Hm, it sounds like you’re claiming that if each pair of x, y, z are pairwise independent conditioned on the third variable, and p(x, y, z) =/= 0 for all x, y, z with nonzero p(x), p(y), p(z), then ?
I tried for a bit to show this but couldn’t prove it, let alone the general case without strong invariance. My guess is I’m probably missing something really obvious.
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[X|Z]P[Y,Z]=P[X|Y]P[Y,Z]
… then we immediately see that P[X|Z]=P[X|Y] for all X,Y,Z such that P[Y,Z]>0.
So if there are no zero probabilities, then P[X|Z]=P[X|Y] for all X,Y,Z.
That, in turn, implies that P[X|Z] takes on the same value for all Z, which in turn means that it’s equal to P[X]. 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[Y|X]P[Z|X]P[X]
=P[Y]P[Z]P[X]
(A similar argument is in the middle of this post, along with a helpful-to-me visual.)
Hm, it sounds like you’re claiming that if each pair of x, y, z are pairwise independent conditioned on the third variable, and p(x, y, z) =/= 0 for all x, y, z with nonzero p(x), p(y), p(z), then ?
I tried for a bit to show this but couldn’t prove it, let alone the general case without strong invariance. My guess is I’m probably missing something really obvious.
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[X|Z]P[Y,Z]=P[X|Y]P[Y,Z]
… then we immediately see that P[X|Z]=P[X|Y] for all X,Y,Z such that P[Y,Z]>0.
So if there are no zero probabilities, then P[X|Z]=P[X|Y] for all X,Y,Z.
That, in turn, implies that P[X|Z] takes on the same value for all Z, which in turn means that it’s equal to P[X]. 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[Y|X]P[Z|X]P[X]
=P[Y]P[Z]P[X]
(A similar argument is in the middle of this post, along with a helpful-to-me visual.)
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!