# Measurement Error - Multivariate Case

I have a linear regression model, with usual assumptions holding; $$E[xu] = 0$$ and rank condition.

$$y_i = \alpha_0 + \alpha_1x_{1i} + \alpha_2x_{2i} + u_i$$

I observe $$\bar{x}_{2i}$$, where:

$$\bar{x}_{2i} = x_{2i} + e_i$$

My estimated model is:

$$y_i = \tilde{\alpha_o} + \tilde{\alpha_1}x_{1i} + \tilde{\alpha_2}x_{2i} + \tilde{u_i}$$

I want to derive the plim of $$\tilde{\alpha_1}$$ and $$\tilde{\alpha_2}$$

My approach:

I substituted for my observation, and evaluated the following:

plim $$\tilde{\alpha_2} = \frac{cov(\bar{x}_{2i}, y)}{var(\bar{x}_{2i})} = \frac{\alpha_{2}*var(x_{2i})}{var(x_{2i} + e_i)} = \frac{\alpha_{2}*var(x_{2i})}{var(x_{2i}) + var(e_i)}$$

plim $$\tilde{\alpha_1} = \frac{cov({x}_{1i}, y)}{var({x}_{1i})} = \frac{\alpha_{1}*var(x_{1i})}{var(x_{1i})} = \alpha_1$$

Is this correct? I am a little worried about $$\tilde{\alpha_1}$$, because shouldn't this be biased?

To derive this you'll want to use the Frisch-Waugh-Lovell theorem.

Using the true variable, $$x_2$$, let $$\widetilde{x_2}$$ be the residual from a regression of $$x_2$$ on $$x_1$$,

$$x_2 = \delta_0 +\delta_1 x_1 +\widetilde{x_2}$$ We thus have,

$$\bar{x_2} = \delta_0 +\delta_1 x_1 +e +\widetilde{x_2}$$

The residual from a regression of $$\bar{x_2}$$ on $$x_1$$ is $$(e +\widetilde{x_2})$$.

By the Frisch-Waugh-Lovell theorem, the OLS estimate of the coefficient for $$x_2$$ in your model of estimation will be the same as the OLS estimate from

$$y_i = \alpha_0 +\alpha_2 (e_i +\widetilde{x_{2i}}) + u_i$$

So we have $$\hat{\alpha_2} = \frac{Cov(y_i, (e_i +\widetilde{x_{2i}}))}{Var(e_i +\widetilde{x_{2i}})}$$

You will plug in for $$y_i = \alpha_0 +\alpha_{1i} x_1 +\alpha_2 x_{2i}+u_i$$, and note that $$Cov(x_{1i}, \widetilde{x_{2i}})=0$$ because $$\widetilde{x_{2i}}$$ is the residual from an OLS regression with $$x_1$$ as a regressor. To proceed, you will need to make an assumption regarding $$Cov(x_{1i}, e_i))$$ and $$Cov(u_{i}, e_i))$$.

To estimate the effect of $$x_1$$ on $$y$$, we need to consider a regression of $$x_1$$ on $$x_2$$.

$$x_1 = a_0 +a_1 x_2 + \widetilde{x_1}$$

Using the mismeasured version of $$x_2$$,

$$x_1 = a_0 +a_1 \bar{x_2} - a_1e + \widetilde{x_1}$$

The residual is $$(- a_1e + \widetilde{x_1})$$.

We apply the Frisch-Waugh-Lovell theorem to know the estimate for the coefficient of $$x_1$$ is the same as the estimate from, $$y_i = \alpha_0 +\alpha_1 (- a_1e + \widetilde{x_1}) + u_i$$

This is $$\hat{\alpha_1} = \frac{Cov(y_i, (- a_1e + \widetilde{x_1}))}{Var(- a_1e + \widetilde{x_1})}$$

Analogous to before, you will plug in for $$y_i = \alpha_0 +\alpha_{1i} x_1 +\alpha_2 x_{2i}+u_i$$, and note that $$Cov(x_{2i}, \widetilde{x_{1i}})=0$$ because $$\widetilde{x_{1i}}$$ is the residual from an OLS regression with $$x_2$$ as a regressor. To proceed, you will need to make an assumption regarding $$Cov(x_{1i}, e_i))$$ and $$Cov(u_{i}, e_i))$$.

• Thanks for your answer. For some reason I am unable to comment. What about: $Cov(x_{1i}, \widetilde{x_{1i}})$? Commented Sep 25, 2022 at 8:57
• Plug in that $x_1 = a_0 +a_1 \bar{x_2} -a_1e +\widetilde{x_1}$, then derive. Commented Sep 26, 2022 at 7:29