In path analysis (i.e., DAGS as linear structural equation models), where all relationships between variables are assumed to be linear, you can compute the association between two variables by multiplying the path coefficients corresponding to each open path between them and take the sum over all those paths.

For example, if my model is $A \leftarrow B \rightarrow C$, the correlation between $A$ and $C$ can be computed as $Corr(A,B) * Corr(B,C)$.

When we have a collider, i.e., the model is $A \rightarrow B \leftarrow C$, the path between $A$ and $C$ is closed, and the correlation between $A$ and $C$ is 0. However, when we condition on the collider, we open up a path between $A$ and $C$, inducing a correlation between them.

I want to know, in terms of $Corr(A,B)$ and $Corr(B,C)$, how to compute this induced correlation. It is not equal to $Corr(A,B) * Corr(B,C)$ in some simple simulations I have run, and I have struggled to use matrix algebra to derive the correlation (or covariance or joint PMF for binary variables). I have not seen this discussed in the literature, though maybe Pearl has something to say about it. Any help would be appreciated.


1 Answer 1


Let the variables be stardadized to unit variance and let's use $\sigma_{ab}$ to denote the covariance of $A$ and $B$. You want the correlation of $A$ and $C$ when conditioning on the collider $B$. Thus, you can write the partial correlation $A$ and $C$ given $B$ as:

$$ \rho_{ac.b} = \frac{\sigma_{ac} - \sigma_{ab}\sigma_{bc}}{\sqrt{1 - \sigma_{ab}^2} \sqrt{1 - \sigma_{bc}^2}} $$

This holds regardless of the graph. In your case since $\sigma_{ac}=0$ this simplifies to,

$$ \rho_{ac.b} = \frac{-\sigma_{ab}\sigma_{bc}}{\sqrt{1 - \sigma_{ab}^2}\sqrt{1 - \sigma_{bc}^2}} $$

Finally, if you want the expression in terms of path coefficients, let the coefficient for $A\rightarrow B$ be $\lambda_{ab}$ and the coefficient for $C \rightarrow B$ be $\lambda_{cb}$. Then,

$$ \rho_{ab.c} = \frac{-\lambda_{ab}\lambda_{bc}}{\sqrt{1 - \lambda_{ab}^2}\sqrt{1 - \lambda_{bc}^2}} $$


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