4
$\begingroup$

I estimate 2 models in OLS.

$Y=\hat{\beta} X+e$ and $Y=\tilde{\beta} X+\gamma W +u$

The inclusion of the $W$ variable decreases the size of $\beta$ but does not change the $Var(\beta)$. $X$ and $W$ are not very collinear, which may explain why the standard errors did not change, but why then does the coefficient on $X$ decrease?

I read this post and was a bit confused on whether it explained my question or not: How are standard errors affected in a multivariate regression?

Edit: To be more specific: The sample size is 86,000. The variance decreases by less than 1% while the coefficient decreases by more than 50%.

$\endgroup$
2
  • $\begingroup$ What is your sample size? In small samples, virtually anything can happen. Also, by "does not change", do you really mean not at all, or not by much? $\endgroup$ Mar 24 '15 at 13:43
  • $\begingroup$ @Christoph-Hanck The sample size is 86,000. The variance increases by less than 1% while the coefficient decreases by 50%. $\endgroup$
    – peru
    Mar 24 '15 at 16:32
1
$\begingroup$

The most plausible candidate seems omitted variable bias in that case. If both $X$ and $W$ have a positive effect on $Y$ and are positively correlated, then regressing $Y$ on $X$ only will force $X$ to also "do the work" for $W$, i.e. also reflect its positive impact on $Y$. Once you also include $W$, $X$ can concentrate on its own effect, which brings down its coefficient.

That the standard errors hardly chance in such a large sample reflects the fact that coefficients are estimated very precisely (although very precisely wrong in the first case).

Consider this little example in which the regressors in x are generated as mean zero multivariate normals with positive covariance of 0.4. $X$ has a positive coefficient of 0.3, while that of $W$ is 0.6. The results roughly replicate your findings.

library(mvtnorm)
n <- 86000
sigma <- matrix(c(1,.4,.4,1), ncol=2)
x <- rmvnorm(n=n, mean=c(1,2), sigma=sigma)
y <- .3*x[,1]+.6*x[,2]+rnorm(n,1)
summary(lm(y~x[,1]))
summary(lm(y~x))
$\endgroup$
1

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.