"by" or "on" ? PreviousNext
Mplus Discussion > Confirmatory Factor Analysis >
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 Tin-chi Lin posted on Friday, October 03, 2008 - 8:15 am
Dear Dr. Muthen,

I just start to learn MPLUS for social research and was wondering if I could seek your clarification for distinguishing "by" and "on", especially when they imply the same causal direction, which has been bothering me for a while.

In the User Manual's example 5.6 (Ch5, p.59), I am wondering what would be different if we change the last line from

"f5 by f1-f4" ....(A)

to

"f1-f4 on f5"? .... (B),


especially in terms of

(i) the substantive interpretation for the relationship between f5 and f1-f4.

(ii) the way MPLUS estimate the model, and

(iii) the default MPLUS makes for the estimation (for example, in (A), by default MPLUS wouldn't estimate the covariance between fi and fj for i,j = 1..4).

I understand that both (A) and (B) would imply the same causal direction----that is, f1-f4 is predicted by f5 (or f5 can be measured by f1-f4). The degree of freedom may be different, because in scenario (A) MPLUS would set one of the coefficients to be one while in (B) MPLUS wouldn't. But what else would be different?

Also, in the language of Bollen’s book (page 319-320 at “Structural Equations with Latent Variables”, 1989), will (A) and (B) have the same “beta”, “gamma”, and “lambda”?
 Linda K. Muthen posted on Friday, October 03, 2008 - 10:59 am
A and B are the same other than the fact that the first factor loading is fixed to one in A. If you change B as follows, they are exactly the same:

f1 ON f5@1;
f2-f4 ON f5;

Note that the ON construction will not work unless the factors are first defined using BY, for example,

f1 BY;

BY is used to define factors as a way to assign names the factors. This has no implications to model estimation or the interpretation of the results. Defaults for BY and ON may be different but these can be overridden. The BY construction will use the lambda matrix. The ON construction will use the beta matrix. The same values will be estimated irrespective of which matrix they are in.
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