Suppose starting with $0<p,q<1$, mapping $f(p,q)$ that maps $(p,q)$ to $f(p,q) = \left(\frac{p}{1-(1-p)q}, \frac{qp}{1-(1-p)q}\right)$.
Is there an invariant for this mapping $f$, such as $g$, so that $g(p,q) = g(f(p,q))$?
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It looks like a logical operator that would check that p and q remain between 0 and 1 is an invariant of the map, sign(p) for example, or sign(1-p). I take it you are looking for something less trivial? – user2309840 Oct 27 '16 at 03:47
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yes, i'm looking for something less trivial. – athos Oct 27 '16 at 03:48
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OK, I found it, replace $q$ with $1-q$, the symmetry will show. – athos Oct 27 '16 at 11:04
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It is perfectly OK to post an answer to your own question. – dxiv Oct 28 '16 at 21:59
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@dxiv I replied in another post: http://math.stackexchange.com/a/1987471/26632 – athos Oct 28 '16 at 22:32
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Thanks, though that iteration looks a bit different (there is no $p$ factor in the numerator of the second coordinate of the transform, for one thing). Also, it's not clear to me which is the invariant $g(p,q)$ in that case, either. – dxiv Oct 29 '16 at 00:55