Generating a binary probability combination function [closed]

Question

I have been trying to develop a function that can combine two probabilities using the rules:

$f(x,y)\in C^\infty (\mathbb{R}^{2})$

$f(x,y)=f(y,x)$

$f(x,1-x)=\frac{1}{2}$

$f(1-x,1-y)=1-f(x,y)$

$f(0,x)=0$

$f(x,\frac{1}{2})=x$

$f(x,1)=1$

$f_x(x,y)\geq 0$

$f(0,1)$ does not exist. All other points $(x,y) \in [0,1]\times [0,1]$ should be defined.

I do not believe any polynomial solution exists. I am wondering if a solution exists and if so how to find it. I believe that if such a solution exists, it would be of the form, or of a similar form, to $a+b^{c}$ where $a$, $b$, and $c$ are linear or quadratic functions of $x$ and $y$, and maybe quartic at worst.

The intent of this function is to make an iterative solver of the game binario based solely on probability. I am aware that this function would not be fully able to solve the game but I am still interested in its existence.

Answer 1

If you want all your conditions to hold for all real $x,y$, then that is impossible. Indeed, if $f(x,1-x)=\frac12$ and $f(0,x)=0$ for all real $x$, then $0=f(0,1)=f(0,1-0)=\frac12$, which is a contradiction.

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Also, the identities $f(0,x)\equiv0$ and $f(x,1)\equiv1$ contradict each other, regarding the value of $f(0,1)$.

Other contradictions can be obtained using the symmetry $f(x,y)\equiv f(y,x)$.

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Looking back at the title of your post about "a binary probability combination function", let me make a wild guess: You actually want your function $f$ to be the joint cumulative distribution function (cdf) of some pair $(X,Y)$ of real-valued random variables (r.v.'s), right? If so, then your conditions present much more trouble for $f$ than just being impossible to define at the two points, $(1,0)$ and $(0,1)$. Indeed, if $f$ were the joint cdf of a pair $(X,Y)$ of r.v.'s, then we would have $$P(1/2<X\le1,1/2<Y\le1)=f(1,1)-f(1,1/2)-f(1/2,1)+f(1/2,1/2) \\ =1-1-1+1/2=-1/2<0,$$ which contradicts the nonnegativity of any probability.