# Pearl, Causal Inference in Statistics Q3.5.1 (Backdoor criterion)

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

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:

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

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?

• I do not think you can rule out $A$ or $D$ based on independence (notice that $Z$ is not independent either). Jul 17 at 21:22
• @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.
– Cryo
Jul 17 at 21:37
• 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. Jul 18 at 3:52