Mark Relyea
I am trying to specify a lavaan CFA that has some factors with three indicators, one factor with two indicators, and two factors with one indicator each. For the single indicator factors, I followed advice to specify the variance of the observed variable to 0 so the latent variable will account for all of the variance in the observed variable. I saw other posts in this group saying setting the variance to 0 may be unnecessary now as this is the default behavior. Either way I run it, I still get a covariance matrix that is not positive definite and a model where the latent variance does not equal the variance for the single items. What is the correct way to model single indicator factors? I've created a reproducible example below.
Simulated data with a positive definite covariate matrix.Â
library(MASS)
mu <- c(2.5,3,3.1,3.5,2.1,1.5,2,4,4.2,4.5)
Sigma <- matrix(.5, nrow=10, ncol=10) + diag(10)*.5
set.seed(1527)
rawvars <- mvrnorm(n=200, mu=mu, Sigma=Sigma)
Convert latent responses to positive ordered categories (to create likert-type items).
i1 = findInterval(rawvars[,1], vec=c(-Inf,2,2.75,3.5,4.25,Inf))
i2 = findInterval(rawvars[,2], vec=c(-Inf,2,2.75,3.5,4.25,Inf))
i3 = findInterval(rawvars[,3], vec=c(-Inf,2,2.75,3.5,4.25,Inf))
i4 = findInterval(rawvars[,4], vec=c(-Inf,2,2.75,3.5,4.25,Inf))
i5 = findInterval(rawvars[,5], vec=c(-Inf,2,2.75,3.5,4.25,Inf))
i6 = findInterval(rawvars[,6], vec=c(-Inf,2,2.75,3.5,4.25,Inf))
i7 = findInterval(rawvars[,7], vec=c(-Inf,2,2.75,3.5,4.25,Inf))
i8 = findInterval(rawvars[,8], vec=c(-Inf,2,2.75,3.5,4.25,Inf))
i9 = findInterval(rawvars[,9], vec=c(-Inf,2,2.75,3.5,4.25,Inf))
i10 = findInterval(rawvars[,10], vec=c(-Inf,2,2.75,3.5,4.25,Inf))
df <- data.frame(cbind(i1,i2,i3,i4,i5,i6,i7,i8,i9,i10))
Confirm all items have small to moderate positive correlations.
> round(cor(df),2)
i1 i2 i3 i4 i5 i6 i7 i8 i9 i10
i1 1.00 0.53 0.55 0.50 0.50 0.42 0.51 0.48 0.51 0.38
i2 0.53 1.00 0.51 0.48 0.39 0.45 0.42 0.50 0.55 0.40
i3 0.55 0.51 1.00 0.46 0.49 0.37 0.43 0.52 0.48 0.42
i4 0.50 0.48 0.46 1.00 0.34 0.43 0.35 0.50 0.42 0.42
i5 0.50 0.39 0.49 0.34 1.00 0.37 0.43 0.36 0.35 0.39
i6 0.42 0.45 0.37 0.43 0.37 1.00 0.34 0.38 0.39 0.32
i7 0.51 0.42 0.43 0.35 0.43 0.34 1.00 0.40 0.46 0.37
i8 0.48 0.50 0.52 0.50 0.36 0.38 0.40 1.00 0.39 0.35
i9 0.51 0.55 0.48 0.42 0.35 0.39 0.46 0.39 1.00 0.43
i10 0.38 0.40 0.42 0.42 0.39 0.32 0.37 0.35 0.43 1.00
Built model constraining two indicator factors to be equal and specifying the variance on the single indicator factor variables (i1 and i10) to 0.
> library(lavaan)
fa.mod <- '
f1=~ i1
f2=~ i2 + i3 + i4
f3=~ i5 + i6 + i7
f4=~ ai8 + ai9
f5=~ i10
i1~~ 0*i1
i10~~ 0*i10'
Model fit was not positive definite and the latent variance does not equal the variance for the single items.
> fa.fit<- sem(fa.mod,data=df)
Warning message:
In lav_object_post_check(object) :
lavaan WARNING: covariance matrix of latent variables
is not positive definite;
use inspect(fit,"cov.lv") to investigate.
> inspect(fa.fit,"cov.lv")
f1 f2 f3 f4 f5
f1 1.312
f2 0.755 0.766
f3 0.595 0.555 0.451
f4 0.565 0.614 0.419 0.383
f5 0.394 0.465 0.360 0.356 0.828
> summary(fa.fit,fit.measures=T,standardized=T)
lavaan (0.5-23.1097) converged normally after 39 iterations
Number of observations 200
Estimator ML
Minimum Function Test Statistic 26.478
Degrees of freedom 28
P-value (Chi-square) 0.547
Model test baseline model:
Minimum Function Test Statistic 760.907
Degrees of freedom 45
P-value 0.000
User model versus baseline model:
Comparative Fit Index (CFI) 1.000
Tucker-Lewis Index (TLI) 1.003
Loglikelihood and Information Criteria:
Loglikelihood user model (H0) -2560.036
Loglikelihood unrestricted model (H1) -2546.797
Number of free parameters 27
Akaike (AIC) 5174.071
Bayesian (BIC) 5263.126
Sample-size adjusted Bayesian (BIC) 5177.587
Root Mean Square Error of Approximation:
RMSEA 0.000
90 Percent Confidence Interval 0.000 0.051
P-value RMSEA |z|) Std.lv Std.all
f1 =~
i1 1.000 1.146 1.000
f2 =~
i2 1.000 0.875 0.722
i3 0.971 0.100 9.709 0.000 0.850 0.716
i4 0.911 0.103 8.856 0.000 0.798 0.653
f3 =~
i5 1.000 0.671 0.633
i6 0.736 0.106 6.947 0.000 0.494 0.579
i7 0.911 0.121 7.551 0.000 0.612 0.641
f4 =~
i8 (a) 1.000 0.619 0.601
i9 (a) 1.000 0.619 0.646
f5 =~
i10 1.000 0.910 1.000
Covariances:
Estimate Std.Err z-value P(>|z|) Std.lv Std.all
f1 ~~
f2 0.755 0.107 7.053 0.000 0.753 0.753
f3 0.595 0.091 6.517 0.000 0.774 0.774
f4 0.565 0.078 7.250 0.000 0.797 0.797
f5 0.394 0.079 4.996 0.000 0.378 0.378
f2 ~~
f3 0.555 0.089 6.256 0.000 0.945 0.945
f4 0.614 0.081 7.541 0.000 1.132 1.132
f5 0.465 0.078 5.976 0.000 0.584 0.584
f3 ~~
f4 0.419 0.065 6.430 0.000 1.008 1.008
f5 0.360 0.065 5.515 0.000 0.589 0.589
f4 ~~
f5 0.356 0.059 6.047 0.000 0.631 0.631
Variances:
Estimate Std.Err z-value P(>|z|) Std.lv Std.all
.i1 0.000 0.000 0.000
.i10 0.000 0.000 0.000
.i2 0.703 0.083 8.422 0.000 0.703 0.478
.i3 0.688 0.081 8.498 0.000 0.688 0.488
.i4 0.854 0.095 9.020 0.000 0.854 0.573
.i5 0.673 0.079 8.547 0.000 0.673 0.599
.i6 0.485 0.054 8.985 0.000 0.485 0.665
.i7 0.536 0.063 8.467 0.000 0.536 0.589
.i8 0.678 0.080 8.527 0.000 0.678 0.639
.i9 0.535 0.068 7.899 0.000 0.535 0.583
f1 1.312 0.131 10.000 0.000 1.000 1.000
f2 0.766 0.137 5.608 0.000 1.000 1.000
f3 0.451 0.099 4.561 0.000 1.000 1.000
f4 0.383 0.075 5.133 0.000 1.000 1.000
f5 0.828 0.083 10.000 0.000 1.000 1.000
Christopher David Desjardins
Hi Mark,
Increasing your N fixes the problem
# from
rawvars <- mvrnorm(n=200, mu=mu, Sigma=Sigma)
# to
rawvars <- mvrnorm(n=20000, mu=mu, Sigma=Sigma)
You can also try with a different seed and it may or may not give that message.
replicate(10, {
rawvars <- mvrnorm(n=200, mu=mu, Sigma=Sigma)
i1 = findInterval(rawvars[,1], vec=c(-Inf,2,2.75,3.5,4.25,Inf))
i2 = findInterval(rawvars[,2], vec=c(-Inf,2,2.75,3.5,4.25,Inf))
i3 = findInterval(rawvars[,3], vec=c(-Inf,2,2.75,3.5,4.25,Inf))
i4 = findInterval(rawvars[,4], vec=c(-Inf,2,2.75,3.5,4.25,Inf))
i5 = findInterval(rawvars[,5], vec=c(-Inf,2,2.75,3.5,4.25,Inf))
i6 = findInterval(rawvars[,6], vec=c(-Inf,2,2.75,3.5,4.25,Inf))
i7 = findInterval(rawvars[,7], vec=c(-Inf,2,2.75,3.5,4.25,Inf))
i8 = findInterval(rawvars[,8], vec=c(-Inf,2,2.75,3.5,4.25,Inf))
i9 = findInterval(rawvars[,9], vec=c(-Inf,2,2.75,3.5,4.25,Inf))
i10 = findInterval(rawvars[,10], vec=c(-Inf,2,2.75,3.5,4.25,Inf))
df <- data.frame(cbind(i1,i2,i3,i4,i5,i6,i7,i8,i9,i10))
fa.fit<- sem(fa.mod,data=df)
})
I ran that a few times and got between 3 and 10 warnings. So sometimes it converges without a warning.
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