What would be a valid argument not to include firm fixed effects in a regression for a panel dataset?
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If you think there's no unobserved, unit-specific, time-stable confounding (spoiler: rarely is this the case). But see this paper here. -
What if I include year fixed effects, would you say it makes sense to additionally include firm fixed effects? The goal is to examine the change of leverage from each firm over time.Comment
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According to the paper I've linked here, Stata (and R for that matter) do incorrect/misleading matrix algebraic calculations when we use both time and year dummies. The idea is that which set of FE you include changes the causal question you're answering and the identification of the coefficients. Quoting them directly,
We are also the first to show both formally and with simulations that if we begin with the assumptions that many researchers bring to the two-way FE model—that it represents a single estimate of X on Y while accounting for unit-level heterogeneity and time shocks—this specification is in fact statistically unidentified. This concerning fact about twoway FE models has probably remained hidden for so long because statistical software packages like R and Stata employ hidden matrix processing to work around the unidentifiability. We urge researchers to choose between models by taking into consideration the way in which parameters must be interpreted. A correct answer to the wrong question is as useless as the wrong answer to the right question.Comment
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Benjamin:
are you talking about -regress- or -xtreg-?Kind regards,
Carlo
(StataNow 18.5)Comment
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I'm talking about -xtreg-Comment
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Benjamin:
the following one is a trivial example that proves the unusefulness of including -i.code- if you go -xtreg,fe- (perfect collinearity with the -panelid):
Code:. use "https://www.stata-press.com/data/r17/nlswork.dta" (National Longitudinal Survey of Young Women, 14-24 years old in 1968) . xtreg ln_wage i.year i.idcode if idcode<=3, fe note: 2.idcode omitted because of collinearity. note: 3.idcode omitted because of collinearity. Fixed-effects (within) regression Number of obs = 39 Group variable: idcode Number of groups = 3 R-squared: Obs per group: Within = 0.5446 min = 12 Between = 0.2670 avg = 13.0 Overall = 0.3678 max = 15 F(14,22) = 1.88 corr(u_i, Xb) = -0.0356 Prob > F = 0.0897 ------------------------------------------------------------------------------ ln_wage | Coefficient Std. err. t P>|t| [95% conf. interval] -------------+---------------------------------------------------------------- year | 69 | .208967 .3918928 0.53 0.599 -.6037689 1.021703 70 | -.2747772 .3439816 -0.80 0.433 -.9881514 .4385969 71 | -.3613911 .326316 -1.11 0.280 -1.038129 .3153467 72 | -.2056973 .326316 -0.63 0.535 -.8824352 .4710406 73 | -.0310461 .326316 -0.10 0.925 -.707784 .6456917 75 | .0416271 .326316 0.13 0.900 -.6351107 .718365 77 | .0358937 .326316 0.11 0.913 -.6408441 .7126316 78 | .2433199 .326316 0.75 0.464 -.4334179 .9200578 80 | .2726139 .326316 0.84 0.412 -.4041239 .9493518 82 | .1747839 .3439816 0.51 0.616 -.5385903 .8881581 83 | .2924489 .326316 0.90 0.380 -.3842889 .9691868 85 | .3712589 .326316 1.14 0.267 -.305479 1.047997 87 | .2960361 .326316 0.91 0.374 -.3807017 .972774 88 | .3038639 .326316 0.93 0.362 -.3728739 .9806018 | idcode | 2 | 0 (omitted) 3 | 0 (omitted) | _cons | 1.659677 .2833366 5.86 0.000 1.072073 2.247281 -------------+---------------------------------------------------------------- sigma_u | .24956596 sigma_e | .27711004 rho | .44784468 (fraction of variance due to u_i) ------------------------------------------------------------------------------ F test that all u_i=0: F(2, 22) = 9.64 Prob > F = 0.0010 .Kind regards,
Carlo
(StataNow 18.5)Comment
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this is a very good thread. i'm quite sure i've been making the mis-steps for years.
for what it's worth - on my stata 17se on a 7 year old pc laptop: the single took less than 1 second to run the model that Carlo Lazzaro presented without i.firm effects, and about 10 minutes to run with i.firm effects. the results were identical, despite the effects from idcode 2/3 not in the model..
regarding Jared Greathouse's advice of when to include a firm fixed effect "If you think there's no unobserved, unit-specific, time-stable confounding (spoiler: rarely is this the case)" - is there a way this can be determined from the data that does not involve running the models with and without that second-level effect, which is so computing-time-intensive?
given the 2-3 magnitude difference in computing time, it would be very efficient to know when to (avoid needing to) include second-level fixed effects.
i'm far from any expert and ask the question with deference.
thanks!Last edited by George Hoffman; 11 Apr 2022, 19:38.Comment
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George:
probably the confusion stems form the fact that we can go -fe- with -regress- too, provided that -panelid- is plugged in as a categorical predictor in the right-hand side of the panel data regression equation:
As expected, the coefficinets of teh shared predictors are identical no matter the approach we follow.Code:use "https://www.stata-press.com/data/r17/nlswork.dta" . xtreg ln_wage c.age##c.age i.year if idcode<=3, fe Fixed-effects (within) regression Number of obs = 39 Group variable: idcode Number of groups = 3 R-squared: Obs per group: Within = 0.7404 min = 12 Between = 0.4068 avg = 13.0 Overall = 0.4014 max = 15 F(16,20) = 3.57 corr(u_i, Xb) = -0.8560 Prob > F = 0.0042 ------------------------------------------------------------------------------ ln_wage | Coefficient Std. err. t P>|t| [95% conf. interval] -------------+---------------------------------------------------------------- age | .0773019 .2865219 0.27 0.790 -.5203723 .6749761 | c.age#c.age | -.0045583 .0012212 -3.73 0.001 -.0071057 -.002011 | year | 69 | .3367906 .4335876 0.78 0.446 -.5676572 1.241238 70 | .2089384 .6771373 0.31 0.761 -1.203545 1.621422 71 | .3144116 .9610926 0.33 0.747 -1.690392 2.319216 72 | .5888124 1.253657 0.47 0.644 -2.02627 3.203894 73 | .8912873 1.550825 0.57 0.572 -2.343676 4.126251 75 | 1.246958 2.152898 0.58 0.569 -3.243908 5.737823 77 | 1.560689 2.761762 0.57 0.578 -4.200247 7.321624 78 | 1.941522 3.068213 0.63 0.534 -4.458659 8.341703 80 | 2.34498 3.684737 0.64 0.532 -5.341247 10.03121 82 | 2.698954 4.315145 0.63 0.539 -6.30228 11.70019 83 | 2.994437 4.618087 0.65 0.524 -6.638723 12.6276 85 | 3.538578 5.245889 0.67 0.508 -7.404154 14.48131 87 | 3.965153 5.878139 0.67 0.508 -8.296429 16.22674 88 | 4.40786 6.407149 0.69 0.499 -8.957218 17.77294 | _cons | 1.465543 5.342682 0.27 0.787 -9.679096 12.61018 -------------+---------------------------------------------------------------- sigma_u | .54258328 sigma_e | .21942548 rho | .85944136 (fraction of variance due to u_i) ------------------------------------------------------------------------------ F test that all u_i=0: F(2, 20) = 10.43 Prob > F = 0.0008 . reg ln_wage c.age##c.age i.year i.idcode if idcode<=3 Source | SS df MS Number of obs = 39 -------------+---------------------------------- F(18, 20) = 4.86 Model | 4.21278813 18 .234043785 Prob > F = 0.0005 Residual | .962950828 20 .048147541 R-squared = 0.8139 -------------+---------------------------------- Adj R-squared = 0.6465 Total | 5.17573896 38 .136203657 Root MSE = .21943 ------------------------------------------------------------------------------ ln_wage | Coefficient Std. err. t P>|t| [95% conf. interval] -------------+---------------------------------------------------------------- age | .0773019 .2865219 0.27 0.790 -.5203723 .6749761 | c.age#c.age | -.0045583 .0012212 -3.73 0.001 -.0071057 -.002011 | year | 69 | .3367906 .4335876 0.78 0.446 -.5676572 1.241238 70 | .2089384 .6771373 0.31 0.761 -1.203545 1.621422 71 | .3144116 .9610926 0.33 0.747 -1.690392 2.319216 72 | .5888124 1.253657 0.47 0.644 -2.02627 3.203894 73 | .8912873 1.550825 0.57 0.572 -2.343676 4.126251 75 | 1.246958 2.152898 0.58 0.569 -3.243908 5.737823 77 | 1.560689 2.761762 0.57 0.578 -4.200247 7.321624 78 | 1.941522 3.068213 0.63 0.534 -4.458659 8.341703 80 | 2.34498 3.684737 0.64 0.532 -5.341247 10.03121 82 | 2.698954 4.315145 0.63 0.539 -6.30228 11.70019 83 | 2.994437 4.618087 0.65 0.524 -6.638723 12.6276 85 | 3.538578 5.245889 0.67 0.508 -7.404154 14.48131 87 | 3.965153 5.878139 0.67 0.508 -8.296429 16.22674 88 | 4.40786 6.407149 0.69 0.499 -8.957218 17.77294 | idcode | 2 | -.4183815 .0918256 -4.56 0.000 -.6099263 -.2268366 3 | .6579353 1.834332 0.36 0.724 -3.168414 4.484284 | _cons | 1.341224 4.651269 0.29 0.776 -8.361153 11.0436 ------------------------------------------------------------------------------ .
That said, while what above is useful to show the way -fe- machinery works, I do prefer -xtreg- when it comes to panel data analysis.Kind regards,
Carlo
(StataNow 18.5)Comment
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If \(T\) is large, you can install reghdfe from SSC and absorb both the firm and time indicators. A test of whether time effects should be included in the model involves comparing the the log-likelihood for the model absorbing only firm indicators with that of the model absorbing both time and firm indicators within which it is nested. Such a likelihood ratio test is equivalent to the Wald test of the joint significance of the time indicators. You need to do this test manually where the degrees of freedom in the test equal the number of absorbed year indicators.
You have that
$$LR= 2(\ell_{1}-\ell_{0})$$
and the test statistic is distributed as \(\chi^2(T-1)\) under the null hypothesis, where \(T\) is the number of time periods and \(\ell_{1}\) and \(\ell_{0}\) are the log-likelihoods from the model with both sets of indicators and model with firm dummies only, respectively. With large \(T\), the LR test and the Wald test of joint significance of the year dummies will be equivalent. Compare:
Res.:Code:webuse nlswork, clear xtset idcode year xtreg ln_wage wks_work tenure i.year, fe testparm i.year reghdfe ln_wage wks_work tenure, absorb(idcode) local ll0= e(ll) reghdfe ln_wage wks_work tenure i.year, absorb(idcode year) local ll1= e(ll) qui levelsof year if e(sample), local(years) local Tminus1 = wordcount("`years'")-1 local LR= 2*(`ll1'-`ll0') display " LR chi2(`Tminus1') = `:di %4.2f `LR''; Prob > chi2 `:di %4.3f chi2tail(`Tminus1',`LR')'"
Code:. testparm i.year ( 1) 69.year = 0 ( 2) 70.year = 0 ( 3) 71.year = 0 ( 4) 72.year = 0 ( 5) 73.year = 0 ( 6) 75.year = 0 ( 7) 77.year = 0 ( 8) 78.year = 0 ( 9) 80.year = 0 (10) 82.year = 0 (11) 83.year = 0 (12) 85.year = 0 (13) 87.year = 0 (14) 88.year = 0 F( 14, 22718) = 49.53 Prob > F = 0.0000 . . local LR= 2*(`ll1'-`ll0') . . display " LR chi2(`Tminus1') = `:di %4.2f `LR''; Prob > chi2 `:di %4.3f chi2tail(`Tminus1',`LR')'" LR chi2(14) = 807.01; Prob > chi2 0.000
Having showed the above, with a large \(T\) dimension, you would almost always find that the time indicators are significant, so one need not bother doing the test.Comment
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