X = [7,-5,-4,-2]
Y = [1,-1,2,-2]
Are the two random variables independent?
P(X=x) = P(Y=y) = .25 for each value of x and y.
P(X=x|Y=y) is either 0 or .25 for each x and y.
The probability of each ordered pair (i.e. (7,1), (-5,-1), (-4,2),(-2,-2) ) is also .25.
I see no relationship between X and Y; can I say they are independent?
How in general do I approach vectors like this in problems about independent random variables?
The last question is too broad without additional info about the distributions of each random variable.
Given:
To show: independence of $X$ and $Y$ $$\forall x \in \{7,-5,-4,-2\}, \forall y \in \{1,-1,2,-2\}, P(X = x \mid Y = y) = \dfrac14 \stackrel{(1)}{=} P(X = x)$$
Suppose not. \begin{align} & \exists x \in \{7,-5,-4,-2\}, \exists y \in \{1,-1,2,-2\}, P(X = x \mid Y = y) \ne \dfrac14 \\ \stackrel{(2)}{\implies}& \exists x \in \{7,-5,-4,-2\}, \exists y \in \{1,-1,2,-2\}, P(X = x \mid Y = y) = 0 \end{align}
However, \begin{align} & \sum_{x' \in \{7,-5,-4,-2\}} P(X = x' \mid Y = y) \\ &= \sum_{x' \in \{7,-5,-4,-2\} \setminus \{x\}} P(X = x' \mid Y = y) \tag{*} \label1 \\ &= 1 \end{align}
Note that \eqref{1} is a sum of three nonnegative numbers, so as least one of them is greater or equal to $\dfrac13$. This contradicts (2). Hence $X$ and $Y$ are independent.