# Model Identification

Could somebody please explain why this model is "just identified" As I see it, there are 5 * 4 / 2 = 10 variances/covariances, 4 observed means, giving 14 available degrees of freedom

5 DF are used on the 5 paths

2 intercepts are estimated

2 error variances are estimated

2 endogenous variances are estimated

2 exogenous variances are estimated

Giving 13 used DF. What am I missing ?

## 1 Answer

Instead of "2 endogenous variances" you should have "2 exogenous means". Also, the missing DF is the covariance between exogenous variables.

Let me elaborate:

There are k(k+3)/2 = 14 available DF as you said.

These are used as follows

• 5 paths
• 1 covariance between the two exogenous variables hs and col
• 2 variances of the exogenous variables
• 2 means for the exogenous variables
• 2 residual variances for the endogenous variables
• intercepts for the endogenous variables

= 14 DF used, so the model is just identified.

The point of confusion here is that, in the path diagram, there is no curved 2-headed arrow between the exogenous variables. Mplus assumes that exogenous variables are correlated. Furthermore, the estimated means and variances of these are also not shown in the output, so the output shown in the link is somewhat unhelpful for calculating the model degrees of freedom.

MODEL RESULTS

Two-Tailed
Estimate       S.E.  Est./S.E.    P-Value

GRE      ON
HS                 0.309      0.065      4.756      0.000
COL                0.400      0.071      5.625      0.000

GRAD     ON
HS                 0.372      0.075      4.937      0.000
COL                0.123      0.084      1.465      0.143
GRE                0.369      0.078      4.754      0.000

Intercepts
GRE               15.534      2.995      5.186      0.000
GRAD               6.971      3.506      1.989      0.047

Residual Variances
GRE               49.694      4.969     10.000      0.000
GRAD              59.998      6.000     10.000      0.000


If we explicity add the covariance term to the model:

Model:
gre on hs col;
grad on hs col gre;
hs with col;


...then the output becomes

                                                    Two-Tailed
Estimate       S.E.  Est./S.E.    P-Value

GRE      ON
HS                 0.309      0.065      4.756      0.000
COL                0.400      0.071      5.626      0.000

GRAD     ON
HS                 0.372      0.075      4.937      0.000
COL                0.123      0.084      1.465      0.143
GRE                0.369      0.078      4.755      0.000

HS       WITH
COL               63.297      8.106      7.809      0.000

Means
HS                52.230      0.723     72.223      0.000
COL               52.645      0.661     79.670      0.000

Intercepts
GRE               15.534      2.995      5.186      0.000
GRAD               6.971      3.506      1.989      0.047

Variances
HS               104.597     10.460     10.000      0.000
COL               87.329      8.733     10.000      0.000

Residual Variances
GRE               49.694      4.969     10.000      0.000
GRAD              59.998      6.000     10.000      0.000


and from this it is clear that 14 parameters have been estimated, and so the model is just identified.

Note that the estimates of all the other parameters are unchanged