# Does an instrumental variable require independence between the instrument and treatment?

An instrumental variable Z is a legitimate instrument for T (the treatment) if the following hold:

1. Z has a causal effect on Y that is fully mediated by T (i.e. no direct effect from Z to Y, and the flow of causation must at least run through T)
2. Z has no backdoor paths to Y.
3. Monotonicity (higher values in the instrument lead to higher values of the treatment being uptaken)
4. A 1st stage actually exists (there is a nonzero correlation between Z and T).

What happens, however, if we have the following DAG?

C is a confounding variable between the treatment and instrument. UC is a set of unobserved confounders. Is Z still a legitimate instrument for T? I was under the impression that the answer is no, but according to https://dagitty.net/dags.html#

...both C and Z are legitimate instruments for T. I can see C being a legitimate instrument since all causal pathways run through T (either directly or mediated by Z). How can Z still be an instrument for T without conditioning on C first, however? I can jump to T through C. Most sources seem to suggest that T and Z must be d-separated, but just wanted to verify this.

• Naturally, DAGs can be much more complicated than this, but for this particular DAG, no fancy anything appears to be required: there's no confounding between $T$ and $Y.$ Commented May 26 at 2:09
• @AdrianKeister Ah, I forgot to include the confounders between T and Y. Let me fix that. Commented May 26 at 2:22
• I try to get a mental picture: Y is a patient's blood pressure, T is the medication dose, and Z is the reading of the blood pressure monitor (sphygmomanometer)? If so, what is C? And UC? Commented May 27 at 6:02
• Besides the discussion already provided, isn't your question's title misleading? It seems to be more about 4., i.e. whether a first stage exists Commented May 27 at 8:50

Edit: I maybe take back the answer below and refer readers to Levis, Kennedy and Keele (2024), who discuss identification of IV effects in great detail. In the article, the assumption in question is assumption 2(b) and $$X$$ is empty. Under some assumptions, the IV estimate can be computed without assuming 2(b), but under others, it is required.

You'll note that $$Z$$ still satisfies 1) and 2) even when its relationship with the treatment is confounded. $$T$$ still fully mediates the relationship between $$Z$$ and $$Y$$; that is, conditional on $$T$$, there is no association between $$Z$$ and $$Y$$. And there are no backdoor paths between $$Z$$ and $$Y$$ (again, conditional on $$T$$). So, yes, $$Z$$ is still a valid instrument. The reason we usually like $$Z$$ to be randomized is because without doing so, it is possible for there to be a common cause of $$Z$$ and $$Y$$, which would violate 2). But it is fine for there to be a common cause of $$Z$$ and $$T$$. If you can find a source that suggests $$Z$$ and $$T$$ must be d-separated, please include it.

• +1. Maybe another angle on basically the same point: while confounding in the first stage will imply that the first stage will not estimate "true causal" relationship between $Z$ and $T$, that is no problem because that is also not what we are after. Commented May 27 at 7:52
• One question/comment...conditional on T, there would be an association between Z and Y through the unobserved confounders no? Commented May 27 at 17:16
• Here are two (unofficial, mind you) sources that state d-separation between T and Z: donskerclass.github.io/EconometricsII/ControlandIV.html Commented May 27 at 17:26
• math.mcgill.ca/dstephens/InstrumentalVariables-Full.pdf Commented May 27 at 17:27
• @ChristophHanck there may be a typo there, sorry! I would recommend reading the paper. It is a bit technical but the sections about the assumptions is relatively straightforward.
– Noah
Commented May 27 at 18:51

Here is an illustration for a linear and homogenous (i.e., No treatment effect heterogeneity, which may be relevant, see the answer by Noah) case in which the treatment is generated by correlated $$C$$ and $$Z$$ and yet, only $$Z$$ is used to instrument the treatment $$Tr$$:

library(mvtnorm)
library(ivreg)

n <- 100
theta.Z <- 1
theta.C <- 1.5
beta <- 2

ZC <- rmvnorm(n, sigma = matrix(c(1, .5, .5, 1), ncol=2))
errors <- rmvnorm(n, sigma = matrix(c(1, .75, .75, 1), ncol=2)) # correlated, so that OLS of Y on Tr would not be consistent

Z <- ZC[, 1] # included instrument
C <- ZC[, 2] # the "ignored, first-stage confounding" instrument

Tr <- theta.Z*Z + theta.C*C + errors[, 2]
Y <- Tr*beta + errors[, 1]  # the "structural" equation

ivreg(Y~Tr|Z) # typically close to true value beta


From a more classical angle, if (to be fair, that both are is the crux of the question) we have two valid IVs and only one endogenous variable, we will get consistent estimators from picking either IV (or both).

I'll throw my two cents in to note that in the case of constant effects illustrated by Christoph (which makes Assumption 3 redundant), we can show analytically why a consistent first stage isn't a necessary assumption for a consistent IV estimator

\begin{align*} Y &= \beta T + u &\quad &(\text{Structural equation}) \\ T &= \theta Z + v &\quad &(\text{First stage})\\ Y &= \phi Z + \varepsilon &\quad &(\text{Reduced form}) \end{align*}

where $$\varepsilon = \beta v + u$$ and $$\phi = \beta\theta$$. Assume

\begin{align*} Cov(Z, u) &= 0 &\quad &(\text{Assumptions 1 and 2}) \\ \theta &\neq 0 &\quad &(\text{Assumption 4}) \end{align*}

From the probability limits of the OLS estimators for the reduced form and first stage

$$\hat{\phi}_{OLS} \overset{p}{\to} \frac{Cov(Z,Y)}{Var(Z)} = \frac{Cov(Z, \beta\theta Z + \beta v + u)}{Var(Z)} = \beta\theta+ \beta \frac{Cov(Z,v)}{Var(Z)}$$

$$\hat{\theta}_{OLS} \overset{p}{\to} \frac{Cov(Z,T)}{Var(Z)} = \frac{Cov(Z, \theta Z + v)}{Var(Z)} = \theta + \frac{Cov(Z,v)}{Var(Z)}$$

we see that, although we cannot consistently estimate the causal effect of the instrument on the treatment or the causal effect of the instrument on the outcome, the IV estimator for the causal effect of the treatment on the outcome is nevertheless consistent.

$$\hat{\beta}_{IV} = \frac{\hat{\phi}_{OLS}}{\hat{\theta}_{OLS}} \overset{p}{\to} \frac{\beta\left(\theta Var(Z) + Cov(Z,v)\right)/Var(Z)}{\left(\theta Var(Z) + Cov(Z,v)\right)/Var(Z)} = \beta$$

The reason is similar to why an instrument with (classical) measurement error can still be valid. The bias in the reduced form is cancelled by the same bias in the first stage.