Range of predicted values in fractional response

I am using fractional response regression for dependent variable that can take values anywhere between 0 and 1, inclusive, in the same spirit as the Papke/Wooldridge. My setup looks like:

 E(y|X) = G(XB) = 1 / (1 + exp(XB))


where

 0 <= y <= 1.


Now, when the predictions, y_hat, are made, it looks like the predicted values range from (0,1), and not [0,1] because the function G above is specified as the logistic function.

Is this right?

It was not clear in the Papke/Wooldridge paper because the references were made to the unit interval. I am not sure if the unit interval means (0,1) or [0,1] for the range of the predicted values.

The paper I am referring to is

ECONOMETRIC METHODS FOR FRACTIONAL RESPONSE VARIABLES WITH AN APPLICATION TO 401 (K) PLAN PARTICIPATION RATES

LESLIE E. PAPKE AND JEFFREY M. WOOLDRIDGE

JOURNAL OF APPLIED ECONOMETRICS, VOL. 11, 619-632 (1996)

In the processing editing my initial question, I found the answer to my question in p.612 of the paper, which was

 This ensures that the predicted values of y lie in the interval (0,1).

• Please give a complete reference and quote the relevant section or sections in context Sep 16, 2016 at 6:13
• In the process of finding the quote in the paper, my question was answered and posted above. Thanks. Sep 16, 2016 at 14:46
• Hi David, Thanks. If I reopen your question could you post a brief answer? Sep 16, 2016 at 22:24
• Sure, I can do that, but do you mean that I answer my own question? Sep 19, 2016 at 13:24
• yes, I mean exactly that. It's fine to answer your own question; often the asker is best placed to do so. Sep 19, 2016 at 14:57

To elaborate on my copied and pasted line from the original paper ---

 "This ensures that the predicted values of y lie in the interval (0,1)"


This fact can be seen from the following:

 First, let E(y|X) = G(XB) = 1 / (1 + exp(XB)).

Second, if XB -> \infty, then, G(XB) approaches 0, since G approaches
1 / (1 + a large number approaching \infty) -> 0.
But, it never takes on the value 0.

Third, if XB -> \minus \infty, then, G(XB) approaches 1, since G approaches
1 / (1 + a small number approaching 0) -> 1.
But, it never takes on the value 1.