How can I
compute the Chow test statistic?
| Title |
|
Computing the Chow statistic |
| Author |
William Gould, StataCorp |
| Date |
January 1999; updated July 2011
|
You can include the dummy variables in a regression of the full model and
then use the test
command on those dummies. You could also run each of the models and then
write down the appropriate numbers and calculate the statistic by
hand—you also have access to functions to get appropriate
p-values.
Here is a longer answer:
Let’s start with the Chow test to which many refer. Consider the
model,
y = a + b*x1 + c*x2 + u
and say we have two groups of data. We could fit that model on
the two groups separately,
y = a1 + b1*x1 + c1*x2 + u for group == 1
y = a2 + b2*x1 + c2*x2 + u for group == 2
and we could fit a single, pooled regression
y = a + b*x1 + c*x2 + u for both groups
In the last regression, we are asserting that a1==a2, b1==b2,
and c1==c2. The formula for the “Chow test” of this
constraint is
ess_c - (ess_1+ess_2)
---------------------
k
---------------------------------
ess_1 + ess_2
---------------
N_1 + N_2 - 2*k
and this is the formula to which people refer. ess_1 and ess_2
are the error sum of squares from the separate regressions, ess_c is
the error sum of squares from the pooled (constrained) regression, k
is the number or estimated parameters (k=3 in our case), and
N_1 and N_2 are the number of observations in the two groups.
The resulting test statistic is distributed F(k, N_1+N_2-2*k).
Let’s try this. I have created small datasets:
clear
set obs 100
set seed 1234
generate x1 = uniform()
generate x2 = uniform()
generate y = 4*x1 - 2*x2 + 2*invnormal(uniform())
generate group = 1
save one, replace
clear
set obs 80
generate x1 = uniform()
generate x2 = uniform()
generate y = -2*x1 + 3*x2 + 8*invnormal(uniform())
generate group = 2
save two, replace
use one, clear
append using two
save combined, replace
The models are different in the two groups, the residual variances are
different, and so are the number of observations. With this dataset, I can
carry forth the Chow test. First, I run the separate regressions:
. regress y x1 x2 if group==1
Source | SS df MS Number of obs = 100
---------+------------------------------ F( 2, 97) = 36.10
Model | 328.686307 2 164.343154 Prob > F = 0.0000
Residual | 441.589627 97 4.55247038 R-squared = 0.4267
---------+------------------------------ Adj R-squared = 0.4149
Total | 770.275934 99 7.78056499 Root MSE = 2.1337
------------------------------------------------------------------------------
y | Coef. Std. Err. t P>|t| [95% Conf. Interval]
---------+--------------------------------------------------------------------
x1 | 5.121087 .728493 7.03 0.000 3.67523 6.566944
x2 | -3.227026 .7388209 -4.37 0.000 -4.693381 -1.760671
_cons | -.1725655 .5698273 -0.30 0.763 -1.303515 .9583839
------------------------------------------------------------------------------
. regress y x1 x2 if group==2
Source | SS df MS Number of obs = 80
---------+------------------------------ F( 2, 77) = 5.02
Model | 544.11726 2 272.05863 Prob > F = 0.0089
Residual | 4169.24211 77 54.1460014 R-squared = 0.1154
---------+------------------------------ Adj R-squared = 0.0925
Total | 4713.35937 79 59.6627768 Root MSE = 7.3584
------------------------------------------------------------------------------
y | Coef. Std. Err. t P>|t| [95% Conf. Interval]
---------+--------------------------------------------------------------------
x1 | -1.21464 2.9578 -0.41 0.682 -7.104372 4.675092
x2 | 8.49714 2.688249 3.16 0.002 3.144152 13.85013
_cons | -2.2591 1.91076 -1.18 0.241 -6.06391 1.545709
------------------------------------------------------------------------------
Then I run the combined regression:
. regress y x1 x2
Source | SS df MS Number of obs = 180
---------+------------------------------ F( 2, 177) = 2.93
Model | 176.150454 2 88.0752272 Prob > F = 0.0559
Residual | 5316.21341 177 30.035104 R-squared = 0.0321
---------+------------------------------ Adj R-squared = 0.0211
Total | 5492.36386 179 30.683597 Root MSE = 5.4804
------------------------------------------------------------------------------
y | Coef. Std. Err. t P>|t| [95% Conf. Interval]
---------+--------------------------------------------------------------------
x1 | 2.692373 1.41842 1.90 0.059 -.1068176 5.491563
x2 | 2.061004 1.370448 1.50 0.134 -.6435156 4.765524
_cons | -1.380331 1.017322 -1.36 0.177 -3.387973 .62731
------------------------------------------------------------------------------
For the Chow test,
ess_c - (ess_1+ess_2)
---------------------
k
---------------------------------
ess_1 + ess_2
---------------
N_1 + N_2 - 2*k
here are the relevant numbers copied from the output above:
ess_c = 5316.21341 (from combined regression)
ess_1 = 441.589627 (from group==1 regression)
ess_2 = 4169.24211 (from group==2 regression)
k = 3 (we estimate 3 parameters)
N_1 = 100 (from group==1 regression)
N_2 = 80 (from group==2 regression)
So, plugging in, we get
5316.21341 - (441.589628+4169.24211) 705.38167
------------------------------------ ---------
3 3
----------------------------------------- = ---------------
441.589628 + 4169.24211 4610.8317
----------------------- ---------
100+80-2*3 174
235.12722
= ----------
26.499033
= 8.8730491
The Chow test is F(k,N_1+N_2-2*k) = F(3,174), so our
test statistic is F(3,174) = 8.8730491.
Now I will do the same problem by running one regression and using
test to test certain coefficients equal to zero. What I want to do
is fit the model
y = a3 + b3*x1 + c3*x2 + a3'*g2 + b3'*g2*x1 + c3'*g2*x2 + u
where g2=1 if group==2 and g2=0 otherwise. I can do this by
typing
. generate g2 = (group==2)
. generate g2x1 = g2*x1
. generate g2x2 = g2*x2
. regress y x1 x2 g2 g2x1 g2x2
Think about the predictions from this model. The model says
y = a3 + b3*x1 + c3*x2 + u when g2==0
y = (a3+a3') + (b3+b3')*x1 + (c3+c3')*x2 + u when g2==1
Thus the model is equivalent to fitting the separate models
y = a1 + b1*x1 + c1*x2 + u for group == 1
y = a2 + b2*x1 + c2*x2 + u for group == 2
The relationship being
a1 = a3 a2 = a3 + a3'
b1 = b3 b2 = b3 + b3'
c1 = c3 c2 = c3 + c3'
Some of you may be concerned that in the pooled model (the one estimating
a3, b3, etc.), we are constraining the var(u) to be the same
for each group, whereas, in the separate-equation model, we estimate
different variances for group 1 and group 2. This does not matter, because
the model is fully interacted. That is probably not convincing, but what
should be convincing is that I am about to obtain the same
F(3,174) = 8.87 answer and, in my concocted data, I
have different variances in each group.
So, here is the result of the alternative test coeffiecients against 0 in a
pooled specification:
. generate g2 = (group==2)
. generate g2x1 = g2*x1
. generate g2x2 = g2*x2
. regress y x1 x2 g2 g2x1 g2x2
Source | SS df MS Number of obs = 180
---------+------------------------------ F( 5, 174) = 6.65
Model | 881.532123 5 176.306425 Prob > F = 0.0000
Residual | 4610.83174 174 26.499033 R-squared = 0.1605
---------+------------------------------ Adj R-squared = 0.1364
Total | 5492.36386 179 30.683597 Root MSE = 5.1477
------------------------------------------------------------------------------
y | Coef. Std. Err. t P>|t| [95% Conf. Interval]
---------+--------------------------------------------------------------------
x1 | 5.121087 1.757587 2.91 0.004 1.652152 8.590021
x2 | -3.227026 1.782504 -1.81 0.072 -6.745139 .2910877
g2 | -2.086535 1.917507 -1.09 0.278 -5.871102 1.698032
g2x1 | -6.335727 2.714897 -2.33 0.021 -11.6941 -.9773583
g2x2 | 11.72417 2.59115 4.52 0.000 6.610035 16.8383
_cons | -.1725655 1.374785 -0.13 0.900 -2.885966 2.540835
------------------------------------------------------------------------------
. test g2 g2x1 g2x2
( 1) g2 = 0
( 2) g2x1 = 0
( 3) g2x2 = 0
F( 3, 174) = 8.87
Prob > F = 0.0000
Same answer.
This definition of the “Chow test” is equivalent to pooling the
data, fitting the fully interacted model, and then testing the group 2
coefficients against 0.
That is why I said, “Chow Test is a term I have heard used by
economists in the context of testing a set of regression coefficients being
equal to 0.”
Admittedly, this leaves a lot unsaid.
The issue of the variance of u being equal in the two groups is subtle, but
I do not want that to get in the way of understanding that the Chow test is
equivalent to the “pool the data, interact, and test” procedure.
They are equivalent.
Concerning variances, the Chow test itself is testing against a pooled,
uninteracted model and so has buried in it an assumption of equal variances.
It is really a test that the coefficients are equal and variance(u) in the
groups are equal. It is, however, a weak test of the equality of variances
because that assumption manifests itself only in how the pooled coefficient
estimates are manufactured. Because the Chow test and the “pool the
data, interact, and test” procedure are the same, the same is true of
both procedures.
Your second concern might be that in the “pool the data, interact, and
test” procedure there is an extra assumption of equality of variances
because everything comes from the pooled model. As shown, this is not true.
It is not true because the model is fully interacted, so the assumption
of equal variances never makes a difference in the calculation of the
coefficients.
In Stata 12, you can also use the
contrast
command with factor variables to perform the same test:
. regress y c.x1##i.g2 c.x2##i.g2
Source | SS df MS Number of obs = 180
-------------+------------------------------ F( 5, 174) = 6.65
Model | 881.532123 5 176.306425 Prob > F = 0.0000
Residual | 4610.83174 174 26.499033 R-squared = 0.1605
-------------+------------------------------ Adj R-squared = 0.1364
Total | 5492.36386 179 30.683597 Root MSE = 5.1477
------------------------------------------------------------------------------
y | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
x1 | 5.121087 1.757587 2.91 0.004 1.652152 8.590021
1.g2 | -2.086535 1.917507 -1.09 0.278 -5.871102 1.698032
|
g2#c.x1 |
1 | -6.335727 2.714897 -2.33 0.021 -11.6941 -.9773583
|
x2 | -3.227026 1.782504 -1.81 0.072 -6.745139 .2910877
|
g2#c.x2 |
1 | 11.72417 2.59115 4.52 0.000 6.610035 16.8383
|
_cons | -.1725655 1.374785 -0.13 0.900 -2.885966 2.540835
------------------------------------------------------------------------------
. contrast g2 g2#c.x1 g2#c.x2,overall
Contrasts of marginal linear predictions
Margins : asbalanced
------------------------------------------------
| df F P>F
-------------+----------------------------------
g2 | 1 1.18 0.2780
|
g2#c.x1 | 1 5.45 0.0208
|
g2#c.x2 | 1 20.47 0.0000
|
Overall | 3 8.87 0.0000
|
Residual | 174
------------------------------------------------
An additional example can be found in the
“Chow tests” section of [R] contrast.
|
