The standard scheme of instrumental variable in terms of causality (->) is:

Z -> X -> Y

Where Z is an instrument, X an endogenous variable, and Y a response.

Is it possible, that following relations:

Z <- X ->Y

Z <-> X ->Y

are also valid?

While correlation between instrument and variable is satisfied, how may I think of exclusion restriction in such cases?

NOTE: The notation <-> is not explicit and might lead to different understandings of the problem. Still, the answers highlight this issue and use it to show important aspects of the problem. When reading, please proceed with caution about this part of the question.


2 Answers 2


Yes, the direction matters. As pointed in this answer, to check whether $Z$ is an instrument for the causal effect of $X$ on $Y$ conditional on a set of covariates $S$, you have two simple graphical conditions:

  1. $(Z \not\perp X|S)_{G}$
  2. $(Z\perp Y|S)_{G_{\overline{X}}}$

The first condition requires $Z$ to be connected to $X$ in the original DAG. The second condition requires $Z$ to not be connected to $Y$ if we intervene on $X$ (represented by the DAG $G_{\overline{X}}$, where you remove the arrows pointing to $X$). Thus,

Z -> X -> Y : here Z is a valid instrument.

Z <-> X -> Y: here Z is a valid instrument (assuming that a bidirected edge represents an unobserved common cause, as it does in semi-Markovian models).

Z <- X -> Y : here Z is not a valid instrument.

PS: jsk's answer is not correct, let me show you how Z <-> X is a valid instrument.

Let the structural model be:

$$ Z = U_1 + U_z\\ X = U_1 + U_2 + U_x\\ Y = \beta X + U_{2} + U_y $$

Where all the $U$'s are unobserved mutually independent random variables. This corresponds to the DAG z <--> x -->y with also x<-->y. Thus,

$$ \frac{cov(Y, Z)}{cov(X, Z)} = \frac{\beta cov(X, Z)}{cov(X,Z)} = \beta $$

  • $\begingroup$ I think this highlights the need to be very clear clear about what exactly $X<->Z$ actually means. In your revised example, I would argue that X and Z are driven by a third variable, which seems different than my understanding of the notation $X<->Z$. $\endgroup$
    – jsk
    Commented Jun 9, 2019 at 22:40
  • $\begingroup$ @jsk this is standard notation for semi-Markovian models. $\endgroup$ Commented Jun 9, 2019 at 23:35
  • 3
    $\begingroup$ Not standard to everyone. Just read a paper by Pearl and Greenland in which they say that SOME authors use the notation in this way. There's nothing in the OP's question to suggest his interpretation of the notation, though he may very well agree with you. $\endgroup$
    – jsk
    Commented Jun 10, 2019 at 0:19
  • $\begingroup$ What if $Y = \beta X + U_1 + U_y$? Would it not then be the case that $Z<->X$ but then Z would be correlated with the omitted variable and hence not be valid instrument? $\endgroup$ Commented Jun 10, 2019 at 13:12
  • $\begingroup$ @JesperHybel If you have U1 in the structural equation of Y, this means the error terms of Z and Y are dependent. Thus you have an extra bidirected edge Z<—>Y and no case works, be it Z—>X or Z<—>X. The graphical conditions are explicitly stated there. $\endgroup$ Commented Jun 10, 2019 at 14:25

Yes, direction does matter.

According to Hernan and Robins' new causal inference book https://cdn1.sph.harvard.edu/wp-content/uploads/sites/1268/1268/20/hernanrobins_v2.17.21.pdf

the following three conditions must be met:

$i.$ $Z$ is associated with $X$.

$ii.$ $Z$ does not affect $Y$ except through its potential effect on $X$.

$iii.$ $Z$ and $Y$ do not share common causes.

Condition $(iii)$ rules out relations such as $X$ - > $Z$ or $X$ < - > $Z$ because $X$ cannot have a causal effect on both $Z$ and $Y$

Edit: whether or not $X<->Z$ is acceptable for an instrument depends on the definition of $X<->Z$. If it means that they are correlated because of a third variable, like in Carlos's example, then it's ok. If it suggests a feedback loop where a causal arrow can be drawn from X to Z as well then Z is not a valid instrument.

  • $\begingroup$ (-1) This is wrong, Z<—>X is fine for an instrument. $\endgroup$ Commented Jun 9, 2019 at 20:57
  • 1
    $\begingroup$ These conditions posited by Hernan and Robins are not precise, they say that themselves---read further the chapter. Also see a trivial counterexample to your claim in the edit of my answer. $\endgroup$ Commented Jun 9, 2019 at 21:19

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