$P(x,z)P(x',z)=P(x,z,x')P(z)$? Why is $P(x,z)P(x',z)=P(x,z,x')P(z)$? I have been able to show this identity using the idea that $P(x|z)P(x'|z)=P(x,x'|z)$. The problem is that I do not understand the intuition behind $P(x,z)P(x',z)=P(x,z,x')P(z)$. Also I am interested in a mathematical proof of this without using conditional probabilities(which I did in my proof). Can you help with the intuition and/or proof?
I am working under the assumptions that $x'$ and $x$ are independent but $x$ and $z$ are dependent and $x'$ and $z$ are dependent.
Thanks!
 A: This "identity" is false. For example, let $X$ be the event that a fair coin toss comes up heads. Let $X'$ be the event where that same coin toss comes up tails instead. (That means exactly one of $X$ and $X'$ will happen.) Let $Z$ be the event that a second, independent fair coin toss comes up heads.
In this scenario, $X$ and $X'$ are both independent of $Z$, but $X$ and $X'$ are very much not independent of each other. We have:

*

*$P(X,Z) = \frac 1 2 \cdot \frac 1 2 = \frac 1 4$

*$P(X',Z) = \frac 1 2 \cdot \frac 1 2 = \frac 1 4$

*$P(X,X',Z) = 0$

*$P(Z) = \frac 1 2$
In other words, the LHS of your identity is $\frac 1 {16}$ but the RHS is $0$.
If you add some more independence assumptions you could create a true version of the identity. But anyway this explains why you had trouble finding an intuitive proof! Probably you made some algebra mistake in your attempted proof-by-formulas. But the good news is that you didn't fall too far astray due to your very valuable reflex of feeling uncomfortable anytime you see a result "proved" without an intuitive explanation.
