I have an output that puzzles me: I think the coef
for sat
in my eivreg
should be lower than in my reg
, but it is higher since .0024138 > .0019311
Let us assume that:
$$
\newcommand{\Var}{{\rm Var}}
\newcommand{\CoV}{{\rm CoV}}
\newcommand{\ability}{{\rm ability}}
\newcommand{\colgpa}{{\rm colgpa}}
\newcommand{\Var}{{\rm Var}}
\frac{\Var(\ability)}{\Var(\ability)+\Var(e)} = 0.8
$$
. eivreg colgpa sat, r(sat .8)
assumed Errors-in-variables regression
variable reliability
---------------------------- Number of obs = 4137
sat 0.8000 F( 1, 4135) = 873.03
* 1.0000 Prob > F = 0.0000
R-squared = 0.2088
Root MSE = .58592
------------------------------------------------------------------------------
colgpa | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
sat | .0024138 .0000817 29.55 0.000 .0022537 .002574
_cons | .1656496 .0846635 1.96 0.050 -.0003365 .3316356
------------------------------------------------------------------------------
. reg colgpa sat
Source | SS df MS Number of obs = 4137
-------------+------------------------------ F( 1, 4135) = 829.26
Model | 299.712725 1 299.712725 Prob > F = 0.0000
Residual | 1494.48295 4135 .36142272 R-squared = 0.1670
-------------+------------------------------ Adj R-squared = 0.1668
Total | 1794.19567 4136 .433799728 Root MSE = .60118
------------------------------------------------------------------------------
colgpa | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
sat | .0019311 .0000671 28.80 0.000 .0017996 .0020625
_cons | .6630568 .0697213 9.51 0.000 .5263656 .799748
------------------------------------------------------------------------------
Here is why I think the coef
should be lower.
Our model first model is:
$$
\colgpa = \beta_0 + \beta_1 \ability + u \tag{1}
$$
where $u$ is the random error term, and then we add $\ability = sat + e \quad (\gamma)$ so the true ability is affected by some noise $e$. So our model could be written
\begin{align}
\colgpa &= \beta_0 + \beta_1 (sat + e) + u \Leftrightarrow \\
\colgpa &= \beta_0 + \beta_1 sat + v \tag{2}
\end{align}
where $v = u - \beta_1 e$.
We then make the classical errors in variables assumption (CEV) that $$ \CoV(\ability, e)=0. $$
We can see that model $(1)$ is computed with the reg-command, whereas $(2)$ is computed with the eivreg
-command.
From theory we know that this assumptions leads to two things.
- Firstly, The standard errors go up when comparing $(1)$ to $(2)$, because we add the fact that $\ability = sat + e$ and with this we add two more assumptions of which the most important one is $\CoV(\ability, e)=0$. If we were to regress only model $(1)$ then we could get a certain standard deviation of our estimates. Later on we add more information - model $(2)$ - and get a higher standard deviation. This can be seen by the formulas since $\Var(\hat \beta_1) = \Var(v) / SSTx$ and Model 2 variance = $\Var(v) = \Var(u - \beta_1 e) = \Var(u) + \beta_1^2 \Var(e) + 0 > \Var(u)$ = model 1 variance.
- Secondly, the estimated
coef
. for sat score is biased, so that $$ \operatorname{plim} \hat \beta_1 = \beta_1 + \frac{\CoV(sat, v)}{\Var(sat)} = \dots = \beta_1 \left( \frac{\Var(\ability)}{\Var(\ability)+\Var(e)} \right). $$ The scalar in front of $\beta_1$ is less than 1 so our new estimate is lower (with enough variance in e the slope tends to zero which is very unpleasant).
Update 151003:
I now noticed that $$ .0024138*.8 = 0.00193104 ≈ .0019311 $$ in other words: model2coef*0.8 = model1coef and $\frac{\Var(\ability)}{\Var(\ability)+\Var(e)} = 0.8$