# Construction of instrumental variables

Consider the linear regression model $$Y_i=X_i^\top \beta+U_i.$$

Suppose some regressors are not orthogonal to $$U_i$$, i.e., $$E(X_i U_i)\neq 0$$. Then, the OLS estimator is not consistent (Hayashi, chapter 2). The usual way to proceed consists of finding instruments $$Z_i$$ such that $$E(Z_i U_i)=0$$ and such instruments should explain some variations in $$X_i$$.

The answer to the question here suggests that, although rare and perhaps tricky, we might be able to construct instruments just by taking appropriate transformations of $$X_i$$. For instance, take $$X_i$$ scalar with $$E(X_i U_i)\neq 0$$. Consider the function $$f(X_i)=X_i^k$$ with $$k$$ even. Suppose $$(X_i,U_i)$$ are symmetric around zero. Then, even if $$E(X_iU_i)\neq 0$$, it holds that $$E(X^k_i U_i)=0.$$ Hence, we could set $$Z_i\equiv X^k_i$$. However, I've never found such a discussion in any texbook. Hence, I wonder whether there is something fundamental I'm missing here. Perhaps, $$Z_i$$ so defined would be a very weak instrument? Could you help me understand?

• Any polynomial of an endogenous regressor is still going to be endogenous, so that's not a winning strategy. The types of transformations you might be thinking of are more involved: one currently en vogue synthetic IV method is the Bartik shift-share instrument. For a graduate-level introduction, see Goldsmith-Pinkham et al. (2020) Commented Apr 24 at 15:10

The method of instrumental variables chooses a variable $$Z$$ (that satisfies certain conditions), uses it to fit a regression of $$X$$ w.r.t. $$Z$$, let's call it $$\hat X(Z)$$, and then uses $$\hat X(Z)$$ to get an estimate of the effect of $$X$$ on $$Y$$.

You suggest $$Z=X^k$$ with $$k$$ being even, e.g. $$k=2$$. That means we have to regress $$X$$ on $$X^2$$. This would be a very bad regression because for each $$X^2$$ there are two possible values $$X$$. This in turn would result in the method of instrumental variables not working well.

Also, for $$Z$$ to be usable as instrumental variable, certain conditions have to be satisfied. In particular: $$(Z \perp\kern-5pt\perp Y)_{G_{\bar X}},$$ meaning that in the graph $$G_{\bar X}$$, which is the original graph $$G$$ with all the arrows going into $$X$$ being cut off, there must not be any d-separation open path between $$Z$$ and $$Y$$. But the original graph $$G$$ looks like this:

and $$G_{\bar X}$$ then is:

and the path $$X^2 \leftarrow X \to Y$$ is clearly an open path. Thus, this condition for $$X^2$$ being an instrumental variable is violated.

The regression of $$X$$ by $$X^2$$:

> x <- -10:10
> df <- data.frame(x = x, z = x^2)
> m <- lm(x ~ z, data = df)
> summary(m)

Call:
lm(formula = x ~ z, data = df)

Residuals:
Min     1Q Median     3Q    Max
-10     -5      0      5     10

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)  8.697e-16  2.088e+00       0        1
z           -2.372e-17  4.250e-02       0        1

Residual standard error: 6.366 on 19 degrees of freedom
Multiple R-squared:  2.005e-32, Adjusted R-squared:  -0.05263
F-statistic: 3.81e-31 on 1 and 19 DF,  p-value: 1

> plot(z, x, panel.first = grid())
> lines(z, m$fitted.values, col = 'red') >  I think it is evident, without any additional metrics, that this regression is bad. • About regressing$X$on$X^2$, when you write "This in turn would result in the method of instrumental variables not working well.": formally, which property would it be violated? – Star Commented Jul 31, 2022 at 10:07 • Even if all the requirements for the method of instrumental variables are satisfied, of course, the quality of the result still depends on the quality of the regression$Z\to X$. Commented Jul 31, 2022 at 10:28 • How do you enstablish the quality of the regression Z on X? – Star Commented Jul 31, 2022 at 10:31 • You have to find a regression of$X$on$Z$(i.e.$Z\to X$), not$Z$on$X\$. Commented Jul 31, 2022 at 10:34
• Yes sorry. How do you enstablish the quality of that?
– Star
Commented Jul 31, 2022 at 10:45