<p>To derive this you'll want to use the Frisch-Waugh-Lovell theorem.</p>
<p>Using the true variable, <span class="math-container">$x_2$</span>, let <span class="math-container">$\widetilde{x_2}$</span> be the residual from a regression of <span class="math-container">$x_2$</span> on <span class="math-container">$x_1$</span>,</p>
<p><span class="math-container">$$x_2 = \delta_0 +\delta_1 x_1 +\widetilde{x_2}$$</span>
We thus have,</p>
<p><span class="math-container">$$\bar{x_2} = \delta_0 +\delta_1 x_1 +e +\widetilde{x_2}$$</span></p>
<p>The residual from a regression of <span class="math-container">$\bar{x_2}$</span> on <span class="math-container">$x_1$</span> is <span class="math-container">$(e +\widetilde{x_2})$</span>.</p>
<p>By the Frisch-Waugh-Lovell theorem, the OLS estimate of the coefficient for <span class="math-container">$x_2$</span> in your model of estimation will be the same as the OLS estimate from</p>
<p><span class="math-container">$$y_i = \alpha_0 +\alpha_2 (e_i +\widetilde{x_{2i}}) + u_i$$</span></p>
<p>So we have <span class="math-container">$$\hat{\alpha_2} = \frac{Cov(y_i, (e_i +\widetilde{x_{2i}}))}{Var(e_i +\widetilde{x_{2i}})}$$</span></p>
<p>You will plug in for <span class="math-container">$y_i = \alpha_0 +\alpha_{1i} x_1 +\alpha_2 x_{2i}+u_i$</span>, and note that <span class="math-container">$Cov(x_{1i}, \widetilde{x_{2i}})=0$</span> because <span class="math-container">$\widetilde{x_{2i}}$</span> is the residual from an OLS regression with <span class="math-container">$x_1$</span> as a regressor.  To proceed, you will need to make an assumption regarding <span class="math-container">$Cov(x_{1i}, e_i))$</span> and <span class="math-container">$Cov(u_{i}, e_i))$</span>.</p>
<p>To estimate the effect of <span class="math-container">$x_1$</span> on <span class="math-container">$y$</span>, we need to consider a regression of <span class="math-container">$x_1$</span> on <span class="math-container">$x_2$</span>.</p>
<p><span class="math-container">$$x_1 = a_0 +a_1 x_2 + \widetilde{x_1}$$</span></p>
<p>Using the mismeasured version of <span class="math-container">$x_2$</span>,</p>
<p><span class="math-container">$$x_1 = a_0 +a_1 \bar{x_2} - a_1e + \widetilde{x_1} $$</span></p>
<p>The residual is <span class="math-container">$(- a_1e + \widetilde{x_1})$</span>.</p>
<p>We apply the Frisch-Waugh-Lovell theorem to know the estimate for the coefficient of <span class="math-container">$x_1$</span> is the same as the estimate from,
<span class="math-container">$$y_i = \alpha_0 +\alpha_1 (- a_1e + \widetilde{x_1}) + u_i$$</span></p>
<p>This is <span class="math-container">$$\hat{\alpha_1} = \frac{Cov(y_i, (- a_1e + \widetilde{x_1}))}{Var(- a_1e + \widetilde{x_1})}$$</span></p>
<p>Analogous to before, you will plug in for <span class="math-container">$y_i = \alpha_0 +\alpha_{1i} x_1 +\alpha_2 x_{2i}+u_i$</span>, and note that <span class="math-container">$Cov(x_{2i}, \widetilde{x_{1i}})=0$</span> because <span class="math-container">$\widetilde{x_{1i}}$</span> is the residual from an OLS regression with <span class="math-container">$x_2$</span> as a regressor.  To proceed, you will need to make an assumption regarding <span class="math-container">$Cov(x_{1i}, e_i))$</span> and <span class="math-container">$Cov(u_{i}, e_i))$</span>.</p>

CC BY-SA 4.0
2022-09-25T08:57:53.773Z
Thanks for your answer. For some reason I am unable to comment. What about: $Cov(x_{1i}, \widetilde{x_{1i}})$?
Wooldridge
CC BY-SA 4.0
2022-09-26T07:29:22.163Z
Plug in that $x_1 = a_0 +a_1 \bar{x_2} -a_1e +\widetilde{x_1}$, then derive.
Michael Gmeiner
CC BY-SA 4.0
2022-09-22T15:30:05.927Z
2022-09-22T15:30:05.927Z
Michael Gmeiner
<p>I have a linear regression model, with usual assumptions holding; <span class="math-container">$E[xu] = 0$</span> and rank condition.</p>
<blockquote>
<p><span class="math-container">$y_i = \alpha_0 + \alpha_1x_{1i}  + \alpha_2x_{2i}  + u_i$</span></p>
</blockquote>
<p>I observe <span class="math-container">$\bar{x}_{2i}$</span>, where:</p>
<blockquote>
<p><span class="math-container">$\bar{x}_{2i} =  x_{2i} + e_i$</span></p>
</blockquote>
<p>My estimated model is:</p>
<blockquote>
<p><span class="math-container">$y_i = \tilde{\alpha_o} + \tilde{\alpha_1}x_{1i} + \tilde{\alpha_2}x_{2i} + \tilde{u_i}$</span></p>
</blockquote>
<p>I want to derive the plim of <span class="math-container">$\tilde{\alpha_1}$</span> and <span class="math-container">$\tilde{\alpha_2}$</span></p>
<p>My approach:</p>
<p>I substituted for my observation, and evaluated the following:</p>
<p>plim  <span class="math-container">$\tilde{\alpha_2} = \frac{cov(\bar{x}_{2i}, y)}{var(\bar{x}_{2i})} = \frac{\alpha_{2}*var(x_{2i})}{var(x_{2i} + e_i)} = \frac{\alpha_{2}*var(x_{2i})}{var(x_{2i}) + var(e_i)}$</span></p>
<p>plim  <span class="math-container">$\tilde{\alpha_1} = \frac{cov({x}_{1i}, y)}{var({x}_{1i})} = \frac{\alpha_{1}*var(x_{1i})}{var(x_{1i})} = \alpha_1$</span></p>
<p>Is this correct? I am a little worried about <span class="math-container">$\tilde{\alpha_1}$</span>, because shouldn't this be biased?</p>

<econometrics><regression><variance>
Measurement Error - Multivariate Case
Wooldridge
# Measurement Error - Multivariate Case

(Asked by: username_0 on 2022-09-21)

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?

**Tags:** econometrics, regression, variance

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## 1 answers

### Answer by username_1 on 2022-09-22. Score: 3 | Accepted answer

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 $Cov(u\_{i}, e\_i))$.

> **Comments:**
> **username_0**: Thanks for your answer. For some reason I am unable to comment. What about: $Cov(x_{1i}, \widetilde{x_{1i}})$?  
> **username_1**: Plug in that $x_1 = a_0 +a_1 \bar{x_2} -a_1e +\widetilde{x_1}$, then derive.  

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