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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.

  1. 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.
  2. 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 $\frac{\Var(\ability)}{\Var(\ability)+\Var(e)} = 0.8$

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  • 1
    $\begingroup$ There is no question here, so my answer might be missing the point…In any case, in the classical errors-in-variables model, the OLS estimate has attenuation bias, ie it is biased towards 0. if the true coefficient is positive, than $\beta_{\text{OLS}}<\beta_{\text{CEV}}$, and vice versa if the true coefficient is negative. I see no problem in your models. By the way, in your equation (2), $v=u\color{red}{+}\beta_{1}e$. $\endgroup$
    – Matthijs
    Commented Oct 4, 2015 at 15:56
  • $\begingroup$ @Matthijs, if a tag isn't incorrect & you aren't at the limit, we usually defer to the OP. Software tags can engage syntax highlighting & help readers know how to interpret the code provided. $\endgroup$ Commented Oct 4, 2015 at 16:15

1 Answer 1

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Small errors in the model-setup and in thinking of which beta should be the biggest one (the output is in fact in line with theory).


Our model 1 is: $$colgpa = \beta_0 + \beta_1 ability + u \quad (1) \\ Cov(ability, u) = 0$$ where $u$ is the random error term.

This is our first model. But then we make some changes to id. Start by introducing $ability = sat - e \quad (\gamma)$ so the true ability is affected by some random noise $e$. So our model could be written $colgpa = \beta_0 + \beta_1 (sat - e) + u \Leftrightarrow colgpa = \beta_0 + \beta_1 sat + v$ where $v = u - \beta_1 e$.

Let's further make the trivial assumption that $E[e]=0$. We then make the crucial assumption - the classical errors in variables assumption (CEV) - that $CoV(ability, e)=0.$ So to conclude, we have our model 2: $$colgpa = \beta_0 + \beta_1 sat + v \quad (2) \\ CoV(ability, e)=0.$$

From theory we know that this leads to two things.

  1. (Std.Err up). Firstly, The standard errors go up when comparing (1) to (2). This can be seen by the formulas since $Var(\hat \beta_1) = Var(v) / SSTx$ and $Var(v) = Var(u - \beta_1 e) = Var(u) + \beta_1^2 Var(e) + 0 > Var(u)$
  2. (Biased coef) Secondly, the estimated coef. The 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 higher. Define $r = \text{reliability ratio} = \frac{Var(ability)}{Var(ability)+Var(e)}$. Then[^2] we get $model2coef \cdot r = model1coef$ or more formally $$\beta_{CEV}^{sat} \cdot r = \beta_{OLS}^{sat}.$$

[^2]: This follows if we look at the equation above containing $\operatorname{plim}$ and reminding ourselves of the fact that model1coef is unbiased.

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