Parameters are tf.Variables while data is not but feed in as constant tensors (or converted to them). RooFit allows to interchange them (think of a gaussian, mu or x can be interchanged, not in TF). Asked Rafa about it, he cannot think of a use-case where interchangeability could be useful. You have something in mind?
Yes, I agree! But it's a difference to RooFit, so just document it. If no one cares, even better. One use case, that implicitly won't work with that, is that we can't integrate over a parameter.
Well, we can always define a different function with parameter a as dimension and explicit dependence. So instead of f(x; a, b) just f(x, a; b) -> integrate over a -> f(x; b). This should work out.
Not sure if I understood the issue in here. I think from the point of view of our current fits, unless I'm missing something, I don't see the use case. In principle people may want to have it, but from my side I wouldn't put this as top priority....
@mayou36 commented on Tue Oct 02 2018
Documentation of limitations
@apuignav commented on Tue Oct 02 2018
Mmm, I am not sure I follow this... What's the idea of this limitation?
@mayou36 commented on Tue Oct 02 2018
Parameters are
tf.Variableswhile data is not but feed in as constant tensors (or converted to them). RooFit allows to interchange them (think of a gaussian, mu or x can be interchanged, not in TF). Asked Rafa about it, he cannot think of a use-case where interchangeability could be useful. You have something in mind?@apuignav commented on Tue Oct 02 2018
Ah ok! Sure! However, I wouldn't say this is a limitation, it's almost by construction of a likelihood function, isn't it?
@mayou36 commented on Tue Oct 02 2018
Yes, I agree! But it's a difference to RooFit, so just document it. If no one cares, even better. One use case, that implicitly won't work with that, is that we can't integrate over a parameter.
@apuignav commented on Tue Oct 02 2018
Mmm, that seems bad... @rsilvaco?
@mayou36 commented on Tue Oct 02 2018
Well, we can always define a different function with parameter a as dimension and explicit dependence. So instead of f(x; a, b) just f(x, a; b) -> integrate over a -> f(x; b). This should work out.
@rsilvaco commented on Tue Oct 02 2018
Not sure if I understood the issue in here. I think from the point of view of our current fits, unless I'm missing something, I don't see the use case. In principle people may want to have it, but from my side I wouldn't put this as top priority....
@rsilvaco commented on Tue Oct 02 2018
Yes @mayou36, is probably not strictly the same as RooFit does, but this should naively work ..
@apuignav commented on Tue Oct 02 2018
Fine, we'll accept requests afterwards :-)