6
$\begingroup$

Initially we have this regression:

$$ \text{hourly wage} = 12.69 + 5.44\text{CollegeEducation} - 2.64\text{Female} $$

with standard errors of $0.21$ on $\text{CollegeEducation}$ and $0.20$ on $\text{Female}$. The variable $\text{Age}$ is added to the regression, which has a coefficient $0.29$, but the standards errors on college education and female stay the same. Dummies for four locations are then added, they have coefficients ranging from $-0.2$ to $0.6$, the standard errors on college education and female stay the same.

The questions are:

  • Why don't the standard errors change?
  • I have read somewhere that added uncorrelated controls can increases accuracy and lower standard errors, is this true?
$\endgroup$
1
  • $\begingroup$ How did you code (score) female ? are u talking about dichotomous - male versus female? $\endgroup$ – Subhash C. Davar May 17 '14 at 10:11
9
$\begingroup$

Yes, adding controls can increase the power of your statistical tests and make standard errors smaller. To see this, consider the following two regressions for comparison: $$ \begin{align} Y_i &= \alpha + \beta D_i + X'_i\gamma +e_i \newline Y_i &= \mu + \pi D_i + u_i \end{align} $$ Assume that $X,D$ are uncorrelated with the error terms $e$ and $u$ and that we have homoscedasticity. Then you can show that: $$ \begin{align} \sqrt{n}(\widehat{\beta} - \beta) &\stackrel{d}\rightarrow N\left(0, \frac{E(e^2)}{\text{Var}(D_i)(1-R^2_{D,X})} \right) \newline \sqrt{n}(\widehat{\mu} - \mu) &\stackrel{d}\rightarrow N\left( 0,\frac{E(u^2)}{\text{Var}(D_i)} \right) \end{align} $$ where $\stackrel{d}\rightarrow$ denotes convergence in distribution and $R^2_{D,X}$ is the $R^2$ from the regression of $D_i$ on $X_i$. I'm not going to prove this unless you explicitly request it because the main point of interest is the next result which uses the variances from these two distributions. The ratio of the two asymptotic variances is: $$ \frac{1-R^2_{Y,(D,X)}}{1-R^2_{Y,D}}\cdot \frac{1}{1-R^2_{D,X}} $$ where again $R^2_{Y,(D,X)}$ and $R^2_{Y,D}$ are the $R^2$s from the first and the second regression, respectively.

What does this ratio tell you?

  1. It shows the trade-off in asymptotic variances when going from the short to the long regression. The first term is smaller than (or equal to) one since $R^2$ increases when you add $X_i$ to the regression. It will be much smaller than one if $X_i$ explains a lot of the variation in $Y_i$. So this is how your standard errors decrease.
  2. The second term will be larger than (or equal to) one depending on the correlation between $D_i$ and $X_i$. If the two are strongly correlated, then the $R^2$ from the regression of $D_i$ on $X_i$ will be large and hence this second term will be large which is why your standard errors increase in this case.

If $D_i$ and $X_i$ are uncorrelated (e.g. if $D_i$ comes from a randomized experiment), then $R^2_{D,X} = 0$. This is the case when adding control variables is very preferable because they soak up the residual variance and increase the power of your statistical tests on $D_i$ which is great if this is your variable of interest.

So why don't your standard errors change? It's probably because the two counteracting effects from adding controls to your regression balance each other.

$\endgroup$
3
  • 1
    $\begingroup$ Your answer is really great. Do you have any references to textbooks, or papers, which elaborate a bit more, perhaps including some of the steps which you omitted? $\endgroup$ – hoyem May 10 '14 at 22:21
  • $\begingroup$ Wooldridge (2008) for the intuition and his advanced textbook Wooldridge (2010) "Econometric Analysis of Cross-Section and Panel Data" for the maths. If you want to try it yourself: start with the 1st regression and apply the Frisch-Waugh theorem to find $\widehat{\beta}$. Then use the Slutsky lemma and continuous mapping theorem to get the asymptotic variance. $\endgroup$ – Andy May 10 '14 at 22:57
  • 1
    $\begingroup$ +1, nice answer. I numbered the things the ratio tells you--I think it looks a little cleaner this way. If you don't like it, feel free to roll it back w/ my apologies. $\endgroup$ – gung - Reinstate Monica May 13 '14 at 2:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.