A discrete random vector $(X,Y)$ has pmf given by
$p_{X,Y}(1,−1) = p_{X,Y}(−1,1) = \frac{1}{2}$.
(a) Check whether the random variables $X$ and $Y$ are independent.
In this question as the $X$ and $Y$ are only taking the value $1$ and $-1$ so if we can show that for all four scenario pair $\left\{(X=1, Y = 1), (X= 1, Y= -1),(X=-1,y =-1) and (X=-1,Y = 1)\right\}$ the joint PMF of the random variable $p_{X,Y}(x,y) = p_{X}(x)p_{Y}(y)$ then it can be shown as independent. But Shall I take the each value and check for all four cases or is there any smart way to that?
Best Answer
it is easy to show that $X$ and $Y$ are not independent. This because using the given data you get that
$$P_{XY}(-1;-1)=P_{XY}(1;1)=0$$
these are 2 sufficient counterexamples against independence
In fact, using the given data you can derive the following contingency table
where you can observe that, for both marginals, you have $P(1)=P(-1)=0.5$ but, as an example, if $X=-1$, the rv Y can take only the value 1, being $P(Y=1|X=-1)=1$
This contraddicts the definition of independence and the probability of Y depends about X occurrence