How to calculate this joint PMF? Consider three random variables X, Y, and Z, associated with the same experiment. The random variable X is geometric with parameter p∈(0,1). If X is even, then Y and Z are equal to zero. If X is odd, (Y,Z) is uniformly distributed on the set S={(0,0),(0,2),(2,0),(2,2)}. The figure below shows all the possible values for the triple (X,Y,Z) that have X≤8. (Note that the X axis starts at 1 and that a complete figure would extend indefinitely to the right.)

Use x, p express:
1) If x is odd and (y,z)∈{(0,0),(0,2),(2,0),(2,2)}, pX,Y,Z(x,y,z)= 
2) If x is even and (y,z)=(0,0), pX,Y,Z(x,y,z)= 
3) If x is odd, pX,Y(x,2)= 
4) pY(2)=
5) var(Y+Z∣X=5)=
I can't really understand this question,  when X is geometric with parameter p∈(0,1), how can I join it with  (Y,Z)? Doesn't it mean X is odd or even with p and 1-p? 
 A: 
1) If $x$ is odd and $(y,z)∈\{(0,0),(0,2),(2,0),(2,2)\}$, $p_{X,Y,Z}(x,y,z)=$

We are given the distribution of $(Y,Z)$ when $X$ is odd.   It's uniform over four points.
Next, if $X$ is odd then the conditional probability of $X=x$ (where $x$ is a particular odd value) will be the probability of a success occurring on the $(x+1)/2$ -th odd value (given it does not occur on any even ones).  Thus the conditioned value is Geometrically Distributed.
$$\begin{align}
p_{_{Y,Z}}(y,z\mid \operatorname{Odd}(X)) & = \frac 1 4 \;\mathbf 1_{(y,z)\in \{(0,0),(0,2),(2,0),(2,2)\}}
\\[2ex]
p_{_X}(x\mid \operatorname{Odd}(X)) & = p(1-p)^{(x-1)/2} \mathbf 1_{x\in \Bbb Z^+\setminus \Bbb 2Z}
\\[2ex]
p_{X,Y,Z}(x,y,z\mid \operatorname{Odd}(X))
& =p_{_X}(x\mid \operatorname{Odd}(X)) \; p_{_{Y,Z}}(y,z\mid \operatorname{Odd}(X))
\\[1ex] & = \frac 1 4 p(1-p)^{(x-1)/2} \mathbf 1_{x\in \Bbb Z^+\setminus \Bbb 2Z}\;\mathbf 1_{(y,z)\in \{(0,0),(0,2),(2,0),(2,2)\}} 
\end{align}$$

2) If $X$ is even and $(Y,Z)=(0,0)$, $p_{X,Y,Z}(x,y,z)=$ 

We are given the distribution of $(Y,Z)$ when $X$ is even; it's a certainty at one point.
$$p_{Y,Z}(y,z \mid \operatorname{Even}(X)) = \mathbf 1_{y=0, z=0}$$
All we need is the conditional distribution of $X$ given that it is even.
As before we argue that if $N= X/2 \mid_{X\in 2\Bbb Z^+}$ then $N$ will be geometrically distributed.
$$p_X(x\mid \operatorname{Even}(X)) = p(1-p)^{x/2-1}$$
Put these together.

3) If $X$ is odd, $p_{X,Y}(x,2\mid \operatorname{Odd}(X))=$

Similar.

4) $p_Y(2)=$

$p_Y(2\mid \operatorname{Odd}(X)) = \frac 1 2
\\[2ex]
p_Y(2) = p_Y(2\mid \operatorname{Odd}(x)) \; \mathsf P(\operatorname{Odd}(x))
$
Now, what is: $\operatorname P(\operatorname{Odd}(X))\;$?

5) $\mathsf{Var}(Y+Z\mid X=5)=$

We know $\mathsf P_{Y,Z}(y,z\mid X=5) = \frac 1 4 \;\mathbf 1_{(y,z)\in \{(0,0), (0,2), (2,0), (2,2)\}}$ so then:
$$\mathsf{Var}(Y+Z\mid X=5)= \frac 1 {\color{red}{4}}(0+4+4+16) - \frac 1{16}(0+2+2+4)^2 
\\ = 2$$
