Fama MACbeth Standard Errors in R

Fama MacBeth standard errors in R

The plm package can estimate Fama-MacBeth regressions and SEs.

require(foreign)
require(plm)
require(lmtest)
test <- read.dta("http://www.kellogg.northwestern.edu/faculty/petersen/htm/papers/se/test_data.dta")
fpmg <- pmg(y~x, test, index=c("year","firmid")) ##Fama-MacBeth

> ##Fama-MacBeth
> coeftest(fpmg)

t test of coefficients:

            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 0.031278   0.023356  1.3392   0.1806    
x           1.035586   0.033342 31.0599   <2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

However note that this method works only if your data can be coerced to a pdata.frame. (It will fail if you have "duplicate couples (time-id)".)

For further details see:

• Fama-MacBeth and Cluster-Robust (by Firm and Time) Standard Errors in R

Newey-West standard errors with Mean Groups/Fama-MacBeth estimator

Currently this is impossible with plm package.

However, you could just create them yourself.

Suppose you have:

fpmg <- pmg(y~x, test, index = c('year', 'firmid'))
fpmg.coefficients <- fpmg$coefficients
# (Intercept)            x 
#  0.03127797   1.03558610 

coeftest(fpmg)
#             Estimate Std. Error t value Pr(>|t|)    
# (Intercept) 0.031278   0.023356  1.3392   0.1806    
# x           1.035586   0.033342 31.0599   <2e-16 ***

Then you can simply create the estimators yourself like in:

the.years <- unique(test$year)
a.formula <- y ~ x

first.step <-  lapply(the.years, function(a.year) {
                temp.data <- test[test$year == a.year, ]
                an.lm <- lm(a.formula, data = temp.data)
                the.coefficients <- an.lm$coef
                the.results <- as.data.frame(cbind(a.year, t(the.coefficients)))
                the.results
                }) 

first.step.df <- do.call('rbind', first.step)

second.step.coefficients <- apply(first.step.df[, -1], 2, mean)
second.step.coefficients
# (Intercept)           x 
#  0.03127797  1.03558610 

identical(fpmg.coefficients, second.step.coefficients)
# [1] TRUE

Check that they are identical both ways just in case.
Last, you can obtain the Newey-West (1987) with one lag adjusted t-statistics for the means with:

library(sandwich)
second.step.NW.sigma.sq <- apply(first.step.df[, -1], 2, 
                             function(x) sqrt(NeweyWest(lm(x ~ 1), 
                               lag = 1, prewhite = FALSE)['(Intercept)',       
                                 '(Intercept)']))
second.step.NW.sigma.sq
#  (Intercept)            x 
#   0.02438398   0.02859447
t.statistics.NW.lag.1 <- second.step.coefficients / second.step.NW.sigma.sq

t.statistics.NW.lag.1
# (Intercept)           x 
#    1.282726   36.216301

Update

In my answer, I had only included the "manual" calculation of the t-statistic, because it is computationally faster.
A more generic solution is to calculcate the Newey-West corrected t-statistics and their p-values with the coeftest() function of the lmtest package.

coeftest(lm(first.step.df$'(Intercept)' ~ 1), vcov = NeweyWest(lm(first.step.df$'(Intercept)' ~ 1), lag = 1, prewhite = FALSE))
# t test of coefficients:
#             Estimate Std. Error t value Pr(>|t|)
# (Intercept) 0.031278   0.024384  1.2827   0.2316
coeftest(lm(first.step.df$x ~ 1), vcov = NeweyWest(lm(first.step.df$x ~ 1), lag = 1, prewhite = FALSE))
# t test of coefficients:
#             Estimate Std. Error t value  Pr(>|t|)    
# (Intercept) 1.035586   0.028594  36.216 4.619e-11 ***

Fama Macbeth Regression in Python (Pandas or Statsmodels)

An update to reflect the library situation for Fama-MacBeth as of Fall 2018. The fama_macbeth function has been removed from pandas for a while now. So what are your options?

1. If you're using python 3, then you can use the Fama-MacBeth method in LinearModels: https://github.com/bashtage/linearmodels/blob/master/linearmodels/panel/model.py

2. If you're using python 2 or just don't want to use LinearModels, then probably your best option is to roll you own.

For example, suppose you have the Fama-French industry portfolios in a panel like the following (you've also computed some variables like past beta or past returns to use as your x-variables):

In [1]: import pandas as pd
        import numpy as np
        import statsmodels.formula.api as smf

In [4]: df = pd.read_csv('industry.csv',parse_dates=['caldt'])
        df.query("caldt == '1995-07-01'")

In [5]: Out[5]: 
      industry      caldt    ret    beta  r12to2  r36to13
18432     Aero 1995-07-01   6.26  0.9696  0.2755   0.3466
18433    Agric 1995-07-01   3.37  1.0412  0.1260   0.0581
18434    Autos 1995-07-01   2.42  1.0274  0.0293   0.2902
18435    Banks 1995-07-01   4.82  1.4985  0.1659   0.2951

Fama-MacBeth primarily involves computing the same cross-sectional regression model month by month, so you can implement it using a groupby. You can create a function that takes a dataframe (it will come from the groupby) and a patsy formula; it then fits the model and returns the parameter estimates. Here is a barebones version of how you could implement it (note this is what the original questioner tried to do a few years ago ... not sure why it didn't work although it's possible back then statsmodels result object method params wasn't returning a pandas Series so the return needed to be converted to a Series explicitly ... it does work fine in the current version of pandas, 0.23.4):

def ols_coef(x,formula):
    return smf.ols(formula,data=x).fit().params

In [9]: gamma = (df.groupby('caldt')
                .apply(ols_coef,'ret ~ 1 + beta + r12to2 + r36to13'))
        gamma.head()

In [10]: Out[10]: 
            Intercept      beta     r12to2   r36to13
caldt                                               
1963-07-01  -1.497012 -0.765721   4.379128 -1.918083
1963-08-01  11.144169 -6.506291   5.961584 -2.598048
1963-09-01  -2.330966 -0.741550  10.508617 -4.377293
1963-10-01   0.441941  1.127567   5.478114 -2.057173
1963-11-01   3.380485 -4.792643   3.660940 -1.210426

Then just compute the mean, standard error on the mean, and a t-test (or whatever statistics you want). Something like the following:

def fm_summary(p):
    s = p.describe().T
    s['std_error'] = s['std']/np.sqrt(s['count'])
    s['tstat'] = s['mean']/s['std_error']
    return s[['mean','std_error','tstat']]

In [12]: fm_summary(gamma)
Out[12]: 
               mean  std_error     tstat
Intercept  0.754904   0.177291  4.258000
beta      -0.012176   0.202629 -0.060092
r12to2     1.794548   0.356069  5.039896
r36to13    0.237873   0.186680  1.274230

Improving Speed

Using statsmodels for the regressions has significant overhead (particularly given you only need the estimated coefficients). If you want better efficiency, then you could switch from statsmodels to numpy.linalg.lstsq. Write a new function that does the ols estimation ... something like the following (notice I'm not doing anything like checking the rank of these matrices ...):

def ols_np(data,yvar,xvar):
    gamma,_,_,_ = np.linalg.lstsq(data[xvar],data[yvar],rcond=None)
    return pd.Series(gamma)

And if you're still using an older version of pandas, the following will work:

Here is an example of using the fama_macbeth function in pandas:

>>> df

                y    x
date       id
2012-01-01 1   0.1  0.4
           2   0.3  0.6
           3   0.4  0.2
           4   0.0  1.2
2012-02-01 1   0.2  0.7
           2   0.4  0.5
           3   0.2  0.1
           4   0.1  0.0
2012-03-01 1   0.4  0.8
           2   0.6  0.1
           3   0.7  0.6
           4   0.4 -0.1

Notice, the structure. The fama_macbeth function expects the y-var and x-vars to have a multi-index with date as the first variable and the stock/firm/entity id as the second variable in the index:

>>> fm  = pd.fama_macbeth(y=df['y'],x=df[['x']])
>>> fm

----------------------Summary of Fama-MacBeth Analysis-------------------------

Formula: Y ~ x + intercept
# betas :   3

----------------------Summary of Estimated Coefficients------------------------
     Variable          Beta       Std Err        t-stat       CI 2.5%      CI 97.5%
          (x)       -0.0227        0.1276         -0.18       -0.2728        0.2273
  (intercept)        0.3531        0.0842          4.19        0.1881        0.5181

--------------------------------End of Summary---------------------------------

Note that just printing fm calls fm.summary

>>> fm.summary

----------------------Summary of Fama-MacBeth Analysis-------------------------

Formula: Y ~ x + intercept
# betas :   3

----------------------Summary of Estimated Coefficients------------------------
     Variable          Beta       Std Err        t-stat       CI 2.5%      CI 97.5%
          (x)       -0.0227        0.1276         -0.18       -0.2728        0.2273
  (intercept)        0.3531        0.0842          4.19        0.1881        0.5181

--------------------------------End of Summary---------------------------------

Also, note the fama_macbeth function automatically adds an intercept (as opposed to statsmodels routines). Also the x-var has to be a dataframe so if you pass just one column you need to pass it as df[['x']].

If you don't want an intercept you have to do:

>>> fm  = pd.fama_macbeth(y=df['y'],x=df[['x']],intercept=False)

How can I use Newey-West Standard Errors in modelplot(), in R?

The modelplot documentation notes that the vcov argument can be either a list of functions, or a list of matrices, or a vector of strings. This means that both of these commands should work:

library(modelsummary)
library(sandwich)

mod <- lm(mpg ~ hp, mtcars)

modelplot(mod, vcov = c("HC3", "NeweyWest"))

Alternatively,

modelplot(mod, vcov = list(vcovHC, NeweyWest))
modelplot(mod, vcov = list(vcovHC(mod), NeweyWest(mod)))

The error message you get suggests that the specific model you are trying to summarize is unsupported by the sandwich package. You can check if it is supported by calling:

sandwich::NeweyWest(mod)
#>             (Intercept)           hp
#> (Intercept)  5.93124548 -0.032929737
#> hp          -0.03292974  0.000221307

In my example, mod was a lm model, but you did not supply a MINIMAL REPRODUCIBLE EXAMPLE, so your problem is impossible to diagnose conclusively.

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