Let $X,A,Y,B,C,D$ be random binary variables. $D$ is independent from $X,A,C$ and $C$ is independent from $Y,B,D$.

Is it true that:

If $I(Y:B|D=0)\leq \epsilon$ then $I(X\oplus Y:A\oplus B|C=0,D=0)\leq \epsilon \times I(X:A|C=0)$.

This result seems intuitive and I managed to show it for $\epsilon=0$ and $\epsilon=1$, but maybe in the general case we should have a function $f(\epsilon)$ on the right hand side and not simply $\epsilon$. If the result is true, how would you show it using the data processing inequality, the chain rule etc.?

Thanks for your help.


For the case $\epsilon=0$, assume $A=X$ is uniform, and $B=Y$, and $C$ and $D$ are deterministically $0$.

Then you say if $H(B) \le 0$ then $H(A\oplus B)\le 0$.

But $H(A\oplus B) = H(A) = 1 \neq 0$, which contradicts your claim.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.