I might misunderstand something or made a mistake and I’m not gonna try to figure it out since the post is old and maybe not alive anymore, but isn’t the following a counter-example to the claim that the method of constructing S described above does what it’s supposed to do?
Let X and Y be independent coin flips. Then S will be computed as follows:
X=0, Y=0 maps to uniform distribution on {{0:0, 1:0}, {0:0, 1:1}} X=0, Y=1 maps to uniform distribution on {{0:0, 1:0}, {0:1, 1:0}} X=1, Y=0 maps to uniform distribution on {{0:1, 1:0}, {0:1, 1:1}} X=1, Y=1 maps to uniform distribution on {{0:0, 1:1}, {0:1, 1:1}}
But since X is independent from Y, we want S to contain full information about X (plus some noise perhaps), so the support of S given X=1 must not overlap with the support of S given X=0. But it does. For example, {0:0, 1:1} has positive probability of occurring both for X=1, Y=1 and for X=0, Y=0. i.e. conditional on S={0:0, 1:1}, X is still bernoulli.
I might misunderstand something or made a mistake and I’m not gonna try to figure it out since the post is old and maybe not alive anymore, but isn’t the following a counter-example to the claim that the method of constructing S described above does what it’s supposed to do?
Let X and Y be independent coin flips. Then S will be computed as follows:
X=0, Y=0 maps to uniform distribution on {{0:0, 1:0}, {0:0, 1:1}}
X=0, Y=1 maps to uniform distribution on {{0:0, 1:0}, {0:1, 1:0}}
X=1, Y=0 maps to uniform distribution on {{0:1, 1:0}, {0:1, 1:1}}
X=1, Y=1 maps to uniform distribution on {{0:0, 1:1}, {0:1, 1:1}}
But since X is independent from Y, we want S to contain full information about X (plus some noise perhaps), so the support of S given X=1 must not overlap with the support of S given X=0. But it does. For example, {0:0, 1:1} has positive probability of occurring both for X=1, Y=1 and for X=0, Y=0. i.e. conditional on S={0:0, 1:1}, X is still bernoulli.