conditional probability which their condition have $XOR and Z=X+Y$

Question

Let $X$ and $Y$ be two independent binary random variable with the same alphabet {$0,1$},ie,$Pr(0)=Pr(1)=\frac{1}{2}$

Define $I(X;Y|Z)=H(X|Z)-H(X|Y,Z)$

$1.$Let $Z=X+Y$,Find $I(X;Y|Z)$

$2.$Let $Z=X ⊕Y$, Find $I(X;Y|Z)$. ⊕ means XOR operation

For these two questions,i don't know how to find their probability,can anyone teach me

By the way, i know the $I$(information) and $H$(entropy) are not the probability,$Pr(X;Y|Z)$,but i need the probability before calculating them

Answer 1

You asked for the probabilities. These are:

X    Y   X+Y  X⊕Y  Prob
0    0    0    0    1/4 
0    1    1    1    1/4
1    0    1    1    1/4
1    1    2    0    1/4

For $\mathbb P(X=x \mid Z=z)$, in Part 1 you have, as spaceisdarkgreen says,

• $\mathbb P(X=0 \mid X+Y=0)=1$, $\,\mathbb P(X=1 \mid X+Y=0)=0$,
• $\mathbb P(X=0 \mid X+Y=1)=\frac12$, $\mathbb P(X=1 \mid X+Y=1)=\frac12$,
• $\mathbb P(X=0 \mid X+Y=2)=0$, $\,\mathbb P(X=1 \mid X+Y=1)=1$

In Part 2 you have

• $\mathbb P(X=0 \mid X⊕Y=0)=\frac12$, $\mathbb P(X=1 \mid X⊕Y=0)=\frac12$,
• $\mathbb P(X=0 \mid X⊕Y=1)=\frac12$, $\mathbb P(X=1 \mid X⊕Y=1)=\frac12$

Given both $Y$ and $Z$, the value of $X$ is certain in both parts 1 and 2, since in part 1 you have $X=Z-Y$ while in part 2 you have $X=Z⊕Y$