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For one unit write $M=(M_1,M_2,M_3)$ and $W=(M,Z_1,Z_3)$. Since the graph makes $S$ independent of $M$, the known distributions determine the conditional assignment lawUnder the sharp causal null hypothesis that $A$ has no effect on $Y$, delete $A\to Y$. The D-separation criterion then gives$M_2$ blocks the route through $C$, $Z_3$ blocks the route through $S$, $M_1$ and $M_3$ block the collider-opened detours through $Z_1$ and $Z_3$, and $A$ remains a collider on routes through $U$.

For independent units $i=1,\dots ,n$, choose a test statistic $T(Z_{2,1:n},Y_{1:n},W_{1:n})$ that measures residual association between $Z_2$ and $Y$. Hold $(Y_i,W_i)$ fixed and independently drawRecompute $T^{(b)}$ after each draw. A valid Monte Carlo conditional randomization test useswith a two-sided statistic or absolute value when appropriate.

Under the null, conditional on $(Y_{1:n},W_{1:n})$, the observed $Z_{2,1:n}$ and its $B$ resamples are exchangeable because they have the same product law ${\prod }_{i}q(\cdot \mid W_i)$. The rank of $T^{\mathrm{obs}}$ among the $B+1$ values is therefore uniform after randomized tie breaking and conservative without it. ConsequentlyTaking expectations proves unconditional Type I error control.