The fibring identity #
The proof of the fibring identity, which is a key component of the proof of PFR.
Main statement #
sum_of_rdist_eq
: If $Y_1,Y_2,Y_3,Y_4$ are independent, then $d[Y_1; Y_2] + d[Y_3; Y_4]$ is equal to the sum of $$d[Y_1+Y_3; Y_2+Y_4] + d[Y_1|Y_1+Y_3; Y_2|Y_2+Y_4] $$ and $$ I[Y_1+Y_2 : Y_2 + Y_4 | Y_1+Y_2+Y_3+Y_4].$$
If $Z_1, Z_2$ are independent, then $d[Z_1; Z_2]$ is equal to $$ d[\pi(Z_1);\pi(Z_2)] + d[Z_1|\pi(Z_1); Z_2 |\pi(Z_2)]$$ plus $$I( Z_1 - Z_2 : (\pi(Z_1), \pi(Z_2)) | \pi(Z_1 - Z_2) ).$$
[d[X;Y]\geq d[\pi(X);\pi(Y)].]
The conditional Ruzsa Distance step of sum_of_rdist_eq
The conditional mutual information step of sum_of_rdist_eq
Let $Y_1,Y_2,Y_3$ and $Y_4$ be independent $G$-valued random variables. Then $$d[Y_1-Y_3; Y_2-Y_4] + d[Y_1|Y_1-Y_3; Y_2|Y_2-Y_4] $$ $$ + I[Y_1-Y_2 : Y_2 - Y_4 | Y_1-Y_2-Y_3+Y_4] = d[Y_1; Y_2] + d[Y_3; Y_4].$$
Let $Y_1,Y_2,Y_3$ and $Y_4$ be independent $G$-valued random variables. Then $$d[Y_1+Y_3; Y_2+Y_4] + d[Y_1|Y_1+Y_3; Y_2|Y_2+Y_4] $$ $$ + I[Y_1+Y_2 : Y_2 + Y_4 | Y_1+Y_2+Y_3+Y_4] = d[Y_1; Y_2] + d[Y_3; Y_4].$$