Daniel C comments on johnswentworth’s Shortform

[Daniel C]

(Was in the middle of writing a proof before noticing you did it already)

I believe the end result is that if we have $Y=(Y_1,Y_2)$, $X=(X_1,X_2,X_3)$ with $P\left(Y|X\right)=P(Y_1|X_1,X_3)P(Y_2|X_2,X_3)$ ($X_1$ upstream of $Y_1$, $X_2$ upstream of $Y_2$, $X_3$ upstream of both),

then maximizing $I(X;Y)$ is equivalent to maximizing $I(Y_1;X_1,X_3)$$+I(Y_2;X_2,X_3)$$-I(Y_1;Y_2)$.

& for the proof we can basically replicate the proof for additivity except substituting the factorization $P(X_1,X_2,X_3)=P\left(X_3\right)P\left(X_1|X_3\right)P\left(X_2|X_3\right)$ as assumption in place of independence, then both directions of inequality will result in $I(Y_1;X_1,X_3)$$+I(Y_2;X_2,X_3)$$-I(Y_1;Y_2)$.

[EDIT: Forgot $-I(Y_1;Y_2)$ term due to marginal dependence $P(Y_1,Y_2)\ne P\left(Y_1\right)P\left(Y_2\right)$]