Continuous Time Survival Analysis PreviousNext
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 Craig A. Mason posted on Thursday, September 20, 2007 - 8:40 am
I have a question regarding continuous-time survival analysis in MPlus. I am looking at survival in a sample of elderly over several years, based on gender. I have 3 variables...

FEMALE is coded 0=male, 1=female.
DIED indicates whether the person died (0=living at end of study, 1=not living).
LASTTIME is the years post-baseline at the time of death (DIED=1) or the final assessment (DIED=0).

The MPLUS input is...


I obtain an estimate for LASTTIME ON FEMALE of -0.787

I know that women were less likely to die.

Given that DIED is not coded as in the Mplus users guide (where 0=not right censored, 1=right censored), I want to make certain that I interpret the output correctly.

Is it the log of the odds-ratio for survival for females relative to males? Is it the log of the odds of death for females to males? Is it a predicting years post-baseline at death/final assessment? Or something else?

In essence, (1) should I recode my indicator for time censoring and (2) how should I interpret the estimate for the FEMALE effect.

 Tihomir Asparouhov posted on Thursday, September 20, 2007 - 1:05 pm
Your setup is correct. The regression coefficient is the log odds of incremental probability of death. It is also the log of the ratio of the hazards
the female hazard = the male hazard * Exp(-0.787).

Here are some links you might find useful:
Mplus specific implementation

and some general Survival Analysis definitions
 Sylvana Robbers posted on Tuesday, November 04, 2008 - 4:17 am
I have a following question on the above posts. If the bivariate covariate FEMALE would have been a continuous covariate, what would then be the interpretation of the regression estimate?
 Linda K. Muthen posted on Tuesday, November 04, 2008 - 10:26 am
The regression coefficient is the log odds of incremental probability of death for a one unit change in the continuous covariate.
 Sylvana Robbers posted on Wednesday, November 05, 2008 - 8:25 am
Thanks for your swift reply. I have a following question about the interpretation of the results.
Let's consider my own data in which the continuous covariate is externalizing problems at age 3 and in which the critical event is parental divorce between age 3 and age 12. When I regress divorce (age of child in months at divorce) on ext, then I get an estimate of 0.021 (p=0.002). So, the odds of the probability of divorce is 1,021 times higher when children have a one unit increase in ext. Does this mean that children with 10 units increase have a 1.021^10=1.23=23% more chance that there parents will divorce? Or does it mean that ext is related to EARLIER divorce? Please correct me if my interpretation is wrong. Is the estimate similar to the hazard ratio? (sorry if this is a stupid question, but I am new to survival analysis).
Thanks alot in advance.
 Linda K. Muthen posted on Thursday, November 06, 2008 - 8:42 am
Is .021 a raw coefficient or an exponentiated raw coefficient?
 Sylvana Robbers posted on Friday, November 07, 2008 - 1:37 am
.021 is the raw coefficient (in the 'estimates' column).
e^.021 = 1.021
 Bengt O. Muthen posted on Friday, November 07, 2008 - 5:56 pm
The log odds increase for a 10 unit increase in x (externalizing) is 10*0.021. If you are new to survival analysis, I recommend warmly the pedagogical book by Singer & Willet, Applied Longitudinal Data Analysis. They discuss things not only in terms of formulas but also in carefully worded words, including how to interpret results.
 Sylvana Robbers posted on Thursday, November 13, 2008 - 8:48 am
Thank you.
I got 2 more output-questions:
1) Where can I find the hazard ratio? Is it e^0.021?
2)And how do I get a confidence interval for the hazard ratio?
Thanks alot for your time.
 Bengt O. Muthen posted on Thursday, November 13, 2008 - 10:47 am
1. Yes.

2. There are 2 ways.

You define the NEW parameter, say hr:

Model Constraint:
hr = exp(p1);

where "p1" is the label for the parameter in the model. This gives you the hr estimate and SE from which you can compute the CI you asked for.

Or, you can get CI limits for p1 by +-1.96*SE(p1) and then exponentiate these limits to be used for the hr CI (the latter is better - see the logistic regression literature).
 Reza ELAIDI posted on Wednesday, April 02, 2014 - 3:31 am
Dear Pr Muthen,
I'd like to obtain the confidence interval for the hazard ratio (HR) of a covariate X1 involved in an interaction with another covariate X2. If beta1 and beta2 Cox regression estimates, then the CI for HR is exp(beta1 + beta2 +/-1.96(variance(beta1)+variance(beta2)+2(covariance(beta1*beta2)). I tried to get the covariance using TECH3 output but results are very different from what I obtain using SAS CovB option in PROC PHREG despite the same estimates are obtained with Mplus and SAS. Is TECH3 the right option to get the variance/covariance of estimates ? What could be the reason for difference with SAS ?
Thank you in advance for your assistance
 Bengt O. Muthen posted on Wednesday, April 02, 2014 - 5:06 pm
TECH3 is the right option. Perhaps SAS is using ML SEs and Mplus MLR SEs? If not, please send both outputs, data, and license number to Support.
 Reza ELAIDI posted on Thursday, April 03, 2014 - 5:25 am
Dear Pr Muthen,
You are right, SAS uses ML. The problem I face now is that I'm unable to match the estimates covariances in the TECH3 file with those of the output matrix, despite I used the variable numbering of TECH1. How is TECH3 file organized ? Than k you again for your help.
 Linda K. Muthen posted on Thursday, April 03, 2014 - 6:21 am
TECH3 contains the variances and covariances of the parameters. This will not match the results in the output. The first entry in TECH3 is the variance of parameter 1 from TECH1. The second parameter is the covariance between parameters 1 and 2. The third parameter is the variance of parameter 2, etc.
 Laura Brown posted on Thursday, February 15, 2018 - 3:40 am
Hi, I'm following the advice from Bengt's post (Thurs Nov 13, 2008, 10:47) to calculate CIs around a hazard ratio but the values I get are non-sensical.
The model gives an estimate of 0.485 for the effect of my continuous exposure on duration (SE 0.118, p-value <0.001) and an HR of 1.624 (SE 0.192, p<0.001). Following the two methods suggested in Bengt's post, I get the following CI estimates:

Method 1:
LCI: -1.96*0.192 = -0.376
UCI: +1.96*0.192 = 0.376
Method 2:
LCI: exp(-1.96*0.118) = 0.794
UCI: exp(1.96*0.118) = 1.260

Method 1 can't be right because HRs and their CIs can't be negative values. Method 2 can't be right because the low p-value dictates that the relationship is significant and so the CI shouldn't cross 1.
I sense-checked this by running the same cox model in Stata, which gives the CIs automatically. I got an estimate of 0.485 (SE 0.113, p-value <0.001, LCI 0.263, UCI 0.707) and an HR of 1.624 (SE 0.184, p<0.001, LCI 1.300, UCI 2.028). Neither of the methods provided gives me the correct confidence intervals here either:

Method 1:
LCI: -1.96*0.184 = -0.361
UCI: 1.96*0.184 = 0.361
Method 2:
LCI: exp(-1.96*0.113) = 0.801
UCI: exp(1.96*0.113) = 1.249

Please can you let me know where I'm going wrong?
 Laura Brown posted on Thursday, February 15, 2018 - 3:56 am
Apologies, I have now seen my error, so please ignore the last message. It was a simple case of forgetting to include the estimate in the CI equation!

i.e. it should be
Method 1:
LCI: 1.62-(1.96*0.192) = 1.248
UCI: 1.62+(1.96*0.192) = 2.003
Method 2:
LCI: exp(0.485-(1.96*0.118)) = 1.289
UCI: exp(0.485+(1.96*0.118)) = 2.047
 Tihomir Asparouhov posted on Thursday, February 15, 2018 - 8:13 pm
Just FYI - Method 2 corresponds to the approach you see in Stata.
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