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?

  • $\begingroup$ 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) $\endgroup$
    – Durden
    Commented Apr 24 at 15:10

1 Answer 1


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:

enter image description here

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

enter image description here

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)

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

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

              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')

enter image description here

I think it is evident, without any additional metrics, that this regression is bad.

  • $\begingroup$ 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? $\endgroup$
    – Star
    Commented Jul 31, 2022 at 10:07
  • $\begingroup$ 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 $Z\to X$. $\endgroup$
    – frank
    Commented Jul 31, 2022 at 10:28
  • $\begingroup$ How do you enstablish the quality of the regression Z on X? $\endgroup$
    – Star
    Commented Jul 31, 2022 at 10:31
  • $\begingroup$ You have to find a regression of $X$ on $Z$ (i.e. $Z\to X$), not $Z$ on $X$. $\endgroup$
    – frank
    Commented Jul 31, 2022 at 10:34
  • $\begingroup$ Yes sorry. How do you enstablish the quality of that? $\endgroup$
    – Star
    Commented Jul 31, 2022 at 10:45

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