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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.

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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

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

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

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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

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

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

enter image description here

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

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