<p>Virtually every social science scenario qualifies, in fact it is much harder to find cases when it does not apply. Consider classic examples:</p>
<p>Education allows people to earn higher income. However, people with higher income can more afford to get educated.  This results in <span class="math-container">$E[\epsilon_1,\epsilon_2] \neq 0$</span> since higher income causes education but education causes higher income.</p>
<p>Or for example consider spending on policing. Higher spending on policing should reduce crime. However, countries with higher crime need to spend more on policing. More policing causes lower crime, but higher crime causes more policing.</p>
<p>The two above are just classic examples, virtually every social science problem will have more or less obvious violation of  <span class="math-container">$E[\epsilon_1,\epsilon_2] \neq 0$</span>. It is extremely rare to find social science problems where this is not a case.</p>
<hr />
<p>Mathematically the <span class="math-container">$E[\epsilon_1,\epsilon_2] \neq 0$</span> because of the following:</p>
<p>Suppose we have reverse causality so we have structural model given by:</p>
<p><span class="math-container">$$y_i= \beta_1 x_i+ \gamma_1z_i+\epsilon_1 \\ z_i=\beta_2x_i+\gamma_2y_i+\epsilon_2 $$</span></p>
<p>Now in the first eq by assumption <span class="math-container">$E(z_{i}u_{i})\neq 0$</span>. If we now substitute second equation into first one and solve for <span class="math-container">$z$</span>  assuming that <span class="math-container">$1-\gamma_{1}\gamma_{2}\neq 0$</span> gets us</p>
<p><span class="math-container">$$z_{i}={\frac  {\beta _{2}+\gamma _{2}\beta _{1}}{1-\gamma _{1}\gamma _{2}}}x_{i}+{\frac  {1}{1-\gamma _{1}\gamma _{2}}}\epsilon_1+\frac{\gamma _{2}}{1-\gamma_1\gamma_2}\epsilon_2$$</span></p>
<p>Assuming <span class="math-container">$x$</span> and <span class="math-container">$\gamma$</span> are uncorrelated then we get:</p>
<p><span class="math-container">$$E[z, \epsilon_2]={\frac {\gamma _{2}}{1-\gamma _{1}\gamma _{2}}}\operatorname {E} (\epsilon_1,\epsilon_2)\neq 0 \implies E(\epsilon_1,\epsilon_2)\neq 0 $$</span></p>

CC BY-SA 4.0
2022-10-20T21:58:33.717Z
Thinking if Y1 as education and Y2 as education, why would this imply the errors are corrected? We want E([Y2,W']eps1) = [E(Y2eps1),0], we subsitute in E((W+eps2)eps1) = E(eps2eps1) but how exactly does that lead to this term being non-zero? Which part of eps1 is correlated with which part of eps2? (For example, in OVB we have E(Xeps)=E(X(W+Eps2)) = E(XW) where XW is a nonzero vector because they're correlated) Having a hard time seeing something like this with E(eps1eps2)
financial_physician
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2022-10-20T22:05:10.337Z
@financial_physician oh I thought you are just looking for intuition because you understand the math. Mathematically if income (y) causes education (e) and education causes income we can write $y=a +be + \epsilon_y$ and $e=c + dy + \epsilon_e$ with b and d $\neq 0 \implies E[\epsilon_y, \epsilon_e] \neq 0$. I mean how in detail you want me to go in here? This is the same thing you have in your opening question just simplified, if you want I can add to my answer the whole math but that in itself is not intuitive
1muflon1
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2022-10-20T22:50:48.170Z
If it's not too much hassle, would really appreciate it. I googled a proof earlier and came across [this pdf](https://www.montana.edu/cstoddard/562/Chapter16studentnotes.pdf) but it wasn't satisfying for me because (using the author's notation) the author writes [obvious y2 corr with u1] but if u1 is noise (random) than this doesn't make sense. It can't be noise if they are correlated haha.

I've edited my question to better explain my current understanding which might be very wrong.
financial_physician
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2022-10-21T23:26:34.117Z
if I should open a new question, I'm happy to accept yours and open a new one
financial_physician
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2022-10-21T23:27:20.330Z
@financial_physician hi I am planning to write extension it’s just lot of math and I was very busy at the university today. If you don’t want to wait you can open it as new question
1muflon1
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2022-10-21T23:27:42.807Z
No problem :) Take your time
financial_physician
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2022-10-23T23:16:44.247Z
@financial_physician I have updated the answer
1muflon1
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2022-10-20T20:43:39.647Z
2022-10-23T23:10:01.763Z
2022-10-23T23:10:01.763Z
1muflon1
<p>I'm reading through Asymptotic Theory for Econometricians by Halbert White and am trying to figure out the intuition for situations where the errors would be correlated.</p>
<p>Here's the setup:</p>
<p><span class="math-container">$Y_{t1} = Y_{t2}\alpha_o + \pmb W_{t1}^T \pmb \delta_o + \epsilon_{t1}, E(\pmb W_{t1}\epsilon_{t1}) = 0$</span></p>
<p><span class="math-container">$Y_{t2} = \pmb W_{t2}^T\pmb \gamma_o + \epsilon_{t2}, E(\pmb W_{t2}\epsilon_{t2}) = 0$</span></p>
<p>There's a couple other assumptions made but the one I'm having a hard time reasoning about is <span class="math-container">$E(\epsilon_{t1}\epsilon_{t2}) \neq 0$</span>. What would be some intuitive scenarios for when this would be expected?</p>
<hr />
<p>Edit:</p>
<p>Using a univariate framework, here's how I currently understand simultaneity (could be dead wrong):</p>
<p>Assume the model <span class="math-container">$y_1 = xb + \epsilon$</span>. We already know that
<span class="math-container">\begin{align}
\frac{cov(x,y_1)}{var(x)} = \hat b
\end{align}</span></p>
<p>Substituting in the equation for <span class="math-container">$y_1$</span> and simplifying we get that <span class="math-container">$b = \hat b$</span>.</p>
<p>Now assume that we have an issue with simultaneity. In particular, we discover the model should be
<span class="math-container">$y_1 = y_2a + xb + \epsilon$</span> where <span class="math-container">$y_2 = x b_2 + \epsilon_2$</span>.</p>
<p>Then what our erroneous <span class="math-container">$\hat b$</span> is actually equal to is:</p>
<p><span class="math-container">\begin{align}
\hat b = \frac{cov(x,y_1)}{var(x)} = 
\frac{cov(x,y_2a + xb + \epsilon_2)}{var(x)}
= \frac{a\ cov(y_2,x)}{var(x)} + b
\neq b 
\end{align}</span></p>
<p>which is a pretty simple bias term to me:</p>
<p><span class="math-container">\begin{align}
\text{bias} &amp;= \frac{a\ cov(y_2,x)}{var(x)}\\ &amp;= \frac{a\ cov(x b_2 + \epsilon_2,x)}{var(x)}
\\&amp; = a b_2
\end{align}</span></p>
<p>But this argument for simultaneity doesn't look like <span class="math-container">$cov(\epsilon_1, \epsilon_2) \neq 0$</span> to me. It really just looks like OVB haha</p>

<econometrics><endogeneity><simultaneous-equation-model>
Simultaneity intuition for correlated error terms
financial_physician
# Simultaneity intuition for correlated error terms

(Asked by: username_0 on 2022-10-20)

I'm reading through Asymptotic Theory for Econometricians by Halbert White and am trying to figure out the intuition for situations where the errors would be correlated.

Here's the setup:

$Y\_{t1} = Y\_{t2}\alpha\_o + \pmb W\_{t1}^T \pmb \delta\_o + \epsilon\_{t1}, E(\pmb W\_{t1}\epsilon\_{t1}) = 0$

$Y\_{t2} = \pmb W\_{t2}^T\pmb \gamma\_o + \epsilon\_{t2}, E(\pmb W\_{t2}\epsilon\_{t2}) = 0$

There's a couple other assumptions made but the one I'm having a hard time reasoning about is $E(\epsilon\_{t1}\epsilon\_{t2}) \neq 0$. What would be some intuitive scenarios for when this would be expected?

---

Edit:

Using a univariate framework, here's how I currently understand simultaneity (could be dead wrong):

Assume the model $y\_1 = xb + \epsilon$. We already know that
\begin{align}
\frac{cov(x,y\_1)}{var(x)} = \hat b
\end{align}

Substituting in the equation for $y\_1$ and simplifying we get that $b = \hat b$.

Now assume that we have an issue with simultaneity. In particular, we discover the model should be
$y\_1 = y\_2a + xb + \epsilon$ where $y\_2 = x b\_2 + \epsilon\_2$.

Then what our erroneous $\hat b$ is actually equal to is:

\begin{align}
\hat b = \frac{cov(x,y\_1)}{var(x)} =
\frac{cov(x,y\_2a + xb + \epsilon\_2)}{var(x)}
= \frac{a\ cov(y\_2,x)}{var(x)} + b
\neq b
\end{align}

which is a pretty simple bias term to me:

\begin{align}
\text{bias} &= \frac{a\ cov(y\_2,x)}{var(x)}\\ &= \frac{a\ cov(x b\_2 + \epsilon\_2,x)}{var(x)}
\\& = a b\_2
\end{align}

But this argument for simultaneity doesn't look like $cov(\epsilon\_1, \epsilon\_2) \neq 0$ to me. It really just looks like OVB haha

**Tags:** econometrics, endogeneity, simultaneous-equation-model

---

## 1 answers

### Answer by username_1 on 2022-10-20. Score: 1 | Accepted answer

Virtually every social science scenario qualifies, in fact it is much harder to find cases when it does not apply. Consider classic examples:

Education allows people to earn higher income. However, people with higher income can more afford to get educated. This results in $E[\epsilon\_1,\epsilon\_2] \neq 0$ since higher income causes education but education causes higher income.

Or for example consider spending on policing. Higher spending on policing should reduce crime. However, countries with higher crime need to spend more on policing. More policing causes lower crime, but higher crime causes more policing.

The two above are just classic examples, virtually every social science problem will have more or less obvious violation of $E[\epsilon\_1,\epsilon\_2] \neq 0$. It is extremely rare to find social science problems where this is not a case.

---

Mathematically the $E[\epsilon\_1,\epsilon\_2] \neq 0$ because of the following:

Suppose we have reverse causality so we have structural model given by:

$$y\_i= \beta\_1 x\_i+ \gamma\_1z\_i+\epsilon\_1 \\ z\_i=\beta\_2x\_i+\gamma\_2y\_i+\epsilon\_2 $$

Now in the first eq by assumption $E(z\_{i}u\_{i})\neq 0$. If we now substitute second equation into first one and solve for $z$ assuming that $1-\gamma\_{1}\gamma\_{2}\neq 0$ gets us

$$z\_{i}={\frac {\beta \_{2}+\gamma \_{2}\beta \_{1}}{1-\gamma \_{1}\gamma \_{2}}}x\_{i}+{\frac {1}{1-\gamma \_{1}\gamma \_{2}}}\epsilon\_1+\frac{\gamma \_{2}}{1-\gamma\_1\gamma\_2}\epsilon\_2$$

Assuming $x$ and $\gamma$ are uncorrelated then we get:

$$E[z, \epsilon\_2]={\frac {\gamma \_{2}}{1-\gamma \_{1}\gamma \_{2}}}\operatorname {E} (\epsilon\_1,\epsilon\_2)\neq 0 \implies E(\epsilon\_1,\epsilon\_2)\neq 0 $$

> **Comments:**
> **username_0**: Thinking if Y1 as education and Y2 as education, why would this imply the errors are corrected? We want E([Y2,W']eps1) = [E(Y2eps1),0], we subsitute in E((W+eps2)eps1) = E(eps2eps1) but how exactly does that lead to this term being non-zero? Which part of eps1 is correlated with which part of eps2? (For example, in OVB we have E(Xeps)=E(X(W+Eps2)) = E(XW) where XW is a nonzero vector because they're correlated) Having a hard time seeing something like this with E(eps1eps2)  
> **username_1**: @financial_physician oh I thought you are just looking for intuition because you understand the math. Mathematically if income (y) causes education (e) and education causes income we can write $y=a +be + \epsilon_y$ and $e=c + dy + \epsilon_e$ with b and d $\neq 0 \implies E[\epsilon_y, \epsilon_e] \neq 0$. I mean how in detail you want me to go in here? This is the same thing you have in your opening question just simplified, if you want I can add to my answer the whole math but that in itself is not intuitive  
> **username_0**: If it's not too much hassle, would really appreciate it. I googled a proof earlier and came across [this pdf](https://www.montana.edu/cstoddard/562/Chapter16studentnotes.pdf) but it wasn't satisfying for me because (using the author's notation) the author writes [obvious y2 corr with u1] but if u1 is noise (random) than this doesn't make sense. It can't be noise if they are correlated haha.  I've edited my question to better explain my current understanding which might be very wrong.  
> **username_0**: if I should open a new question, I'm happy to accept yours and open a new one  
> **username_1**: @financial_physician hi I am planning to write extension it’s just lot of math and I was very busy at the university today. If you don’t want to wait you can open it as new question  
> **username_0**: No problem :) Take your time  
> **username_1**: @financial_physician I have updated the answer  

---

