# Significance Test for No Confounding

I am reading Pearl, J. (January 1998). "Why there is no statistical test for confounding, why many think there is, and why they are almost right". UCLA Computer Science Department, Technical Report R-256, as cited under 'Further Reading' in the Wikipedia article on confounding.

Pearl gives the following 'associational criteria' for no confounding where $Y$ is the response to the intervention $X: X$ and $Y$ are not confounded in the presence of $T$, if for every $Z$ in $T$:

$(U_1) P(x|z)=P(x)$; or $(U_2) P(y|z,x)=P(y|x)$.

Surely these conditions will never hold exactly for any empirical data set subject to random effects. Does one then replace equality by "no statistically significant inequality" in these conditions? If so, could some pointers on doing this be provided, or a reference.

• When you say subject to random effects, are you referring to mixed models or are you referring more broadly to data generating mechanisms? Commented Nov 21, 2017 at 3:41
• AAdamO By "random effects", I simply mean subject to stochastic "noise". Commented Nov 21, 2017 at 10:49

That said, testing (conditional) independence such as $P(x|z) = P(z)$ or $P(y|x, z) = P(y|x)$, is a regular statistical procedure. You can use any parametric/non-parametric (conditional) independence test best suited for your application. For example, for the multivariate normal case you could use linear regression and proceed with significance testing or other (better) inferential tools for the task. For categorical data, you could use Chi-squared tests etc.
But please remember that, by itself, testing (conditional) independence does not test for confounding, precisely because of the reasons mentioned by Pearl in the paper you are reading. For example, for the associational criteria to be a necessary condition for unconfoundedness you may further make the causal assumptions that: (i) $z$ is not caused by $x$, (ii) $z$ causes $y$ and (iii) there are no incidental cancellations going on.
• @AdamO If you assume that $z$ is is not caused by $x$, that $z$ affects $y$ and that there's no incidental cancelation, then the associational criteria is necessary for no confounding. Commented Nov 21, 2017 at 4:16