Consider $P(X,Y)$ discrete and two chains $X \rightarrow Z$ and $X \rightarrow Z'$. Does then following inequality hold?
$$ P(X,Y,Z) = P(X,Y,Z') = P(X,Y) $$
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In general, no. We have \begin{align} \mathbb{P}(X,Y,Z) = \mathbb{P}(Z|X,Y)\mathbb{P}(X,Y) \end{align} and \begin{align} \mathbb{P}(X,Y,Z') = \mathbb{P}(Z'|X,Y)\mathbb{P}(X,Y) \end{align} which does not satisfy $\mathbb{P}(Z|X,Y)=\mathbb{P}(Z'|X,Y)$ in general. However, you did not specify all dependencies between $X,Y,Z,Z'$. For equality to hold, we could satisfy $\mathbb{P}(Z|X,Y)=\mathbb{P}(Z'|X,Y)=1$ if $Z=f(X,Y)$ and $Z'=g(X,Y)$ for invertible functions $f,g$. The idea here is that knowing $X$ or $Y$ should give enough information to reconstruct $Z$ and $Z'$ perfectly, and hence, their conditional probabilities are one.
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$\begingroup$ is it possible to Establishment Establishment relationship between $P(X,Y,Z)$ and $P(X,Y,Z')$? $\endgroup$– CesareCommented Dec 9, 2021 at 11:20
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$\begingroup$ At first glance, the two relationships I saw were $\mathbb{P}(X,Y,Z)=\frac{\mathbb{P}(Z|X,Y)}{\mathbb{P}(Z'|X,Y)} \mathbb{P}(X,Y,Z')$ and $|\mathbb{P}(X,Y,Z)-\mathbb{P}(X,Y,Z')|=\mathbb{P}(X,Y)|\mathbb{P}(Z|X,Y)-\mathbb{P}(Z'|X,Y)|$. $\endgroup$– rey_patoCommented Dec 9, 2021 at 12:31
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$\begingroup$ I have managed to specify better the problem I want to address. Would you mind checking this: math.stackexchange.com/questions/4330654/… $\endgroup$– CesareCommented Dec 12, 2021 at 5:34