Given two random variables $X$ and $Y$, does there exist a random variable $Z$ which is independent of $X$ but not $Y$? In other words, given two sigma algebras $\mathcal{F}_1$ and $\mathcal{F}_2$, does there exist a sigma algebra which is independent of one and not the other?
I wanted to use this in this context. Let $$X_1 = \begin{cases} 1 & \text{on } \mathbb{Q}^c \cap [0,1] \\ 0 & \text{on } \mathbb{Q} \cap [0,1] \end{cases}$$ and $$X_2 = \begin{cases} 1 & \text{on }[0,\tfrac{1}{2}] \\ -1 & \text{on } (\tfrac{1}{2}, 1]. \end{cases}$$ I wanted to know whether there exists a random variable independent of $X_2 - X_1$ but not $X_2$ or does there exist a sigma algebra $\mathcal{G}$, which is independent of $\sigma(X_2 - X_1)$ but not $\sigma(X_2)$.
This might help.
$$X_2 - X_1 = \begin{cases} 2 & \text{on } \mathbb{Q}^{c} \cap [0,\tfrac{1}{2}] \\ 1 & \text{on } \mathbb{Q} \cap [0,\tfrac{1}{2}] \\ 0 & \text{on } \mathbb{Q}^{c} \cap (\tfrac{1}{2}, 1] \\ -1 & \text{on } \mathbb{Q} \cap (\tfrac{1}{2}, 1]. \end{cases}$$.
Therefore, $$\sigma(X_2 - X_1) =\{\emptyset, \mathbb{Q} \cap [0,\tfrac{1}{2}], \mathbb{Q}^{c} \cap [0,\tfrac{1}{2}],\mathbb{Q} \cap (\tfrac{1}{2},1],\mathbb{Q}^{c} \cap (\tfrac{1}{2},1],[0,1]\}. $$ $$\sigma(X_2) = \{\emptyset, [0,\tfrac{1}{2}],(\tfrac{1}{2},1],[0,1]\}$$
You cannot do this on the n same sample space. For example if $\Omega=\{1,2\}$, $P(\{1\})=P(\{2\})=\frac 1 2$ and $X(1)=1, X(2)=2$, $Y(1)=2,Y(2)=1$ then the only random variables independent of $X$ are the constants $1$ and $2$. So $Z$ must be a constant, hence independent of $Y$ also.