Hi Chenying,

You might be having an issue with data separation. This means you have a "perfect" predictor, which means that if X=1, then Y always equals 1 (or always 0). In this case, there is no exact solution to the maximum likelihood equation and the computer returns an approximate solution. That could explain why changing the tolerance (epsilon) affects the estimate. HOWEVER, I don't know why the SIGN would change. Maybe the response only has four 1's? Or the predictor only has four 1's?

Example, which borrows some code from Renaud:

dat

I cannot reproduce the problem

> set.seed(123)

> dat

> fm1 fm2 fm3

> rbind(coef(fm1), coef(fm2), coef(fm3))

(Intercept) x

[1,] -1.389146 0.2905372

[2,] -1.389147 0.2905346

[3,] -1.389147 0.2905346

Hi Chenying,

You might be having an issue with data separation. This means you have a "perfect" predictor, which means that if X=1, then Y always equals 1 (or always 0). In this case, there is no exact solution to the maximum likelihood equation and the computer returns an approximate solution. That could explain why changing the tolerance (epsilon) affects the estimate. HOWEVER, I don't know why the SIGN would change. Maybe the response only has four 1's? Or the predictor only has four 1's?

Example, which borrows some code from Renaud:

dat