I am wondering whether $I(X; Z) \geq H(T)$ when the following conditions hold:
- $H(T | X) = H(T)$
- $H(T | Y) = H(T)$
- $H(T | X, Y) = 0$
- $H(Y | Z) = H(T | Z) = 0$
- $X, Y, Z, T$ are discrete.
I know first two conditions happen when $X$ is independent of $T$; and $Y$ is independent of $T$. The third condition happens when $T$ can be determined by the realization of $(X, Y)$. Finally, the random variables meet the fourth condition when $X$ and $Y$ can be determined by the realization of $Z$.
I can think of several examples but are quite similar to the following:
Consider $\Omega = \{ 1, 2, 3, 4\}$ with uniform distribution. Define the following subsets of $\Omega$: $A = \{ 1, 2\} $, $B = \{ 2, 3 \}$ and $C = \{ 3, 1 \}$. Now let $X = I_A$, $Y = I_B$, $T = I_C$ and $Z = (Y, T)$, where $I_E$ denotes the indicator function of the set $E$.
It can be easily checked that this set of random variables meet the conditions. In this case we have:
$$ I(X; Z) = H(T) = 1 $$
Is the $I(X; Z) \geq H(T)$ inequality always valid when the former conditions are met?
Thank you in advance.