Find $H(Y,Z)$ and $I(X,Y)$ given $p(x,y,z)$ Given the joint probability distribution function:
$$p(x, y, z)=p(x)p(y|x)p(z|x)$$
with:
$ X: p(X=0)= p(X=1)= 1/4, \ p(X=2)=1/2 ,$
$ Y: p(Y=0| X=0)= p(Y=1| X=1)= 1, \ p(Y=0| X=2) = p(Y=1|X=2)=1/2 ,$
$ Z: p(Z=0|X=0)= p(Z=0|X=1)=p(Z=1|X=2)=1:$


*

*Question 1:


Is it right that  $H(Z|Y)=H(Y│Z)=0$? I am asking because I want to find $H(Y,Z)$ using:
$$
H(Y,Z)=H(Y)+H(Z|Y)=H(Z)+H(Y│Z).
$$


*

*Question 2:
Is it right that I can find $I(X,Y)$ in this way?:
\begin{split}
I(X,Y)&=Η(Y)-H(Y│X) \\
&=1-\sum_{x\in X }\sum_{y\in Y}p(Y=y|X=x)\log_2(p(Y=y|X=x))\\
&=1-(1 \log_2⁡(1)+0+1/2 \log_2(1⁡/2)+0+1 log_2⁡(1)+ 1/2 \log_2(⁡1/2))\\
&=2.\end{split}


Thank you very much!
 A: Note that if $H(Y|Z)=0$ then $Y$ should be a deterministic function of $Z$ which is not the case here obviously. But indeed it can be shown that $Z$ and $Y$ are independent and hence $H(Z|Y)=H(Z)$ and $H(Y|Z)=H(Y)$. 
To see the Independence, we verify in directly. Note that $p(Y=y)$ is given
$$
p(Y=0)=\frac 14+\frac 12\frac 12=\frac 12\implies p(Y=1)= \frac 12\\
p(Z=0)=\frac 14+\frac 14=\frac 12\implies p(Z=1)=\frac 12.
$$
But since $p(X=0|Z=0)=\frac{1\times p(X=0)}{p(Z=0)}=\frac 12$, $p(X=1|Z=0)=\frac{1\times p(X=1)}{p(Z=0)}=\frac 12$, 
$p(X=2|Z=1)=1$:
$$
p(Y=0|Z=0)=\sum_i p(X=i|Z=0)p(Y=0|X=i)=\frac 12\times 1+\frac 12\times 0=\frac 12\\
p(Y=1|Z=0)=\sum_i p(X=i|Z=0)p(Y=1|X=i)=\frac 12\times 0+\frac 12\times 1=\frac 12\\
p(Y=0|Z=1)=\sum_i p(X=i|Z=1)p(Y=0|X=i)=1\times \frac 12=\frac 12,
p(Y=1|Z=1)=\sum_i p(X=i|Z=1)p(Y=1|X=i)=1\times \frac 12=\frac 12.
$$
So $Y$ and $Z$ are independent.
Question 2: note that 
$$
H(Y|X)=-\sum_{x\in X }\sum_{y\in Y}p(Y=y,X=x)\log_2(p(Y=y|X=x))
$$
So what you write there is wrong both the expression and its sign. See that (why?):
$$
H(Y|X)=p(X=2)H(Y|X=2)=\frac 12\log_2 2=\frac 12.
$$
Hence: $I(X;Y)=1-\frac 12=\frac 12$.
