probability: arithmetic with random variables.

Question

I have a question of using arithmetic on random variables.
Please refer to the following question, to which I will present my solution using the arithmetic (which I thought it's correct but actually not).

$$P(X+Y+Z=1|X-Y=0) => P( [X+Y+Z=1]\text{ AND }[X-Y=0])/P(X-Y=0) $$

for the top part :

$$[X-Y=0] => X=Y$$

so

$$ P( [X+Y+Z=1]\text{ AND }[X-Y=0])=P(X+X+Z=1)=P(2X+Z=1)$$

But this is not correct. Why?

Answer 1

What you can conclude it is the events $$\{X+Y+Z=1\} \cap \{X=Y\} = \{2X+Z=1\} \cap \{X=Y\}$$

$\cap$ means AND.

Note that the event $\{X+Y+Z=1\} \cap \{X=Y\} \neq \{2X+Z=1\}$ in general. For example, let be $w$ such that $X(w)=0$ and $Y(w)=Z(w)=1$. Such $w$ is a possibility in $\{2X+Z=1\}$ but it is not in $\{X+Y+Z=1\} \cap \{X=Y\}$. Then the events $\{X+Y+Z=1\} \cap \{X=Y\}$ and $\{2X+Z=1\}$ aren't equal, so $$P(\{X+Y+Z=1\} \cap \{X=Y\}) \neq P(\{2X+Z=1\})$$

Answer 2

$P([X+Y+Z=1]\,and [X-Y=0]) = P(X=0\,and\,Y=0\,and\,Z=1)= (\frac{1}{2})^3$