Message/Author 

Jonika Hash posted on Tuesday, October 10, 2017  1:09 pm



Hello! I am a PhD student looking for help with predicted probabilities. I have a path model with a latent mediator and a final dichotomous outcome (u2). I would like to calculate predicted probabilities of u2 at different levels of two predictors (x3 & x4), holding all else constant. I am using ML with Monte Carlo integration and the logit link function. The model is: i1  y1@1 y2@1 y3@1; u2 ON i1 x2 x3 x4 x5 x6; i1 ON x1 x2 x3; x1 x2 x5; where x6 is an interaction term = x3*x4 I think my equations are: i1 = a1 + b11*x1 + b12*x2 + b13*x3 u2 = a2 + B21*i1 + b22*x2 + b23*x3 + b24*x4 + b25*x5 + b26*x3*x4 where a2 = u2 threshold estimate in Mplus output a1 = i1 intercept estimate in output b's = regression coefficients for corresponding x's B21 = regression coefficient for u2 ON i1 Then to calculate logits at varying levels of x3 & x4, holding all else constant at 0: logit = a2 + B21*a1 + B21*b13*x3 + b23*x3 + b24*x4 + b26*x3*x4 And predicted probabilities: 1/(1+e^(1(logit))) Wondering if I can get help if this does not look correct? Thank you. 


It is tempting to think of that as the answer, but what's not taken into account is the residual variance in i which has to be "integrated out". With a logistic link and a normal i residual, this calls for numerical integration so not something done by hand. But with a probit link the expression simplifies to using a standard univariate normal distribution function using as argument the expectation that you have on the righthand side of logit, divided by the square root of a residual variance expression. This is shown on page 310 of our book Regression and Mediation Analysis using Mplus. You can also figure it out adding residuals to both of your 2 equations where the probit residual variance is 1. 

Jonika Hash posted on Tuesday, October 10, 2017  3:21 pm



Hi Dr. Muthen, Thank you so much for your help! I am so glad I asked. I am not as familiar with probit regression as I am with logit, so I am hoping I understand. Does this now look correct?: Using the probit link function and taking residuals into account, I think my equations would now be: i1 = a1 + b11*x1 + b12*x2 + b13*x3 + i1residual u2 = a2 + B21*i1 + b22*x2 + b23*x3 + b24*x4 + b25*x5 + b26*x3*x4 + u2residual where i1residual = the i1 residual variance estimate given in my Mplus output u2 residual = 1 (I think this is what you meant by the probit residual variance being 1?) Then, I calculate my probabilities as shown in the Mplus Users guide using the formula: P(u = 1)  x) = F(a + b*x) = F(t + b*x) For example, I would calculate the probability of u2 being 1 when x3 = 1 and x4 = 1 using the formula: P(u = 1  x3 = 1, x4 = 1 ) = F (a2 + B21*a1 + B21*b13*1 + B21*i1residual + b23*1 + b24*1 + b26*1*1 + 1) Would you help clarify if I am not understanding this? Thank you so much for your time! 


No, that's not right. It's a bit much to explain so I refer to the book page I mentioned. 


Dear professors, I'm trying to calculate predicted probabilities from the probit regression y ON x gender age Y dichotomous X continuous (number of events: 08) Gender dichotomous Age continuous Output: Y ON X 0.179 SEX 0.197 AGE 0.009 Thresholds Y$1 1.931 P (y=1x) = F (1.931 + 0.179 * x + 0.197*sex 0.009*age) 1. Sex and age are control variables. Should I put their sample means on the equations? P (y=1x=0) = F (1.931 + 0.179 * 0 + 0.197*0.472 0.009*39.052) = F (2.189) = 0.014 ... P (y=1x=8) = F (1.931 + 0.179 * 8 + 0.197*0.472 0.009*39.052) = F (0.757) = 0.224 2. Is this correct? 3. Is there any way to do these calculations on MPLUS or do I have to do them by hand? 3. Can I plot these predicted probabilities directly on MPLUS? Thank you!!! 


Looks ok but I wouldn't use the mean of gender (what does that mean) but instead consider one at a time. You can let Mplus do this by using Model Constraint with Loop and Plot. For an example, see the first example of our Topic 11 short course video and handout at http://www.statmodel.com/course_materials.shtml 

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