johnswentworth comments on Natural Latents: The Math

[johnswentworth]

Yeah, that’s right.

The secret handshake is to start with ”$X$ is independent of $Y$ given $Z$” and ”$X$ is independent of $Z$ given $Y$”, expressed in this particular form:

$P[X,Y,Z]=P\left[X|Z\right]P[Y,Z]=P\left[X|Y\right]P[Y,Z]$

… then we immediately see that $P\left[X|Z\right]=P\left[X|Y\right]$ for all $X,Y,Z$ such that $P[Y,Z]>0$.

So if there are no zero probabilities, then $P\left[X|Z\right]=P\left[X|Y\right]$ for all $X,Y,Z$.

That, in turn, implies that $P\left[X|Z\right]$ takes on the same value for all Z, which in turn means that it’s equal to $P\left[X\right]$. Thus $X$ and $Z$ are independent. Likewise for $X$ and $Y$. Finally, we leverage independence of $Y$ and $Z$ given $X$:

$P[X,Y,Z]=P\left[Y|X\right]P\left[Z|X\right]P\left[X\right]$

$=P\left[Y\right]P\left[Z\right]P\left[X\right]$

(A similar argument is in the middle of this post, along with a helpful-to-me visual.)

[LawrenceC]

Right, the step I missed on was that P(X|Y) = P(X|Z) for all y, z implies P(X|Z) = P(X). Thanks!