Proof of Wright’s path-tracing rules

Question

I have a path diagram. According to Wright’s path-tracing rules, correlation of B & D = b + a*f. However, when I tried to proof the above result by solving

cov(B, D) = cov(fA + gC, aA + bB)

I didn't get the same result with Wright’s rule. It seems the problem originated from using B = fA + gC instead of B = fA. But I thought B = fA + gC would be correct because variable B is actually affected by both A and C. Even when I applied B = fA,

cov(B, D) = cov(fA, aA + bB)
 = cov(fA, aA) + cov(fA, bB)
 = fa*1 + fb*cov(A, B)
 = fa * (f^2)*b

Can anyone tells me what I've got wrong. Thanks!

Answer 1

I thought B = fA + gC would be correct because variable B is actually affected by both A and C

No, nothing is pointing to B, so nothing "affects" it. In a path diagram, a variable('s variance) is the sum of (the variance of) everything that directly points to it (including residuals; see below). Double-headed arrows are not directed paths; they represent (co)variances.

The b path is a partial regression slope (i.e., slope of B's residual predicting D's residual, after A is partialled out of both B and D). So the reason that fa must be added to b is that B--D's zero-order correlation (i.e., all variance in common between B and D) includes both their partial correlation (b) and the correlation of D with the part of B that is in common with A (i.e., fa).

Here is the correct linear/covariance algebra corresponding to the path diagram:

cov(B, D) = cov(B, aA   + bB + i)
          = cov(B, aA)  + cov(B, bB)  + cov(B, i)
          = a*cov(B, A) + b*var(B)    + 0
          = a*f         + b*1 (because all are standardized)
          = af + b

Loehlin's (2004) SEM book is freely available at this site, if that helps.