Instrumental Variables

Instrumental Variables

In order to overcome the difficulty associated with $E[X U] \neq 0$ (endogeneity). We assume that there is an additional random vector $Z$ taking values in $\mathbf{R}^{l+1}$ with $l+1 \geq k+1$ such that $E[Z U]=0$.

Any exogenous component of $X$ is contained in $Z$ (the so-called included instruments). In particular, we assume the first component of $Z$ is constant equal to one, i.e., $Z=\left(Z_0, Z_1, \ldots, Z_l\right)^{\prime}$ with $Z_0=1$.

We also assume that $E\left[Z X^{\prime}\right]<\infty, E\left[Z Z^{\prime}\right]<\infty$ and that there is no perfect collinearity in $Z$.

In summary, we assume:

1. $E[Z U]=0$: Instrument Exogeneity

2. $E\left[Z X^{\prime}\right]<\infty$

3. $E\left[Z Z^{\prime}\right]<\infty$

4. There is no perfect collinearity in $Z$

5. We further assume the rank of $E\left[Z X^{\prime}\right]$ is $k+1$. This is termed Instrument Relevance or Rank Condition.

   A necessary condition for 5 to be true is $l \geq k$. This is referred to as the Order Condition.

To further understand the IV estimator, we need to know the following:

For regression

$$Y=X^{\prime} \beta+U$$

• $X$: dimension is $(k+1) \times 1$

• $Z$: dimension is $(\ell+1) \times 1$, and $l \geqslant k$

• $Z$ only influence $Y$through $X$, not through $U$. this means that $Z$ only correlated with $X$ but not correlated with $U$.

Here is an example of using IV in real-world empirical analysis:

To analyze the education level of out-of-service military people education level's (e.g., years of schooling) relationship with proximity to the school (e.g., distance to the nearest school or a measure of school accessibility), we can choose draft lottery return status (e.g., whether an individual was likely to be drafted for military service) as an Instrument Variable. Now, we check the relevance and exogenity of the draft lottery return status as an IV.

• Relevance: The draft lottery status could affect an individual's proximity to school, as those with a higher likelihood of being drafted might choose to stay in school longer or closer to educational institutions.

• Exogeneity: The draft lottery status is presumably random and thus not correlated with the unobserved factors affecting education levels directly.

Therefore, the draft lottery return status can serve as a good IV.

Solving For Beta

Using that $U=Y-X^{\prime} \beta$ and $E[Z U]=0$, we see that $\beta$ solves the system of equations

$$E[Z Y]=E\left[Z X^{\prime}\right] \beta$$

Proof:

$E[Z Y]=E\left[Z(U+X^{\prime}\beta)\right] = 0 + E\left[Z X^{\prime}\right] \beta = E\left[Z X^{\prime}\right] \beta$

Note that the invertible of $E\left[Z X^{\prime}\right]$ is not guaranteed. This is because since $l+1 \geq k+1$, this may be an over-determined system of equations. There is more information than we need. Which can be shown in the following:

$$Z=\left(\begin{array}{c} 1 \\ Z_1 \\ \vdots \\ Z_l \end{array}\right)_{(l+1) \times 1} \quad X=\left(\begin{array}{c} 1 \\ X_1 \\ \vdots \\ X_k \end{array}\right)_{(k + 1) \times 1} \quad \beta=\left(\begin{array}{c} 1\\ \beta_1 \\ \vdots \\ \beta_k \end{array}\right)_{(k+1) \times 1}$$

Therefore, in order to solve for $\beta$, we introduce the following lemma:

Lemma

Suppose there is no perfect collinearity in $Z$ and let $\Pi$ be such that $B L P(X \mid Z)=\Pi^{\prime} Z . E\left[Z X^{\prime}\right]$ has rank $k+1$ if and only if $\Pi$ has rank $k+1$. Moreover, the matrix $\Pi^{\prime} E\left[Z X^{\prime}\right]$ is invertible.

Note that if some $X_j$ are exogenous, then we do not need IVs for them.

Solve for Beta

As $\beta \text { solves: } E[Z Y]=E\left[Z X^{\prime}\right] \beta \text { or } \Pi^{\prime} E[Z Y]=\Pi^{\prime} E\left[Z X^{\prime}\right]\beta$, then we can have that using the previous lemma and $\Pi=E\left[Z Z^{\prime}\right]^{-1} E\left[Z X^{\prime}\right]$, we can derive three formulae for $\beta$

Since we have $X=\left(\begin{array}{c}1 \\ X_1 \\ \vdots \\ X_k\end{array}\right) \quad Z=\left(\begin{array}{c}1 \\ Z \\ \vdots \\ Z_l\end{array}\right)$, and $\Pi=\left[\Pi_0, \Pi_1, \cdots, \Pi_k\right]$, the shape of $\Pi$ is $(l+1)\times(k+1)$.

First regression $X$ on instrument variable $Z$:

$$\begin{aligned} X_0&=Z^{\prime} \Pi_0+U_0=1 \\ X_1&=Z^{\prime} \Pi_1+U_1 \\ X_2&=Z^{\prime} \Pi_2+U_2 \\ \vdots \\ X_k&=Z^{\prime} \Pi_k+U_k \end{aligned}$$

Then we can get the BLP of $\Pi_i$s from these equations:

$$\begin{aligned} & \Pi_1=\left(\mathbb{E} Z Z^{\prime}\right)^{-1} \mathbb{E}\left[Z X_1\right] \\ & \Pi_2=\left(\mathbb{E} Z Z^{\prime}\right)^{-1} \mathbb{E}\left[Z X_2\right] . \\ & \vdots \\ & \Pi_k=\left(\mathbb{E} Z Z^{\prime}\right)^{-1} \mathbb{E}\left[Z X_k\right] \end{aligned}$$

Note that if $X_1$ is exogenous, then $Z_1=X_1$, projection $X_1$ on $Z$, we can have that,

$$\Pi_1=\left(\begin{array}{c} 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{array}\right)\text{ where }\beta_1=1$$

This is because, in $X_1$'s regression on $Z$, we can have that $X_1=\beta_0+\beta_1 X_1+\beta_2 Z_2+\cdots+U$, in this regression model, $X_1$ will perfectly explain itself.

Equation 1: IV estination

By $\Pi^{\prime} E[Z Y]=\Pi^{\prime} E\left[Z X^{\prime}\right] \beta$, we have that,

$$\beta=\left[\Pi^{\prime} E\left(Z X^{\prime}\right)\right]^{-1} \Pi^{\prime} E[Z Y]$$

As $\Pi=E\left[Z Z^{\prime}\right]^{-1} E\left[Z X^{\prime}\right]$, and if $l=k$, we can have that:

$$\begin{aligned} \beta & =\left[E(Z X^{\prime})^{\prime}\left(E\left(Z Z^{\prime}\right)\right)^{-1} E(Z X^{\prime})\right]^{-1} E\left(Z X^{\prime}\right)^{\prime} (E(Z Z^{\prime}))^{-1} E(ZY) \\ & =E\left(Z X^{\prime}\right)^{-1} E\left(Z Z^{\prime}\right)\left[E\left(Z X^{\prime}\right)^{\prime}\right]^{-1}E\left(Z X^{\prime}\right)^{\prime} E\left(Z Z^{\prime}\right)^{-1} E (Z Y) \\ & =E\left(Z X^{\prime}\right)^{-1} E (Z Y) \end{aligned}$$

Equation 2: TSLS Version 1

As we can have that:

$$X=\left(\begin{array}{c} 1\\ X_1 \\ \vdots \\ X_k \end{array}\right)=\left(\begin{array}{c} Z^{\prime} \Pi_0 \\ Z^{\prime} \Pi_1 \\ \vdots \\ Z^{\prime} \Pi_k \end{array}\right)+\left(\begin{array}{c} U_0 \\ U_1 \\ \vdots \\ U_k \end{array}\right)=\Pi^{\prime} Z+U .$$

As $\Pi^{\prime}Z$ is the BLP of $X$, we can have that $E[U]=0 \text{ and } E[ZU]=0$. Take this into the formula $\beta=\left[\Pi^{\prime} E\left(Z X^{\prime}\right)\right]^{-1} \Pi^{\prime} E[Z Y]$, we can have that.

$$\begin{aligned} \beta & =\left[\Pi^{\prime} E\left[Z\left(\Pi^{\prime} Z+U\right)^{\prime}\right]\right]^{-1} \Pi^{\prime} E[Z Y] \\ & =\left[\Pi^{\prime} E\left[Z Z^{\prime}\right] \Pi+\Pi^{\prime}E[ZU]\right]^{-1} \Pi^{\prime} E[Z Y] \\ & =\left[\Pi^{\prime} E\left[Z Z^{\prime}\right] \Pi\right]^{-1} \Pi^{\prime} E[Z Y] \end{aligned}$$

Equation 3: TSLS Version 2

As $\Pi^{\prime}$ is a matrix of real numbers, we can put it in the expectation. So $\beta$ can be rewritten as

$$\beta=\left[\mathbb{E}\left[\left(\Pi^{\prime} Z\right)\left(\Pi^{\prime} Z\right)^{\prime}\right]\right]^{-1} \mathbb{E}\left[\left(\Pi^{\prime} Z\right) Y\right]$$

We can denote that $W = \Pi^{\prime}Z$, which stands for the linear combination of instruments. Take this into the formula of $\beta$, we have

$$\beta=\left(E\left[W W^{\prime}\right]\right)^{-1} E[W Y]$$

Interpreting The Rank Condition (Instrument Relevance)

• The rank condition for IV estimation is a technical requirement that ensures the IV or set of IVs provides enough information to identify the model.

• Essentially, it requires that the matrix of instruments $Z$ should have sufficient rank so that the projection matrix $Π$ adequately captures the relevant information in the endogenous variables.

Interpretation: Consider the case where $k=l$ and only $X_k$ is endogenous. Let $Z_j=X_j$ for all $0 \leq j \leq k-1$. In this case,

$$\Pi^{\prime}=\left(\begin{array}{ccccc} 1 & 0 & \ldots & 0 & 0 \\ 0 & 1 & \ldots & 0 & 0 \\ \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & \ldots & 1 & 0 \\ \pi_0 & \pi_1 & \ldots & \pi_{l-1} & \pi_l \end{array}\right)$$

The rank condition therefore requires $\pi_l \neq 0$ : the instrument $Z_l$ must be "correlated with $X_k$ after controlling for $X_0, X_1, \ldots, X_{k-1}$."

1. Strong IV:

   • A strong IV is highly correlated with the endogenous explanatory variable.

   • This strong correlation ensures that the IV effectively captures the variation in the endogenous variable that is not related to the error term in the regression model.

   • Strong IVs lead to more reliable and precise estimates in IV regression.

2. Weak IV:

   • A weak IV has a weak correlation with the endogenous explanatory variable.

   • This weak correlation means that the IV does not effectively capture the variation in the endogenous variable, making it less effective in dealing with endogeneity.

   • Weak IVs can lead to biased estimates and poor inference because they do not provide a good substitute for the endogenous variable.

In our scenario,

• If $π_l​$ is close to zero, $Z_l​$ is a weak IV because it does not add much explanatory power beyond what is already captured by $X_0,X_1,…,X_{k−1}$.

• If $π_l​$ is significantly different from zero, $Z_l​$ is considered a strong IV, as it is meaningfully correlated with the endogenous variable $X_k​$, independent of the other explanatory variables.

• If $Π$ is Near-Singularity, we can have that this is a weak IV estimator that doesn't explain $X$ well. This weak IV estimator problem will lead to:

  • Large Variance: When the instrument is weak, the variance of the IV estimator becomes very large. This leads to wide confidence intervals, making it difficult to draw precise inferences about the parameters.

  • Biased Estimation: In small samples, a weak instrument can lead to biased estimates, and these biases can be as bad or even worse than the OLS estimates that suffer from endogeneity.

  • Inconsistent Estimation: In theory, the IV estimator is consistent as the sample size approaches infinity. However, with a weak instrument, the convergence to the true parameter value can be very slow, leading to practical issues in estimation, even with large samples.

In summary, if the $Z$ doesn't satisfy the rank condition (relevance condition), here are the consequences:

• Weak Instrument Problem: If the IV is weakly correlated (or not correlated) with the endogenous variable, it results in a weak instrument problem. This can lead to biased and inconsistent estimates, similar to or even worse than the original OLS estimates that were affected by endogeneity.

• Inefficiency: The estimates may also have large standard errors, leading to inefficiency and making it difficult to draw reliable inferences.

• Identification Issue: In the extreme case where there is no correlation at all, the model becomes unidentified, meaning that you cannot reliably estimate the coefficients of the endogenous variables.

Interpreting The Exogeneity Condition

Consequence of Violation:

• Biased and Inconsistent Estimates: If the IV is correlated with the error term, the estimates will be biased and inconsistent. This is because the instrument is not isolating only the exogenous variation in the endogenous explanatory variable—it's also capturing some of the effects that should be in the error term.

• Invalid Inferences: Any inferences made about the effect of the endogenous explanatory variable on the dependent variable would be invalid because they are contaminated by the correlation with the error term.

Partition of Beta: Endogenous Components

Note that the IV can also be the variable itself, in following regression:

$$Y=X_1^{\prime} \beta_1+X_2^{\prime} \beta_2+U$$

• If $E\left[X_1 U\right] \neq 0$: we can choose to find an IV $Z_1$ for $X_1$, such that $E[Z_1 U]=0$

• If $E\left[X_2 U\right]=0$: $X_2$ itself can be view as an IV for $X_2$, such as $Z_2 = X_2$

Here, we partition $X$ into $X_1$ and $X_2$, where $X_2$ is exogenous. Partition $Z$ into $Z_1$ and $Z_2$ and $\beta$ into $\beta_1$ and $\beta_2$ analogously.

We have that, in this model:

• $Z_2=X_2$ are included instruments

• $Z_1$ are excluded instruments

We can conveniently re-write this by projecting $(B L P)$ on $Z_2=X_2$. Consider the case $k=l$

$$B L P\left(Y \mid Z_2\right)=B L P\left(X_1 \mid Z_2\right)^{\prime} \beta_1+X_2^{\prime} \beta_2 .$$

Define $Y^*=Y-B L P\left(Y \mid Z_2\right)$ and $X_1^*=X_1-B L P\left(X_1 \mid Z_2\right)$ so that

$$E\left[Z_1 Y^*\right]=E\left[Z_1 X_1^{* \prime}\right] \beta_1+E\left[Z_1 U\right]$$

It follows that

$$\beta_1=\left(E\left[Z_1 X_1^{* \prime}\right]\right)^{-1} E\left[Z_1 Y^*\right]$$

IV Estimator

The IV estimator is used under the case that number of instrument variables is equal to the number of explanatory variables. Means using it in the following condition:

$$\text { Just-identified case: } l=k$$

Denote $P$ the marginal distribution of $(Y, X, Z)$. Let $\left(Y_1, X_1, Z_1\right), \ldots,\left(Y_n, X_n, Z_n\right)$ be an i.i.d. sequence of random variables with distribution $P$.

By analogy with $\beta=\left(E\left[Z X^{\prime}\right]\right)^{-1} E[Z Y]$, the natural estimator of $\beta$ is simply

$$\hat{\beta}=\left(\frac{1}{n} \sum_i Z_i X_i^{\prime}\right)^{-1}\left(\frac{1}{n} \sum_i Z_i Y_i\right) .$$

This estimator is called the instrumental variables (IV) estimator of $\beta$. Note that $\hat{\beta}$ satisfies

$$\frac{1}{n} \sum_i Z_i\left(Y_i-X_i^{\prime} \hat{\beta}\right)=0$$

In particular, $\hat{U}_i=Y_i-X_i^{\prime} \hat{\beta}$ satisfies

$$\frac{1}{n} \sum_i Z_i \hat{U}_i=0 .$$

Insight on the IV estimator: assume $X_0=1$ and $X_1 \in \mathbf{R}$. An interesting interpretation of the IV estimator $\beta_1$ is obtained by multiplying and dividing by $\frac{1}{n} \sum_{i=1}^n\left(Z_{1, i}-\bar{Z}_{1, n}\right)^2$, i.e.,

$$\begin{aligned} \hat{\beta}_1 & =\frac{\frac{1}{n} \sum_{i=1}^n\left(Z_{1, i}-\bar{Z}_{1, n}\right) Y_i / \frac{1}{n} \sum_{i=1}^n\left(Z_{1, i}-\bar{Z}_{1, n}\right)^2}{\frac{1}{n} \sum_{i=1}^n\left(Z_{1, i}-\bar{Z}_{1, n}\right) X_{1, i} / \frac{1}{n} \sum_{i=1}^n\left(Z_{1, i}-\bar{Z}_{1, n}\right)^2} \\ & =\frac{\text { slope of } Y \text { on } Z}{\text { slope of } X \text { on } Z} . \end{aligned}$$

Matrix Notation

This estimator may be expressed more compactly using matrix notation. Define

$$\begin{aligned} & \mathbb{Z}=\left(Z_1, \ldots, Z_n\right)^{\prime} \\ & \mathbb{X}=\left(X_1, \ldots, X_n\right)^{\prime} \\ & \mathbb{Y}=\left(Y_1, \ldots, Y_n\right)^{\prime} \end{aligned}$$

In this notation, we have

$$\hat{\beta}=\left(\mathbb{Z}^{\prime} \mathbb{X}\right)^{-1}\left(\mathbb{Z}^{\prime} \mathbb{Y}\right) .$$