Controlling confounders in a causal diagram. Isn't the backdoor criterion sufficient?

Question

In Judea Pearl's The Book of Why we find the following causal diagram:

where $U_1$ and $U_2$ are unobserved variables.

The diagram is accompanied by a comment that ensures that neither the back door criterion nor the front door criterion are sufficient to figure out the causal effect $P(Y | do(X))$.

However, as I understood the back door criterion, it should be enough to control for $Z$.

The backdoor paths from $X$ to $Y$ are $X\leftarrow Z \leftarrow W \leftarrow U_1 \rightarrow Y$, and $X\leftarrow U_2 \rightarrow W \leftarrow U_1 \rightarrow Y$.

If we block $Z$, then the first path becomes blocked. And the second path is blocked without a need to control variables by the collider at $W$.

So what is going on? Why is my analysis wrong?

Answer 1

Glad to help.

By conditioning on Z, which is a descendant of W, we will be opening the backdoor path $X\leftarrow U_2 \rightarrow W \leftarrow U_1 \rightarrow Y$.

Judea Pearl

ps. Glad this problem made its way to this math-inspired blog; the problems we encounter in graphical modeling and causal inference are indeed intellectually challenging and fun to solve.

The answer to this puzzle was posted yesterday on Twitter.

It is amazing that a graph with just 6 nodes can captivate and resist the best minds in the country, when the solution requires just 7 lines.

I hope you enjoy the game of causal inference. Judea Pearl @yudapearl

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EDIT BY OP: Here I copy the solution to the puzzle that was shared on twitter.

\begin{align*} P(y|do(x)) &= P(y|do(x),do(z),do(w))\\ &=P(y|x,do(z),do(w))\\ &=P(y|x, do(z))\\ &=\sum_w P(y|x,do(z),w)\,P(w|x,do(z))\\ &=\sum_w P(y|x,do(z),w)\,P(x|do(z), w)\,P(w|do(z))/P(x|do(z))\\ &=\sum_w P(y|x, z, w)\,P(x|z,w)\,P(w) \sum_w P(x|w,do(z))\,P(w|do(z))\\ &=\sum_w P(y|x,z,w)\,P(x|z,w)\,P(w) \sum_w P(x|w,z)\,P(w). \end{align*}