Estimating $P(A,B|X)$ given $P(A|X)$, $P(B|X)$ and $P(B|A)$ I have to deal with a situation where I have the following knowledge:


$P(A)$
$P(B)$
$P(A|X)$
$P(B|X)$
$P(B|A)$


I do now want to compute $P(A,B|X)$.
Firstly, I observed that $P(A,B|X) = P(B|A,X) \cdot P(A|X)$. So what I need is a function which takes as input $P(B), P(B|X)$ and $P(B|A)$ and estimates $P(B|A,X)$.

My intuition is that I don't have enough information to logically deduce a solution. So what I want is an estimation which is as sensible as possible.
Therefore, I made a list of necessary conditions for this function.

Firstly, in order to be probability distribution, it is necessary that


P: $0 \leq P(B|A,X) \leq 1$


Then, I checked some special cases, for the values of $P(B|A)$:


A1: $~P(B|A,X) \rightarrow 1~$ for $~P(B|A) \rightarrow 1, ~P(B|X) \not = 0$
A2: $~P(B|A,X) = P(B|X)~$ if $~P(B|A) = P(B)$
A3: $~P(B|A,X) = 0~$ if $~P(B|A) = 0$


And then also for the values of $P(B|X)$:


X1: $~P(B|A,X) \rightarrow 1~$ for $~P(B|X) \rightarrow 1, ~P(B|A) \not = 0$
X2: $~P(B|A,X) = P(B|A)~$ if $~P(B|X) = P(B)$
X3: $~P(B|A,X) = 0~$ if $~P(B|X) = 0$


Furthermore it should hold that


M: $~P(B|A,X)$ is strictly monotonically increasing in $P(B|A)$ and in $P(B|X)$.


I considered the function $P(B|A,X) = \min(1, \frac{P(B|X) \cdot P(B|A)}{P(B)})$ which satisfies P, A2, A3, X2, X3, M but not A1, X1.

Can you come up with a function that satisfies all of these constraints?

Thanks in advance for ideas :)
I found a function which satisfies all of the above constraints and is not too complicated:

$P(B|A,X) = \min \left(1, P(B|X) \cdot P(B|A) \cdot \left[ \frac{(P(B|X)-P(B))^2 \cdot (P(B|A)-P(B))^2}{(1-P(B|X)) \cdot (1-P(B|A))} + \frac{1}{P(B)}  \right] \right)$

This is what the function looks like in a 3D-plot:

x := P(B|X), y := P(B|A), P(B) := 0.8

I cannot give any reasons for its structure, apart from the fact that it fulfils the above necessary constraints. My intuition is that I don't have enough information to logically deduce a function, which is why I only asked for an estimation which is as sensible as possible.

So, before you downvote me, give me suggestions on how this function should be different!
I found a function which satisfies all of the above constraints and is not too complicated:

$P(B|A,X) = \min \left(1, P(B|X) \cdot P(B|A) \cdot \left[ \frac{(P(B|X)-P(B))^2 \cdot (P(B|A)-P(B))^2}{(1-P(B|X)) \cdot (1-P(B|A))} + \frac{1}{P(B)}  \right] \right)$

This is what the function looks like in a 3D-plot:

x := P(B|X), y := P(B|A), P(B) := 0.8

I cannot give any reasons for its structure, apart from the fact that it fulfils the above necessary constraints. My intuition is that I don't have enough information to logically deduce a function, which is why I only asked for an estimation which is as sensible as possible.

So, before you downvote me, give me suggestions on how this function should be different!