e-TA 2: Box-Cox and Partial Residual Plot
Welcome to e-Tutorial, your on-line help to Econ508. The present issue focuses on the basic operations of R. The introductory material presented below is designed to enhance your understanding of the topics and your performance on the homework. This issue focuses on the basic features of Box-Cox transformations and Partial Residual Plots.1
Introduction
In problem set 1, question 1, you are asked to estimate two demand equations for bread using the data set available here (or if you prefer, visit the data set collection at the Econ 508 web page, under the name "giffen"). You can save the data using the techniques suggested on e-Tutorial 1. As a general guideline, I suggest you to script your work.
Parts (i)-(iii) of the problem set involve simple linear regression and hypothesis testing that should be straightforward to solve once you are familiar with Stata. For every hypothesis testing, please make clear what are the null and alternative hypotheses. Please also provide a simple table with the main estimation results. I recommend you to ALWAYS include standard deviations for ALL parameters you estimate. Remember the first rule of empirical paper writing: "All good estimates deserve a standard error". Another useful advice is to summarize your main conclusions. It is strongly encouraged that you structure your problem set as a paper. Finally, graphs are very welcome as long as you provide labels and refer to them on your comments. Don't include any remaining material (e.g., software output or your preliminary regressions) in your report.
Partial Residual Plot
Question 1, part (iv) requires you to compare the plots of the Engel curves for bread in the "short" and "long" versions of the model using partial residual plot for the latter model. As mentioned in Professor Koenker's Lecture 2, "the partial residual plot is a device for representing the final step of a multivariate regression result as a bivariate scatterplot." Here is how you do that:
Theorem: (Gauss-Frisch-Waugh)
Recall the results of the Gauss-Frisch-Waugh theorem in Professor Koenker's Lecture Note 2 (pages 8-9). Here you will see that you can obtain the same coefficient and standard deviation for a given covariate by using partial residual regression. I will show the result using the gasoline demand data available here. In this data set, Y corresponds to the log per capita gas consumption, P is the log gas price, and Z is the log per capita income. The variables are already in the logarithmic form, so that we are actually estimating log-linear models.
At first you run the full model (Model A) and observe the
coefficient and standard deviation of P. Then you run a shorter version
of the model (Model B), excluding P. You get the residuals of this
partial regression and call them "resB". After that you run another
short model (Model C), but in this case you regress the omitted
variable P on the same covariates of model B. You obtain the residuals
of the latter model and call them "resC". Finally, you regress the
residuals of the model B on the residuals of the model C.
insheet using gasnew.dat, clear
describeContains data
obs: 201
vars: 4
size: 3,216
--------------------------------------------------------------
storage display value
variable name type format label variable label
--------------------------------------------------------------
year float %9.0g
lnp float %9.0g ln(p)
lnz float %9.0g ln(z)
lny float %9.0g ln(y)
--------------------------------------------------------------
Sorted by:
Note: dataset has changed since last saved
gen Y=lny
genP=lnp
genZ=lnz
* Model A: Y = \alpha_{0} + \alpha_{1}*P + \alpha_{2}*Z
regress Y P Z
* Model B: Y = \beta_{0} + \beta_{1}*Z
regressY Z
* Residual Model B
predict resB, resid
* Model C: P = \gamma_{0} + \gamma_{1}*Z
regressP Z
* Residual Model C
predictresC, resid
* Gauss-Frisch-Waugh: resB = \theta_{0} + \theta_{1}*resC
regressresB resC
. * Model A: Y = \alpha_{0} + \alpha_{1}*P + \alpha_{2}*Z
. regress Y P Z
Source | SS df MS Number of obs = 201
-------------+------------------------------ F( 2, 198) = 1674.11
Model | 21.0866833 2 10.5433417 Prob > F = 0.0000
Residual | 1.24698236 198 .006297891 R-squared = 0.9442
-------------+------------------------------ Adj R-squared = 0.9436
Total | 22.3336657 200 .111668328 Root MSE = .07936
------------------------------------------------------------------------------
Y | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
P | -.4075322 .0196435 -20.75 0.000 -.4462696 -.3687948
Z | 1.75935 .0419446 41.94 0.000 1.676634 1.842065
_cons | -5.266504 .1121704 -46.95 0.000 -5.487706 -5.045303
------------------------------------------------------------------------------
. * Model B: Y = \beta_{0} + \beta_{1}*Z
. regress Y Z
Source | SS df MS Number of obs = 201
-------------+------------------------------ F( 1, 199) = 923.98
Model | 18.3759978 1 18.3759978 Prob > F = 0.0000
Residual | 3.95766793 199 .019887779 R-squared = 0.8228
-------------+------------------------------ Adj R-squared = 0.8219
Total | 22.3336657 200 .111668328 Root MSE = .14102
------------------------------------------------------------------------------
Y | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
Z | .9735875 .0320289 30.40 0.000 .9104278 1.036747
_cons | -3.106386 .0741504 -41.89 0.000 -3.252608 -2.960165
------------------------------------------------------------------------------
. * Residual Model B
. predict resB, resid
. * Model C: P = \gamma_{0} + \gamma_{1}*Z
. regress P Z
Source | SS df MS Number of obs = 201
-------------+------------------------------ F( 1, 199) = 878.73
Model | 72.0708177 1 72.0708177 Prob > F = 0.0000
Residual | 16.321319 199 .082016678 R-squared = 0.8154
-------------+------------------------------ Adj R-squared = 0.8144
Total | 88.3921367 200 .441960684 Root MSE = .28639
------------------------------------------------------------------------------
P | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
Z | 1.928099 .065043 29.64 0.000 1.799837 2.056361
_cons | -5.300484 .1505815 -35.20 0.000 -5.597424 -5.003544
------------------------------------------------------------------------------
. * Residual Model C
. predict resC, resid
. * Gauss-Frisch-Waugh: resB = \theta_{0} + \theta_{1}*resC
. regress resB resC
Source | SS df MS Number of obs = 201
-------------+------------------------------ F( 1, 199) = 432.59
Model | 2.71068559 1 2.71068559 Prob > F = 0.0000
Residual | 1.24698237 199 .006266243 R-squared = 0.6849
-------------+------------------------------ Adj R-squared = 0.6833
Total | 3.95766796 200 .01978834 Root MSE = .07916
------------------------------------------------------------------------------
resB | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
resC | -.4075322 .0195941 -20.80 0.000 -.446171 -.3688934
_cons | -5.37e-10 .0055835 -0.00 1.000 -.0110104 .0110104
------------------------------------------------------------------------------
Next we can plot those residuals and insert a fitting line:
graph twoway (scatter resB resC, m(oh)) (lfit resB resC), sch(s1mono)
ytitle(Per capita gas consumption) title(Partial Residuals)
Box-Cox Transformation:
For question 2, parts (a)-(d) are also straightforward. You are expected to calculate the estimates in both linear and log-linear form. Besides them, you are expected to run a Box-Cox version of the model, and interpret it. Here I will give you some help by using the same gasoline demand data as above.
Just for a minute, suppose somebody told you that a nice gasoline demand equation should also include two additional covariates: the squared price of gas, and the effect of price times income. You can obtain those variables as follows:
gen Psq = P^2
genPZ = P*Z
Next you are asked to run this extended model, in a traditional log-linear form (remember that all covariates are already in logs). So, the easiest way to do that is as follows:
regress Y P Z Psq PZ
Source | SS df MS Number of obs = 201
-------------+------------------------------ F( 4, 196) = 1499.50
Model | 21.6269476 4 5.4067369 Prob > F = 0.0000
Residual | .706718113 196 .003605705 R-squared = 0.9684
-------------+------------------------------ Adj R-squared = 0.9677
Total | 22.3336657 200 .111668328 Root MSE = .06005
------------------------------------------------------------------------------
Y | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
P | -2.573682 .1858084 -13.85 0.000 -2.940122 -2.207242
Z | 2.571179 .077643 33.12 0.000 2.418056 2.724302
Psq | -.2712707 .0375378 -7.23 0.000 -.3453005 -.1972408
PZ | .7170179 .0593085 12.09 0.000 .6000532 .8339826
_cons | -7.391471 .2032132 -36.37 0.000 -7.792236 -6.990706
------------------------------------------------------------------------------
The log-linear form seems to be a nice attempt to estimate the gasoline demand.
Next, suppose you are so confident on this model that you write a paper about the topic and send it to a journal. Two weeks later you receive a letter from a referee saying she is suspicious about your log-linear equation. She asks you to reestimate the same model but using the dependent variable linearly (i.e., without the logs), and the rest of the equation remaining as before. She asks you to revise and resubmit the paper with your new findings.
In the search for elements that support your original model, you start the following experiment: 1. Run the model suggested by the referee, using a Box-Cox transformation to find the MLE ofλ, 2. Plot the concentrated log-likelihood function, and 3. Reestimate the model conditional on the MLE of λ:
geny = exp(Y)
boxcox y P Z Psq PZ, level(95)
Fitting comparison model
Iteration 0: log likelihood = 110.10193
Iteration 1: log likelihood = 112.76717
Iteration 2: log likelihood = 112.77299
Iteration 3: log likelihood = 112.77299
Fitting full model
Iteration 0: log likelihood = 426.09821
Iteration 1: log likelihood = 458.18797
Iteration 2: log likelihood = 458.18823
Iteration 3: log likelihood = 458.18823
Number of obs = 201
LR chi2(4) = 690.83
Log likelihood = 458.18823 Prob > chi2 = 0.000
------------------------------------------------------------------------------
y | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
/theta | -.050263 .1154772 -0.44 0.663 -.2765942 .1760682
------------------------------------------------------------------------------
Estimates of scale-variant parameters
----------------------------
| Coef.
-------------+--------------
Notrans |
P | -2.624961
Z | 2.658853
Psq | -.2801617
PZ | .7253629
_cons | -7.633947
-------------+--------------
/sigma | .0619262
----------------------------
---------------------------------------------------------
Test Restricted LR statistic P-value
H0: log likelihood chi2 Prob > chi2
---------------------------------------------------------
theta = -1 423.34476 69.69 0.000
theta = 0 458.09427 0.19 0.665
theta = 1 426.09821 64.18 0.000
---------------------------------------------------------
version 6: boxcox y P Z Psq PZ, level(95) graph
As you see, the MLE for λ is very close to zero (-0.050263). The picture is drawn using smoothing spline techniques to help you envisage the log-likelihood function and the MLE of λ. The horizontal line corresponds to the 95% confidence interval. Next we can apply the power transform to y and then fit the revised model:
qui:boxcox y P Z Psq PZ, level(95)
gen ytrans=((y^r(est))-1)/r(est)
regressytrans P Z Psq PZ
Source | SS df MS Number of obs = 201
-------------+------------------------------ F( 4, 196) = 1507.45
Model | 23.7132269 4 5.92830673 Prob > F = 0.0000
Residual | .770805496 196 .003932681 R-squared = 0.9685
-------------+------------------------------ Adj R-squared = 0.9679
Total | 24.4840324 200 .122420162 Root MSE = .06271
------------------------------------------------------------------------------
ytrans | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
P | -2.624961 .1940504 -13.53 0.000 -3.007656 -2.242266
Z | 2.658853 .0810871 32.79 0.000 2.498938 2.818768
Psq | -.2801617 .0392029 -7.15 0.000 -.3574754 -.202848
PZ | .7253629 .0619393 11.71 0.000 .6032099 .8475158
_cons | -7.633947 .2122272 -35.97 0.000 -8.052489 -7.215405
------------------------------------------------------------------------------
The latter regression (conditional on the MLE of λ) provides results
close to your log-linear suggestion. Now you have reasonable support to
write back the referee and defend your original model.
Andrews Test
Finally, for the Econ 508 problem set 1, question 2, you are also required to perform the David Andrews (1971) test. As an example, I use the same gasoline data as above, and follow the routine on Professor Koenker's Lecture Note 2:
- Run the linear model and get the predicted values of y (call this variable \(\hat y\)):
genp=exp(P)gen
z=exp(Z)regress
y p z
Source | SS df MS Number of obs = 201
-------------+------------------------------ F( 2, 198) = 1609.94
Model | 3.70695117 2 1.85347559 Prob > F = 0.0000
Residual | .227951976 198 .001151273 R-squared = 0.9421
-------------+------------------------------ Adj R-squared = 0.9415
Total | 3.93490315 200 .019674516 Root MSE = .03393
------------------------------------------------------------------------------
y | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
p | -.3375906 .0144982 -23.28 0.000 -.3661814 -.3089999
z | .0741067 .0016292 45.49 0.000 .0708939 .0773195
_cons | -.1536729 .0112284 -13.69 0.000 -.1758155 -.1315303
------------------------------------------------------------------------------
predict yhat- Reestimate the augmented model and test (\gamma = 0):
genLY=log(yhat)
genYLY=yhat*LY
regressy p z YLY
Source | SS df MS Number of obs = 201
-------------+------------------------------ F( 3, 197) = 1610.62
Model | 3.78075786 3 1.26025262 Prob > F = 0.0000
Residual | .15414529 197 .000782463 R-squared = 0.9608
-------------+------------------------------ Adj R-squared = 0.9602
Total | 3.93490315 200 .019674516 Root MSE = .02797
------------------------------------------------------------------------------
y | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
p | -.2958436 .0127019 -23.29 0.000 -.3208928 -.2707945
z | .0651617 .0016286 40.01 0.000 .0619501 .0683734
YLY | 1.082626 .1114711 9.71 0.000 .8627956 1.302455
_cons | .2845145 .0460572 6.18 0.000 .1936861 .375343
------------------------------------------------------------------------------
test YLY
( 1) YLY = 0
F( 1, 197) = 94.33
Prob > F = 0.0000
From the test above we can reject the null hypothesis that (\gamma =0). Can you interpret what does this mean?
Please send comments to bottan2@illinois.edu or srmntbr2@illinois.edu