 The Delta Method    Message/Author  Anonymous posted on Saturday, November 23, 2002 - 2:49 pm
In a LGM with four time points, the time scores were fixed at 0 and 1 for time points t1 and t2, while they were freely estimated for t3 and t4. I want to test if the change in the true score of the outcome from t3 to t4 was statistically significant given the slope factor S was statistically significant. Delta method was applied for this test.

Suppose, change in true score of the outcome from t3 to t4 is: f=(I+St4)-(I+St3)=S(t4-t3), thus:
D={(t4-t3) S S}
V={Var(S) Cov(S,t4) Cov(S,t3),
Cov(S,t4) Var(t4) Cov(t4,t3),
Cov(S,t3), Cov(t4,t3), Var(t3)}
The asymptotic variance of f can be estimated as: Var(f)=DVD'

Where D is a vector of the partial derivatives of f with respect to S, t4, and t3; D' is the transposition of D; and V is a variances/covarinaces matrix in which the diagonal elements are the Mplus estimated variances, while the off-diagonal elements are the Mplus estimated covariance of S, t4, and t3.

The LGM was re-run with the time scores set as: 0, *, *, and 1 for t1, t2, t3, and t4 so that the slope factor S would reflect the change in the overall period (i.e., from t1 to t4). The estimated change from t3 to t4 (i.e., S(t4-t3)) from the two models were almost identical. However, its variance estimated from the Delta method was quite different. Is anything wrong in the Delta method I described above? Your help will be appreciated.  bmuthen posted on Sunday, November 24, 2002 - 6:12 am
Looks like you are using the Delta method correctly in principle, except you should have D = {(t4-t3) S -S}, right?

When you instead use 0, *, *, 1, I assume you are doing the Delta method on S(1-t3), where t3 is the new time score, right?  Anonymous posted on Sunday, November 24, 2002 - 11:49 am
Thank you so much for your prompt reply.
Yes, D={(t4-t3)S -S} and S(1-t3) was used when time scores were 0,*,*,1. In addition, var(t4)=0, cov(t4,t3)=0 and cov(t4,S)=0 because t4=1.
The following were the results from the two models:
When time scores were 0,1,*,*:
f=(t4-t3)=-0.151, se(f)=0.094.
When time scores were 0,*,*,1:
f=(t4-t3)=-0.151, se(f)=0.098.
I am wondering why the estimated values of se(f)were not identical while the values of f were?  bmuthen posted on Sunday, November 24, 2002 - 12:39 pm
Perhaps this is simply due to rounding errors in your Delta method computations. How many digits do you work with here?  Anonymous posted on Monday, November 25, 2002 - 10:42 am
Number of decimal places does make a difference.
I did a test for the model with time scores 0 * * 1. I found:

1) With 3 decimal places using the estimates from Mplus output:
se(S(t2-t1))=0.09951

2) With 4 decimal places using the estimates from the results of RUNALL Utility with the original sample only. However, cov(S,t2) was not availble in the RUNALL results, thus its stimate was from Mplus output.

se(S(t2-t1))=0.10303  bmuthen posted on Monday, November 25, 2002 - 2:01 pm
Makes sense.    Topics | Tree View | Search | Help/Instructions | Program Credits Administration