This is a question about backdoor criterion (as per J. Pearl) on finding causal effects. It is linked to a specific exercise in a specific book, but I hope it will be sufficiently generic and self-contained to be of general use.

Problem statement

I am self-studying Pearl, Glymour, Jewell Causal Inference in Statistics, A Primer. Not quite sure about Q3.5.1 b. There we are given a causal diagram

enter image description here

And asked to find z-specific effect of X on Y, i.e.:

$$ P\left[Y=y\, \Big|\,do\left(X=x\right),\,Z=z\right] $$

As soon as we condition on $Z$, we are creating constraint that correlates $B$ and $C$ and thus opens a back-door $XABZCDY$. To estimate effect of $X$ on $Y$, that backdoor path needs to be broken. In my understanding, this can be done by conditioning on any of the $A$, $B$, $C$ or $D$.

Model solution

I also have model solutions (found online). There, only one option is mentioned - to condition on $C$:

$$ P\left[Y=y\, \Big|\,do\left(X=x\right),\,Z=z\right]=\sum_{c} P\left[Y=y\, \Big|\,X=x,\,Z=z,\,C=c\right]\cdot P\left[C=c\right] $$

Attempt to explain the model solution

Is there a reason why conditioning on $C$ is given as a sole solution? I can rule out conditioning on $A$ or $D$ since those are not independent variables. That leaves a question of whether I could condition on $B$.

One way I can think of explaining why conditioning on $B$ would not work is by noting that causal effect corresponds to conditional probability on a modified diagram:

enter image description here

$$ P\left[Y=y\, \Big|\,do\left(X=x\right),\,Z=z\right]=P_m\left[Y=y\, \Big|\,X=x,\,Z=z\right] $$

Now, I can express this as:

$$ \begin{align} P_m\left[Y=y\, \Big|\,X=x,\,Z=z\right] &= \sum_{b} P_m\left[Y=y\, \Big|\,X=x,\,Z=z,\,B=b\right]\cdot P_m\left[B=b\right] \\ &=\sum_{c} P_m\left[Y=y\, \Big|\,X=x,\,Z=z,\,C=c\right]\cdot P_m\left[C=c\right] \end{align} $$

Since $B$ is independent, its probability would not be affected by modification of the diagram, so $P_m\left[B=b\right]=P\left[B=b\right]$, and same for $C$.

When it comes to conditional probability, we can use the fact that with fixed $Z=z$ there is no causal link from $C$ to $X$, thus conditioning on $C$ is the same on both the original and the modified diagram: $P_m\left[Y=y\, \Big|\,X=x,\,Z=z,\,C=c\right]=P\left[Y=y\, \Big|\,X=x,\,Z=z,\,C=c\right]$. This logic would not work for $B$ since $B$ does affect $X$ on the original diagram. Therefore we can only condition on $C$:

\begin{align} P\left[Y=y\, \Big|\,do\left(X=x\right),\,Z=z\right]&=P_m\left[Y=y\, \Big|\,X=x,\,Z=z\right] \\ &=\sum_{c} P_m\left[Y=y\, \Big|\,X=x,\,Z=z,\,C=c\right]\cdot P_m\left[C=c\right] \\ &=\sum_{c} P\left[Y=y\, \Big|\,X=x,\,Z=z,\,C=c\right]\cdot P\left[C=c\right] \end{align}

Does this make sense?

  • 3
    $\begingroup$ I do not think you can rule out $A$ or $D$ based on independence (notice that $Z$ is not independent either). $\endgroup$
    – Alexis
    Commented Jul 17, 2022 at 21:22
  • $\begingroup$ @Alexis, good point. Perhaps a better way to say this would be that for this diagram it makes more sense to condition on $B$ or $C$ since those are independent variables, as opposed to $A$ or $D$ that is. $\endgroup$
    – Cryo
    Commented Jul 17, 2022 at 21:37
  • 1
    $\begingroup$ Honestly I think you’re making this too complicated. In part a) of that problem you’re asked to find the C-specific effect. In that case you’re forced to adjust for Z. My guess is that the solution authors, having already shown that adjustment for C and Z together satisfies the back door criterion, saw no reason to do any more work. Thinking like mathematicians in other words. $\endgroup$
    – einar
    Commented Jul 18, 2022 at 3:52

1 Answer 1


No you were right to begin with, you can control for any variable along the back door path so long as it doesn’t open up new such paths.

You can try it for yourself for the specific diagram here (set Z to adjusted and some other one to see only the causal path remain colored): http://dagitty.net/dags.html?id=331

  • 1
    $\begingroup$ Thanks @einar , the daggity link certainly helps! $\endgroup$
    – Cryo
    Commented Jul 17, 2022 at 21:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.