Why dose I(x;y)>=I(E;y) hold in the `Proof of Theorem 1'

[FlyGreyWolf]

according the data processing inequality, we have: X->Z->Y, I(X;Y)<=I(Z;Y)

In the proof of theorem 1'' of your paper, for x->g(x)->y, is this supposed to be I(x;y)<=I(g(x);y)" instead of `` I(x;y)>=I(E;y)'' ?

[donghao51]

hello, please check the data processing inequality (https://en.wikipedia.org/wiki/Data_processing_inequality), we have X->Y->Z, I(X;Y)>=I(X;Z)

[FlyGreyWolf]

hello, please check the data processing inequality (https://en.wikipedia.org/wiki/Data_processing_inequality), we have X->Y->Z, I(X;Y)>=I(X;Z)

OK, I got it. Thanks.