# The distribution of biased and inconsistent OLS estimator under CEV assumption

We know the result that the OLS estimator with measurement error under the Classical Measurement Error (CEV) assumption is biased and inconsistent, and you can write down the probability limit of $$\hat{\beta}_{OLS}$$. But what is the distribution of this biased and inconsistent OLS estimator?

I compute and obtain the asymptotic variance, but it depends on $$\beta$$, which throws me off. I also believe it is normal, but I don't know. Please assume that the disturbance in the regression equation without measurement error and the measurement error are not necessarily normal, but have zero mean and constant variance.

• That's going to depend upon the distribution of the measurement error, except perhaps asymptotically. Apr 18 at 20:38
• What would be the asymptotic variance? Apr 19 at 3:02
• A lot depends on how much and what you know about the measurement error - its variance, the ratio of its variance to the error variance, or... Apr 19 at 3:35

Asymptotically, we know the distribution of $$\hat\beta_{OLS}$$, because it's just the distribution of $$\hat\beta_{OLS}$$ $$\sqrt{n}(\hat\beta_{OLS}-\beta_{OLS})\stackrel{d}{\to} N(0, V)$$ where $$V=A^{-1}BA$$ and $$A$$ and $$B$$ are the probability limits of $$(X^TX)/n$$, $$X^T(Y-\mu)^2X/n$$ respectively. That's just a fact about the OLS estimator, not depending on the probability model (except for assumptions like finite variance and independence).
Now if $$X=Z+\eta$$ where $$\eta$$ has zero mean and constant finite variance, and if the residuals also have constant variance and are independent of $$\eta$$, the variance simplifies to $$\sigma^2(X^TX)^{-1}$$. The variance of the slope, in particular, simplifies to $$\sigma^2/\mathrm{var}[X]$$. Since by assumption $$\mathrm{var}[X]=\mathrm{var}[Z]+\mathrm{var}[\eta]$$ we've made quite a bit of progress. Especially as you said you already knew what the limiting value $$\beta_{OLS}$$ was.
In the real world, you don't know $$\mathrm{var}[\eta]$$, and $$\eta$$ doesn't have zero mean or constant variance, so things are more difficult, but under the classical assumptions it's not too hard.