An entropy and mutual information problem

Question

Let's suppose we have 4 random variables X,Y,Z and T and that the following equations hold about the entropy: $$H(T|X)=H(T)$$ $$H(T|X,Y)=0$$ $$H(T|Y)=H(T)$$ $$H(Y|Z)=0$$ $$H(T|Z)=0$$

I want to prove the following inequality: $$I(X;Z)\geq H(T)$$

What I have done: \begin{eqnarray*} I(X;Z)\geq H(T) &\Leftrightarrow & H(X)-H(X|Z)\geq H(T)\\ &\Leftrightarrow & H(X)-H(X|Z)\geq H(T|X)\\ &\Leftrightarrow & H(X)-H(X|Z)\geq H(X,T)-H(X)\\ &\Leftrightarrow & 2H(X)\geq H(X,T)+H(X|Z)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\\ \end{eqnarray*}

Also, the following properties generally hold for entropy: $$H(X)\geq H(X|Z)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$ $$H(X)+H(T)\geq H(X,T)\ \ \ \ (3)$$

Adding $(2)$ and $(3)$ side-by-side we get: $$2H(X)+H(T)\geq H(X,T)+H(X|Z)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$$

So if $H(T)=0$, then $(1)$ is always true, but I am stuck here.

Answer 1

$H(Y|Z)=0$ and $H(T|Z)=0 \Rightarrow H(T+Y|Z)=H(T|Z)+H(Y|Z)-I(T;Y|Z)=0$

Then, we can use the fact that if $H(A/B)=0$ then for any $C$ , $I(B;C) \geq I(A;C)$

(intuitive if one considers that B entirely determines A)

proof (not very elegant...) :

$I(B;C)=H(B)+H(C)-H(B+C)=H(A+B)-H(A|B) +H(C)-H(A+B+C)+H(A/B+C)=H(A+B)+H(C)-H(A+B+C)=I(A+B;C) \geq I(A;C)$

So, we have that : $I(X;Z)\geq I(X;Y+T)$

and we can proceed :

$I(X;Z)\geq I(X;Y+T)=H(X)+H(Y+T)-H(X+Y+T)=H(X)+H(Y+T)-H(X+Y)-H(T/X+Y)=H(X)+H(Y)+H(T/Y)-H(X+Y)=H(X)+H(Y)+H(T/Y)-H(X)-H(Y)+I(X;Y)=H(T)+I(X;Y) \geq H(T)$