# Multiple linear regression inconsistency?

I've got a linear model: $y_i=β_1x_{i1}+β_2x_{i2}+ε_i$ where E($ε_i$)=0 and Var($ε_i$)= $σ^2I_n$ for i=1,...,n

Supposed we don't have the data for $x_{i2}$ and we estimate: $y_i=β_1x_{i1}+ε_i$ for i=1,...,n

So far I've shown that $\hat{\beta}_1$ is biased and that $s^2= {ε_i'ε_i}/(n-1)$ is also biased and that such bias is positive.

I now have to explain why I'd expect that the OLS estimator of $\hat{\beta}_1$ is inconsistent and then suggest a suitable estimator giving conditions where it'd be consistent.

Please could someone give me a helping hand, I'm not sure why $\hat{\beta}_1$ is inconsistent?

Thanks

• What you hav writtn os not mak any snse! $\beta_1$ is a number, not a random variable, so what does it mean for it to b inosistnt? something is surely missing hre. – kjetil b halvorsen May 31 '15 at 10:48
• @kjetilbhalvorsen oops most of those β should be β hat. I'm not sure how to amend that! Also added the other equation which I missed out! – Emma May 31 '15 at 11:01
• it your post! to get $\hat{\bta}$ write (within dollar signs) \hat{\beta} – kjetil b halvorsen May 31 '15 at 13:09
• There we go, should be fully correct now! – Emma May 31 '15 at 13:44
• @kjetilbhalvorsen sorry forgot to tag you! – Emma May 31 '15 at 16:28

$\hat{\beta} = (x_1'x_1)^{-1}x_1'y$

Now substitute the true model $y=X\beta + \epsilon$ into $y$

$\hat{\beta} = (x_1'x_1)^{-1}x_1'y = \beta_1 + (x_1'x_1)^{-1}x_1'x_2\beta_2+ (x_1'x_1)^{-1}x_1'\epsilon_i$

Taking the limit sends the last term to $0$, which leaves

$\hat{\beta} \rightarrow \beta_1 + \frac{cov(x_1,x_2)}{var(x_1)}\beta_2$

Which means all you need to do if figure out when $\frac{cov(x_1,x_2)}{var(x_1)}\beta_2$=0

• why is it $cov(x_1,x_2)$ and $var(x_1)$? I get E($x_1x_2'$) and E($x_1x_1'$)? – Emma Jun 2 '15 at 7:50
• or are they the same? – Emma Jun 2 '15 at 7:50
• Well, the issue is you have no constant in your regression, so I'm not sure they would be exactly the same. With a constant, then we know that $\hat{\beta}=\frac{cov(x_1,y)}{var(x_1)}$. Therefore, by the same logic as above we get $\hat{\beta} \rightarrow \beta_1 + \beta_2\frac{cov(x_1,x_2)}{var(x_1)}$. I chose to use covariances in my answer because I think it shows more clearly the conditions you need to satisfy. But, the middle line in the solution is the more general answer. – Greg Jun 2 '15 at 12:48