<p>Note that the question, as you have written it, is a poor way to get you to understand the attenuation bias that arises out of classical measurement error in the independent variable. But before I get into why, let me first answer your question as it is posed. First, to simplify notation, I’m going to assume a model without a constant where <span class="math-container">$\mathbb{E}[y^*_i] = 0$</span>; this is equivalent to your model as stated. We then have</p>
<p><span class="math-container">$$ z_i = \beta_1 y^*_i + \epsilon_i \tag{1} $$</span> and
<span class="math-container">$$ z_i = \delta_1y + \mu. $$</span></p>
<p>We can then compute the following
<span class="math-container">$$
\begin{align*}
   \mathbb{E}[y \mu ] &amp;= \mathbb{E}[y(\beta_1 y^* + \epsilon - \delta_1(y^* + \gamma)] \\
&amp;= \mathbb{E}[(y^* + \gamma)((\beta_1 - \delta_1)y^* - \delta_1 \gamma + \epsilon] \\
&amp;= \mathbb{E}[(y^* + \gamma)((\beta_1 - \delta_1)y^* - \delta_1 \gamma] \\
&amp;= (\beta_1 - \delta_1)\mathbb{V}\text{ar}[y^*] - \delta_1 \mathbb{V}\text{ar}[\gamma].
\end{align*}
$$</span></p>
<p>Note that in the third and fourth equalities, we used our orthogonality assumptions. Now, we can <em>impose</em> <span class="math-container">$\delta_1 = \beta_1$</span> to recover the negative covariance that you needed. This is a little awkward, as there’s no good <em>a priori</em> reason to do impose this equality. Instead, it is better to set the above equality to zero (since that’s what the actual estimator would do in sample!) and recover that</p>
<p><span class="math-container">$$ \delta_1 = \beta_1 \frac{\mathbb{V}\text{ar}(y^*)}{\mathbb{V}\text{ar}(y^*) + \mathbb{V}\text{ar}(\gamma)} $$</span></p>
<p>which best highlights the aspect of measurement error basically crowding out the <em>signal</em> in <span class="math-container">$y^*$</span> with the noise in <span class="math-container">$\gamma$</span>.</p>

CC BY-SA 4.0
2023-02-17T23:58:33.537Z
2023-02-17T23:58:33.537Z
Yashaswi Mohanty
<p>I'm thinking about the following problem:</p>
<p>Suppose we have the equation that describes the true beta parameter for <span class="math-container">$y^{*}$</span></p>
<p><span class="math-container">$$z=\beta_{0}+\beta_{1}y^{*}+\epsilon$$</span></p>
<p>But instead, we only can only use <span class="math-container">$y$</span> to measure <span class="math-container">$y^{*}$</span> as</p>
<p><span class="math-container">$y=y^{*}+\gamma$</span></p>
<p>Then, we estimate the following equation</p>
<p><span class="math-container">$z=\beta_{0}+\beta_{1}y+\mu$</span></p>
<p>Making some assumptions (<span class="math-container">$\beta_{1}&gt;0$</span>, no correlation between <span class="math-container">$y$</span> and <span class="math-container">$\gamma$</span>, between <span class="math-container">$y$</span> and <span class="math-container">$\epsilon$</span> or between <span class="math-container">$\epsilon$</span> and <span class="math-container">$\gamma$</span>.) I have to prove that <span class="math-container">$E(y\mu)&lt;0$</span>.).</p>
<p>I tried by rewriting the last equation as :</p>
<p><span class="math-container">$z=\beta_{0}+\beta_{1}y+\epsilon+\beta_{1}\gamma$</span></p>
<p>where</p>
<p><span class="math-container">$\mu=\epsilon+\beta_{1}\gamma$</span></p>
<p>knowing that <span class="math-container">$\hat{\beta}={cov(y,z)}{v(y)}$</span> I can derive the following expression <span class="math-container">$\hat{\beta}=\beta_{1}+{cov(\gamma, z)}{v(y)}$</span>, but I'm stuck now and cant think about how to proceed to prove that <span class="math-container">$y$</span> and <span class="math-container">$\mu$</span> are negatively correlated.</p>
<p>Any hint?</p>
<p>Thanks</p>

<econometrics>
Misspecification bias
Martin 
# Misspecification bias

(Asked by: username_0 on 2023-02-11)

I'm thinking about the following problem:

Suppose we have the equation that describes the true beta parameter for $y^{\*}$

$$z=\beta\_{0}+\beta\_{1}y^{\*}+\epsilon$$

But instead, we only can only use $y$ to measure $y^{\*}$ as

$y=y^{\*}+\gamma$

Then, we estimate the following equation

$z=\beta\_{0}+\beta\_{1}y+\mu$

Making some assumptions ($\beta\_{1}>0$, no correlation between $y$ and $\gamma$, between $y$ and $\epsilon$ or between $\epsilon$ and $\gamma$.) I have to prove that $E(y\mu)<0$.).

I tried by rewriting the last equation as :

$z=\beta\_{0}+\beta\_{1}y+\epsilon+\beta\_{1}\gamma$

where

$\mu=\epsilon+\beta\_{1}\gamma$

knowing that $\hat{\beta}={cov(y,z)}{v(y)}$ I can derive the following expression $\hat{\beta}=\beta\_{1}+{cov(\gamma, z)}{v(y)}$, but I'm stuck now and cant think about how to proceed to prove that $y$ and $\mu$ are negatively correlated.

Any hint?

Thanks

**Tags:** econometrics


---

## 1 answers

### Answer by username_1 on 2023-02-17. Score: 1

Note that the question, as you have written it, is a poor way to get you to understand the attenuation bias that arises out of classical measurement error in the independent variable. But before I get into why, let me first answer your question as it is posed. First, to simplify notation, I’m going to assume a model without a constant where $\mathbb{E}[y^\*\_i] = 0$; this is equivalent to your model as stated. We then have

$$ z\_i = \beta\_1 y^\*\_i + \epsilon\_i \tag{1} $$ and
$$ z\_i = \delta\_1y + \mu. $$

We can then compute the following
$$
\begin{align\*}
\mathbb{E}[y \mu ] &= \mathbb{E}[y(\beta\_1 y^\* + \epsilon - \delta\_1(y^\* + \gamma)] \\
&= \mathbb{E}[(y^\* + \gamma)((\beta\_1 - \delta\_1)y^\* - \delta\_1 \gamma + \epsilon] \\
&= \mathbb{E}[(y^\* + \gamma)((\beta\_1 - \delta\_1)y^\* - \delta\_1 \gamma] \\
&= (\beta\_1 - \delta\_1)\mathbb{V}\text{ar}[y^\*] - \delta\_1 \mathbb{V}\text{ar}[\gamma].
\end{align\*}
$$

Note that in the third and fourth equalities, we used our orthogonality assumptions. Now, we can *impose* $\delta\_1 = \beta\_1$ to recover the negative covariance that you needed. This is a little awkward, as there’s no good *a priori* reason to do impose this equality. Instead, it is better to set the above equality to zero (since that’s what the actual estimator would do in sample!) and recover that

$$ \delta\_1 = \beta\_1 \frac{\mathbb{V}\text{ar}(y^\*)}{\mathbb{V}\text{ar}(y^\*) + \mathbb{V}\text{ar}(\gamma)} $$

which best highlights the aspect of measurement error basically crowding out the *signal* in $y^\*$ with the noise in $\gamma$.


---

