I have seen his notation to describe the Instrumental Variable framework, and I wish to make sure I understand it. Y is the dependent variable, x is treatment, and z is the instrument:

$y = f(x,\epsilon)$

$x = g(z,\eta)$

and the endogeneity structure is defined as: $cov(\epsilon,\eta)\neq0$, $cov(z,\epsilon)=0$, $cov(z,\eta)=0$

I want to make sure I understand what this is saying.

  1. First, is any variable z that can fit this an instrument?

  2. If I am say approximating these functions with linear equations, that $x = \pi z + \eta$, is this saying we can partition the entire variation of x as the variation explained by z and then all the remaining variation $\eta$, and the endogeneity can be expressed as $cov(\epsilon,\eta)\neq0$? I am confused because usually this is simply expressed as $cov(x,\epsilon)\neq0$, and I am not familiar with writing this all in terms of errors. is this the same since I can just plug in the model of x as $cov(\pi z + \eta,\epsilon) = cov(\eta,\epsilon)$ given the exogeneity of z?

  3. Is this equivalent as saying there exists some subset of variables, $r\in \epsilon$ and $r \in \eta$, i.e. omitted variables that determine x and determine y?

  • $\begingroup$ Do you have a specific reference in which the equations for describing IV are as you have defined it? Specifically, if $f, g$ are not additively separable in $\varepsilon, \eta$, it is typically not enough to only assume uncorrelatedness, $cov(z, \varepsilon) = 0$, as you do. You usually need something stronger like $z \perp \varepsilon$. $\endgroup$ Mar 30, 2021 at 19:45
  • $\begingroup$ Also, in question 3, you ask "Is this equivalent as saying...". What is the "this" referring to there? $\endgroup$ Mar 30, 2021 at 19:47
  • $\begingroup$ I was looking at this paper anderson.ucla.edu/faculty_pages/christian.dippel/IVmediate.pdf figure 1. and for question 3, I was referring to cov($\eta,\epsilon$)=/=0 in the 2nd point- sorry aboutthat $\endgroup$
    – Steve
    Mar 30, 2021 at 23:35

1 Answer 1


Before answering, let me first comment that if you want to model your instrument with a general functional form as you did above instead of in a linear model, it will not suffice to simply assume that $z$ is uncorrelated with the errors as you have done. Correlation only captures the linear relationship between two variables, so if you have a potentially non-linear model like $y = f(x, \epsilon)$, the lack of correlation between $z$ and $\varepsilon$ will not rule out cases where the instrument $z$ still is correlated with the residual in explain $y$, even though it is uncorrelated with the exact error term $\varepsilon$. You will need the stronger condition $z \perp \varepsilon$, $z\perp \eta$ in those cases where you do not have a linear model. In the paper you linked, they also make these stronger assumptions in the general (non-linear) case. With that correction in mind, let me try to address your questions one by one.

  1. While instrumental variables are often developed for trying to conduct causal inference given observational data, it is often useful for intuition to understand them in relation to an actual experiment. Consider the following stylized setting. Suppose $x$ is whether or not somebody takes a given drug, and $y$ is some health outcome. Consider a case where it is either too costly or unethical to conduct a clinical trial to assess the effectiveness of $x$. On the other hand, maybe it is cheaper/more ethical to instead subsidize people if they wanted to buy x, so $z$ is a subsidy to buy $x$. If we control who gets the subsidy, we can conduct an experiment where we vary $z$. In this case, $z \perp \varepsilon$ and $z \perp \eta$ would be by design: that's what randomization of $z$ is! On the other hand, because $\varepsilon$ represents everything else except $x$ that affects $y$ while $\eta$ represents all the other reasons why somebody might use the drug $x$ aside from the subsidy $z$, the assumption that $\eta \not\perp \varepsilon$ captures the idea that people who end up taking up the drug do so for a reason. For example, maybe people who take the are sicker to begin with.
  2. When we are working with the linear model, having assumed that $cov(z,\varepsilon) = 0$, the assumptions $cov(x,\varepsilon) \neq 0$ and $cov(\eta,\varepsilon) \neq 0$ are equivalent. The reason is a simple covariance decomposition: $cov(x, \varepsilon) = cov(\pi z + \eta, \varepsilon) = \pi cov(z,\varepsilon) + cov(\eta ,\varepsilon) = 0 + cov(\eta,\varepsilon)$
  3. I think you're close to having the right intuition here, but I would word it a bit differently. The equation $x = g(z,\eta), z \perp \eta$ is really a tautology that says "$x$ is determined by the instrument $z$ and everything else that is not $z$, which we will just call $\eta$". Similarly, the equation $y = f(x, \varepsilon), z \perp \varepsilon$ is just saying that "$y$ is determined by $x$ and everything that is not $x$, and "everything that is not $x$" is independent of our instrument $z$" (note that because of this second condition, this statement is not a tautology - it places substantive restrictions on what $z$ can be. Finally, the assumption $\eta \not\perp \varepsilon$ is just saying that $x$ on its own would be endogenous. Some of the causal factors influencing $x$ are also influencing $y$, which means that if we directly modeled the statistical relationship between $x$ and $y$, this relationship could not be interpreted as causal (hence why we need the instrument in the first place).

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