<p>Adding to the excellent answer by @Alecos Papadopoulos, here are two simple numerical examples with <span class="math-container">$z=x^2+v$</span>, <span class="math-container">$z^*$</span> is mean-centred <span class="math-container">$z$</span> and <span class="math-container">$v$</span> is independent from <span class="math-container">$u$</span>, in which <span class="math-container">$E(xu)\neq 0$</span>, <span class="math-container">$E(xz^*)\neq 0$</span>, but <span class="math-container">$E(z^*u)=0$</span>.</p>

<p><strong>Example 1 (with <span class="math-container">$x$</span> having a symmetric distribution)</strong></p>

<p><a href="https://i.sstatic.net/Xcp6m.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Xcp6m.jpg" alt="enter image description here"></a></p>

<p><span class="math-container">$E(xu)=1.5$</span>, <span class="math-container">$E(xz^*)=-0.5$</span>, <span class="math-container">$E(z^*u)=0$</span></p>

<p><strong>Example 2 (showing that the distribution of <span class="math-container">$x$</span> need not be symmetric)</strong></p>

<p><a href="https://i.sstatic.net/u7VwH.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/u7VwH.jpg" alt="enter image description here"></a></p>

<p><span class="math-container">$E(xu)=2$</span>, <span class="math-container">$E(xz^*)=-3$</span>, <span class="math-container">$E(z^*u)=0$</span></p>

CC BY-SA 4.0
2020-03-08T21:39:03.703Z
2020-03-08T21:39:03.703Z
Adam Bailey
<blockquote>
  <p>Endogenous variables are correlated with the error terms and z is
  correlated with endogenous variable.</p>
  
  <p>Doesn't this imply that z is correlated with error terms?</p>
</blockquote>

<p>No it doesn't. For mean-centered variables for simplicity, we have for linear models,</p>

<p><strong>ENDOGENEITY</strong>: <span class="math-container">$E(xu) \neq 0$</span>  </p>

<p><strong>RELEVANCE</strong> : <span class="math-container">$E(xz) \neq 0$</span></p>

<p>The above conditions do <strong>not</strong> necessarily imply <strong>non</strong>-Validity (<span class="math-container">$E(zu) \neq 0$</span>).</p>

<p>A <strong>very simple example</strong> to show that we may have a valid instrument.
Assume <span class="math-container">$x, u$</span> are centered on their mean, <span class="math-container">$E(x) = E(u) = 0$</span>, and that <span class="math-container">$x$</span> has a symmetric distribution, <span class="math-container">$E(x^3) = 0$</span>. Assume that  <span class="math-container">$E(u \mid x) = ax$</span>.
Then  </p>

<p><strong>ENDOGENEITY</strong></p>

<p><span class="math-container">$$E(xu) = E\big[E(xu\mid x)\big] =  E\big[xE(u\mid x)\big] = aE(x^2) \neq 0.$$</span></p>

<p>Consider the candidate instrument <span class="math-container">$z = x^2 + v$</span>, with <span class="math-container">$v$</span> independent from <span class="math-container">$u$</span>, but <span class="math-container">$E(xv) \neq 0$</span>. We have</p>

<p><strong>RELEVANCE</strong>
<span class="math-container">$$E(xz) = E(x^3) + E(xv) = 0 + E(xv) \neq 0$$</span>.</p>

<p><strong>VALIDITY</strong></p>

<p><span class="math-container">$$E(zu) = E(x^2u) + E(vu) = E\big[E(x^2u\mid x)\big] + E(v)E(u)$$</span></p>

<p><span class="math-container">$$= E\big[x^2E(u\mid x)\big] + 0 = E\big[x^2a x\big] = aE(x^3) = 0.$$</span></p>

CC BY-SA 4.0
2023-12-13T23:53:27.440Z
For the top rated answer, how can v be independent with u but E(xv) =! 0?
Xinyue Wu
CC BY-SA 4.0
2023-12-14T03:02:37.140Z
@XinyueWu I don;t see where the problem is. Can you elaborate?
Alecos Papadopoulos
CC BY-SA 4.0
2020-03-07T20:51:31.807Z
2020-03-07T20:51:31.807Z
Alecos Papadopoulos
<p>As the title says, I have some confusions about instrumental variable. </p>

<p>According to my lecture note (which is supposed to be reliable), instrumental variable has two big properties: </p>

<ol>
<li><p>Validity: an instrumental variable and error term are not correlated and it implies that the instrumental variable does not directly lead to changes in dependent variable.</p></li>
<li><p>Relevance: an instrumental variable and endogenous variable are correlated. </p></li>
</ol>

<p>So here is my confusion. (let's denote z is an instrumental variable)</p>

<p>Endogenous variables are correlated with the error terms and z is correlated with endogenous variable. </p>

<p>Doesn't this imply that z is correlated with error terms? If so, it violates the Validity (mentioned above).</p>

<p>In case my question isn't clear enough. Here is the case where I am stuck: <span class="math-container">$$
score_i = \beta_0 + \beta_1 course_i + u_i
$$</span>
where <span class="math-container">$course_i$</span> is a binary variable of taking a preparatory course and <span class="math-container">$u_i$</span> is an error term.</p>

<p>Quite obviously, <span class="math-container">$course_i$</span> is correlated with <span class="math-container">$u_i$</span> since the parental income is one of the unobserved variables (so in <span class="math-container">$u_i$</span>) and <span class="math-container">$course_i$</span> and parental income are correlated (I think this is also quite obvious). </p>

<p>So far we know that parental income (I think this is an instrumental variable) is correlated with <span class="math-container">$course_i$</span>. But since parental income was belonged to error term, it violates Validity.</p>

<p>Any help is welcome.</p>

<econometrics><instrumental-variable><endogeneity>
Some confusions about instrumental variable (econometrics)
Math is like Friday
# Some confusions about instrumental variable (econometrics)

(Asked by: username_0 on 2020-03-07)

As the title says, I have some confusions about instrumental variable.

According to my lecture note (which is supposed to be reliable), instrumental variable has two big properties:

1. Validity: an instrumental variable and error term are not correlated and it implies that the instrumental variable does not directly lead to changes in dependent variable.
2. Relevance: an instrumental variable and endogenous variable are correlated.

So here is my confusion. (let's denote z is an instrumental variable)

Endogenous variables are correlated with the error terms and z is correlated with endogenous variable.

Doesn't this imply that z is correlated with error terms? If so, it violates the Validity (mentioned above).

In case my question isn't clear enough. Here is the case where I am stuck: $$
score\_i = \beta\_0 + \beta\_1 course\_i + u\_i
$$
where $course\_i$ is a binary variable of taking a preparatory course and $u\_i$ is an error term.

Quite obviously, $course\_i$ is correlated with $u\_i$ since the parental income is one of the unobserved variables (so in $u\_i$) and $course\_i$ and parental income are correlated (I think this is also quite obvious).

So far we know that parental income (I think this is an instrumental variable) is correlated with $course\_i$. But since parental income was belonged to error term, it violates Validity.

Any help is welcome.

**Tags:** econometrics, instrumental-variable, endogeneity

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## 2 answers

### Answer by username_2 on 2020-03-07. Score: 5 | Accepted answer

> Endogenous variables are correlated with the error terms and z is
> correlated with endogenous variable.
>
> Doesn't this imply that z is correlated with error terms?

No it doesn't. For mean-centered variables for simplicity, we have for linear models,

**ENDOGENEITY**: $E(xu) \neq 0$

**RELEVANCE** : $E(xz) \neq 0$

The above conditions do **not** necessarily imply **non**-Validity ($E(zu) \neq 0$).

A **very simple example** to show that we may have a valid instrument.
Assume $x, u$ are centered on their mean, $E(x) = E(u) = 0$, and that $x$ has a symmetric distribution, $E(x^3) = 0$. Assume that $E(u \mid x) = ax$.
Then

**ENDOGENEITY**

$$E(xu) = E\big[E(xu\mid x)\big] = E\big[xE(u\mid x)\big] = aE(x^2) \neq 0.$$

Consider the candidate instrument $z = x^2 + v$, with $v$ independent from $u$, but $E(xv) \neq 0$. We have

**RELEVANCE**
$$E(xz) = E(x^3) + E(xv) = 0 + E(xv) \neq 0$$.

**VALIDITY**

$$E(zu) = E(x^2u) + E(vu) = E\big[E(x^2u\mid x)\big] + E(v)E(u)$$

$$= E\big[x^2E(u\mid x)\big] + 0 = E\big[x^2a x\big] = aE(x^3) = 0.$$

> **Comments:**
> **username_3**: For the top rated answer, how can v be independent with u but E(xv) =! 0?  
> **username_2**: @XinyueWu I don;t see where the problem is. Can you elaborate?  

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### Answer by username_1 on 2020-03-08. Score: 1

Adding to the excellent answer by @Alecos Papadopoulos, here are two simple numerical examples with $z=x^2+v$, $z^\*$ is mean-centred $z$ and $v$ is independent from $u$, in which $E(xu)\neq 0$, $E(xz^\*)\neq 0$, but $E(z^\*u)=0$.

**Example 1 (with $x$ having a symmetric distribution)**

[![enter image description here](https://i.sstatic.net/Xcp6m.jpg)](https://i.sstatic.net/Xcp6m.jpg)

$E(xu)=1.5$, $E(xz^\*)=-0.5$, $E(z^\*u)=0$

**Example 2 (showing that the distribution of $x$ need not be symmetric)**

[![enter image description here](https://i.sstatic.net/u7VwH.jpg)](https://i.sstatic.net/u7VwH.jpg)

$E(xu)=2$, $E(xz^\*)=-3$, $E(z^\*u)=0$


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