 P-value(F) 0.000000
Dummy VariablesPremium for
lots near the
university
is….
Premium for
having a pool
is…
Premium for
having a
fireplace is …
Δ Price due to
Δ in house
size is . . .
A. $24,500
B. $76
C. $4,377
D. $1.2994
E. $1,649
F. $27,453
A. $24,500
B. $76
C. $4,377
D. $1.2994
E. $1,649
F. $27,453
A. $24,500
B. $76
C. $4,377
D. $1.2994
E. $1,649
F. $27,453
A. $24,500
B. $76
C. $4,377
D. $1.2994
E. $1,649
F. $27,45
Dummy Variables
Based on these regression results, we estimate:
the location premium, for lots near the university, to be
$27,453
the price per square foot to be $89.12 for houses near the
university, and $76.12 for houses in other areas.
that houses depreciate $190.10 per year
that a pool increases the value of a home by $4377.20
that a fireplace increases the value of a home by
$1649.20
Dummy VariablesVariable Coefficient X Units Y Units Marginal Effect
UTOWN (1 = close
to university)
27.453 1,0 1000s 1 x 2.7453 x 1000 =
$27453
SQFT (size in 100s) 7.612 100s of
SQFT
1000s 7.612 ÷ 100 = .07612 x
1000 = $76.12
USQFT ( x SQFT) .01299 1,0 1000s .01299 x 1000 = $12.99
+ $76.12 = $89.11
AGE (years) -.190 Years 1000s 1 x -.19 x 1000 = - $190
FPLACE (1 =
fireplace)
1.649 1,0 1000s 1 x 1.649 x 1000 =
$1,649
POOL (1 = pool) 4.377 1,0 1000s 1 x 4.377 x 1000 =
$4,377
Dummy Variables/Gretl Practice
• We are now going to return to a model we
looked at earlier for women’s labor force
participation
• This model used 1990 data from 50 states.
Dummy Variables/Gretl Practice
Dummy Variables/Gretl Practice
Dummy Variables/Gretl Practice
Gretl Practice
• Here is a model of the determinants of women’s labor force participation for all 50 states.
• WLFP = Participation rate (%) of women > 16 in the labor force
• YF = Annual median earnings by females (000s of $)
• YM = Annual median earnings by males (000s of $)
• EDUC = Female HS grads > 24 (%)
• UE = Unemployment rate (%)
• MR = Marriage rate (%) women over 16
• DR = Divorce rate (%)
• URB = % of state’s population that is urban
• WH = % of state’s female > 16 population who are white
Gretl Practice
• This is another “kitchen sink” model:
• WLFP = b1 + b2YF + b3YM + b4EDUC +
b5UE + b6MR + b7DR + b8URB + b9WH e
Gretl Practice
• Model 1: OLS, using observations 1-50• Dependent variable: wlfp
• Coefficient Std. Error t-ratio p-value• const 44.5096 8.97496 4.9593 0.00001 ***• yf 0.987983 0.407583 2.4240 0.01985 **• ym -0.174345 0.306207 -0.5694 0.57221• educ 0.285129 0.0931647 3.0605 0.00389 ***• ue -1.61058 0.313617 -5.1355 <0.00001 ***• mr -0.0782145 0.173139 -0.4517 0.65383• dr 0.437371 0.258336 1.6930 0.09804 *• urb -0.0926339 0.0333355 -2.7788 0.00820 ***• wh -0.0874916 0.0398446 -2.1958 0.03382 **
• Mean dependent var 57.47400 S.D. dependent var 4.248784• Sum squared resid 193.9742 S.E. of regression 2.175104• R-squared 0.780710 Adjusted R-squared 0.737922• F(8, 41) 18.24590 P-value(F) 2.90e-11• Log-likelihood -104.8395 Akaike criterion 227.6790• Schwarz criterion 244.8872 Hannan-Quinn 234.2319
Gretl Practice
• Model 4: OLS, using observations 1-50• Dependent variable: wlfp
• Coefficient Std. Error t-ratio p-value• const 41.346 5.55984 7.4365 <0.00001 ***• yf 1.06712 0.364515 2.9275 0.00550 ***• educ 0.258172 0.0708648 3.6432 0.00073 ***• ue -1.59099 0.307647 -5.1715 <0.00001 ***• dr 0.391632 0.235404 1.6637 0.10363• urb -0.0876356 0.0311463 -2.8137 0.00742 ***• wh -0.0850871 0.0391115 -2.1755 0.03527 **• ym -0.198418 0.298664 -0.6644 0.51010
• Mean dependent var 57.47400 S.D. dependent var 4.248784• Sum squared resid 194.9397 S.E. of regression 2.154396• R-squared 0.779619 Adjusted R-squared 0.742888• F(7, 42) 21.22554 P-value(F) 6.62e-12• Log-likelihood -104.9636 Akaike criterion 225.9272• Schwarz criterion 241.2234 Hannan-Quinn 231.7521
Gretl Practice
• Model 5: OLS, using observations 1-50• Dependent variable: wlfp
• Coefficient Std. Error t-ratio p-value• const 41.8336 5.47528 7.6405 <0.00001 ***• yf 0.849264 0.158152 5.3699 <0.00001 ***• educ 0.249152 0.0690987 3.6057 0.00080 ***• ue -1.67758 0.276859 -6.0593 <0.00001 ***• dr 0.434104 0.22508 1.9287 0.06039 *• urb -0.0942172 0.0293363 -3.2116 0.00250 ***• wh -0.0960861 0.0352037 -2.7294 0.00916 ***
• Mean dependent var 57.47400 S.D. dependent var 4.248784• Sum squared resid 196.9882 S.E. of regression 2.140355• R-squared 0.777303 Adjusted R-squared 0.746229• F(6, 43) 25.01455 P-value(F) 1.55e-12• Log-likelihood -105.2249 Akaike criterion 224.4499• Schwarz criterion 237.8341 Hannan-Quinn 229.5467
Dummy Variables/Gretl Practice
• We are going to look at data from both 1980
and 1990 because it’s possible there was a
structural change in WLFP over that decade.
• We are going to create a Dummy Variable for
1990, called D90. For 1990, D = 1.
• Now we are going to incorporate interaction
terms, by multiplying D90 by the other
variables.
Dummy Variables/Gretl Practice
• Here is the model we are starting with:
• WLFP = b1 + b2YF + b3YM + b4EDUC +
b5UE + b6MR + b7DR + b8URB + b9WH
q1(D90*YF) + q2(D90*YM) + q3(D90*EDUC)
+ q4(D90*UE) + q5(D90*MR) + q6(D90*DR)
+ q7(D90*URB) + q8(D90*WH) + e
Dummy Variables/Gretl Practice
• Model 2: OLS, using observations 1-100
• Dependent variable: WLFP
• Omitted due to exact collinearity: D90YM
• coefficient std. error t-ratio p-value
• ------------------------------------------------------------
• const 49.6235 10.5465 4.705 1.00e-05 ***
• YF 0.00470565 0.000948974 4.959 3.71e-06 ***
• YM -0.000133492 0.000273021 -0.4889 0.6262
• EDUC 0.286358 0.0586220 4.885 4.97e-06 ***
• UE -1.09155 0.269898 -4.044 0.0001 ***
• MR -0.210187 0.153263 -1.371 0.1739
• DR 0.208349 0.172456 1.208 0.2304
• URB -0.0665164 0.0299182 -2.223 0.0289 **
• WH -0.126810 0.0343940 -3.687 0.0004 ***
• D90 -4.85398 13.6340 -0.3560 0.7227
• D90YF -0.00376523 0.000880843 -4.275 5.08e-05 ***
• D90EDUC -0.00164444 0.111949 -0.01469 0.9883
• D90UE -0.537329 0.396399 -1.356 0.1789
• D90MR 0.127952 0.230950 0.5540 0.5811
• D90DR 0.239853 0.304639 0.7873 0.4333
• D90URB -0.0276884 0.0437808 -0.6324 0.5288
• D90WH 0.0369987 0.0517175 0.7154 0.4764
• R-squared 0.861932 Adjusted R-squared 0.835316
Dummy Variables/Gretl Practice
• Suspecting multicollinearity, we eliminate variables with insignificant coefficients one at a time.
Dummy Variables/Gretl Practice
• Model 3: OLS, using observations 1-100
• Dependent variable: WLFP
• coefficient std. error t-ratio p-value
• ----------------------------------------------------------
• const 47.6366 6.57840 7.241 1.52e-010 ***
• YF 0.00477939 0.000733949 6.512 4.28e-09 ***
• EDUC 0.275070 0.0455059 6.045 3.43e-08 ***
• UE -1.06141 0.245591 -4.322 4.02e-05 ***
• MR -0.207293 0.104894 -1.976 0.0512 *
• DR 0.281618 0.133697 2.106 0.0380 **
• URB -0.0784652 0.0206237 -3.805 0.0003 ***
• WH -0.111495 0.0242421 -4.599 1.40e-05 ***
• D90YF -0.00405375 0.000682124 -5.943 5.36e-08 ***
• D90UE -0.569355 0.327225 -1.740 0.0853 *
• D90MR 0.126361 0.0509756 2.479 0.0151 **
• R-squared 0.858214 Adjusted R-squared 0.842283
• F(10, 89) 53.87052 P-value(F) 1.98e-33
Dummy Variables/Gretl Practice
• Here is the final model:
• WLFP = 47.63 + .00478 YF - .00405 (D90 *
YF) + .275 EDUC – 1.06 UE - .569 (D90 *
UE) - .207 MR + .126 (D90 * MR) + .282 DR
- .078 URB - .111 WH
• Adjusted R2 = .842
Dummy Variables/Gretl Practice
For 1980, set D90 = 0
• WLFP = 47.63
+ .00478YF + .275
EDUC – 1.06 UE - .207
MR +.282 DR - .078 URB
- .111 WH
For 1990, set D90 = to 1 and
combine terms
• WLFP = 47.63
+ .00073 YF + .275
EDUC – 1.63UE - .081
MR + .282 DR - .078
URB - .111 WH
Multiple Regression
• We continue with addressing these three
issues:
– Where do we get data to do this analysis?
– How do we create the model relating the data?
– How do we relate data to on another?
– How do we evaluate these relationships?
Multiple Regression
• Sources of specification errors:
– Choice of variables
– Functional forms (non-linear relationships)
– Structure of the error terms (e’s)
Multiple Regression
• Sources of specification errors:
– Choice of variables
• Omitted Variables
• Irrelevant Variables
• Multicollinearity
– Functional forms (non-linear relationships)
• Non-linear models
– Structure of the error terms (e’s)
Structure of the Error Terms (e’s)
Heteroskedasticity Serial Correlation
Structure of the Error Terms (e’s)
• What about the variance of our estimates?
– All e’s are equally distributed with the same
conditional variance (s2) [Homoskedasticity
(equal scatter]
– e’s are independently distributed; cov (ei, ej) = 0
• Then, among all unbiased, linear combinations
of Ys, our estimates -- b’s -- have the lowest
variance = most efficient.
Heteroskedasticity
• Under heteroskedasticity,
the error term varies with
the value of x. No longer
is the VAR = s2 which is
a constant.
• Now VAR (ei) =
si2
Heteroskedasticity
• The consequences of Heteroskedasticity
• (1) The standard errors usually computed for the least
squares estimator are incorrect. Confidence intervals
and hypothesis tests that use these standard errors ARE
WRONG!!!
• (2) The least squares estimator is still a linear and
unbiased estimator, but it is no longer best. There is
another estimator with a smaller variance.
Heteroskedasticity
• Why? Consider the following equations:
• Remember the standard error (se) = Var 1/2
• And that t = bi
se
Homoskedasticity Heteroskedasticity
Var (e) s2 si2
Var (b) =
Heteroskedasticity
• Thus each observation has a different error
variance.
• Now, we will look at a quick fix for the
problems created by heteroskedasticity.
Heteroskedasticity
• By using our squared residuals (êi2)to estimate
each E(ei2) (=σi
2), we can estimate the
HETEROSKEDASTICITY-ROBUST
VARIANCE
• Var (b) = S [(xi - x̅)2
êi2]
S [(xi – x̅)2]]2
Heteroskedasticity
• When you correct the standard errors in this
way (the White correction)
• Do your parameter estimates (bi,…) change??
Heteroskedasticity
• When you correct the standard errors in this
way (the White correction)
• Do your parameter estimates (b2, b3) change??
• NO!!!
• Only the standard errors change.
Heteroskedasticity
• Correcting Standard Errors (se) Example:
Food Expenditure Model:
y = weekly food expenditures ($0)
x = weekly income ($100)
Heteroskedasticity
• Correcting Standard Errors (se) Example:
Food Expenditure Model:
ˆ 83.42 10.21
(27.46) (1.81) (White se)
(43.41) (2.09) (incorrect se)
i iy x
Heteroskedasticity
Using, heteroskedasticity-robust standard errors is
probably the MOST IMPORTANT part of this lecture!
This is what most practitioners do to address
potential heteroskedasticity.
Why? Because changing estimation (which we will talk
about in a moment) requires specifying the form of
heteroskedasticity.
If we are wrong about the form, we are potentially
introducing error in the estimation.
Heteroskedasticity
Heteroskedasticity
• An observation, continued:
• HOWEVER, we need to talk about form of heteroskedasticity in order to TEST for it, which practitioners want to know about.
• Why? Because if you have evidence of heteroskedasticity, they may want to see that your results are robust to correcting for it in the estimation.
Heteroskedasticity
Therefore, we turn to detecting heteroskedasticity.
• Diagnostic: residual plots
• Forms of heteroskedasticity and testing for them
• Using specific forms of heteroskedasticity to perform Generalized Least Squares (to make the least squares estimator BEST, not just unbiased)
Heteroskedasticity
Residual Plots
Estimate the model using least squares and plot the
least squares residuals.
With more than one explanatory variable, plot the
least squares residuals or squared residuals against
each explanatory variable, or against , to see if those
residuals vary in a systematic way relative to the
specified variable.
ˆiy
Heteroskedasticity
Heteroskedasticity
Residual plot in gretl
In OLS output, click
path
– Graphs
– Residuals plot
– Against x
Squared Residual plot in
gretl
In OLS output, click
path
– Save
– Squared Residuals
– OK
Return to Data
– Choose X-Y Graph
– X = variable; Y = u2
Heteroskedasticity
No Patterns to Residuals Heteroskedastic Pattern
Heteroskedasticity
• Summary of key points on heteroskedasticity:
• 1) Implies OLS standard errors are WRONG, so t-stats
(hypothesis tests) and confidence intervals based on OLS
are WRONG
– Correct with Heteroskedasticity-robust (White)
standard errors
• 2) Detecting (testing for) specific forms of heteroskedasticity: B-P, White,
Harvey
• 3) Implies OLS estimator is not minimum variance (efficient / BLUE).
Can obtain BLUE estimator for specific form of heteroskedasticity by using
Generalized Least Squares (OLS on “transformed data,” obtained by
dividing all data by estimated standard deviation of observation i, σ^i)
Serial Correlation
• Serial correlation (or autocorrelation) violates
the 6th Assumption that underpins Gauss
Markov: cov (ei, ej) = 0
• With serial correlation cov (ei, ej) 0
Structure of the Error Terms (e’s)
Heteroskedasticity Serial Correlation
Serial Correlation
• We are going to concern ourselves with First
Order Serial Correlation.
• This is often found in time series data.
• A period’s error term is related to the error
term of the previous period.
• A primary cause is the existence of long-term
cycles and trends in the data.
Serial Correlation
• Here is a formal statement using a simple linear
regression:
• Model:
• Error term:
• Because r is the coefficient of the error term lagged
one period, this is called the first order
autocorrelation coefficient and the process is called
the first order autoregressive process or AR(1).
1 2t t ty x e b b
1t t te e v r 1 1 r
Serial Correlation
• The implication of AR(1) is:
• The least squares estimator is still a linear and
unbiased estimator, but it is no longer best. There
is another estimator with a smaller variance.
• The standard errors usually computed for the
least squares estimator are incorrect.
Confidence intervals and hypothesis tests that
use these standard errors may be misleading.
Serial Correlation
• The most common test for first order serial
correlation is the Durbin-Watson (DW) test.
• The DW statistic is calculated from the value
of the residuals.
t=n
• d = St=2 (êt – êt-1)2
S(ê2)
Serial Correlation
• Working out the d statistic yields, approximately:
• d 2(1 - r) r is the estimate of r
• Because r can range from -1 to 1, the range for d
is 0 – 4.
• Durbin-Watson critical values are upper (du) and
lower (dl) bounds. N = sample size and k’ =
number of coefficients NOT COUNTING THE
CONSTANT!
Serial Correlation
• Rules of Thumb: because if r is 0, d = 2, then a
DW statistic around 2 (1.75 – 2.25) means there is
no first order serial correlation.
• A r close to 1 means a d close to 0, which means
strong positive serial correlation.
• Similarly, a r close to -1 means a d close to 4,
which indicates strong negative serial correlation.
Serial Correlation
Serial Correlation
The Durbin-Watson bounds test.
•
•
• if the test is inconclusive.
(Lower d values only)
Serial Correlation
What can you do about serial correlation?
• First differences
• Change the model specification
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