How to know when two random variables are independent? A pair of r.v (X,Y) is equally likely to be of any of these pairs of values $(0,1), (1,0), (-1,0), (0,-1)$. Both X and Y have mean = 0. 


*

*$E[XY]=$? 

*Is $Var(X+Y) = Var(X) + Var (Y)$?


I know that if X and Y are independent, $E[XY] = E[X]E[Y] = µxµy$. But my doubt is that nowhere in the question above it is stated that X and Y are independent. Same with the second question, I know that it is valid only if X and Y are independent (Please correct me if I am wrong). 
So how do I know if they are independent only using the info from the question above?
 A: To solve this one, let's go back to first principles: what does it actually mean for two events $A, B$ to be independent?
There're many ways to formulate this, but we'll go with this one: the probability of event $A$ conditioning on event $B$ is equal to the probability of event $A$. That is: $P(A|B) = P(A)$, or alternatively, $P(A \cap B) = P(A)P(B)$. For two random variables to be independent, we treat each assignment to $k$ variables as $k$ events. This guarantee must hold for all value assignments to the random variables for them to be independent.
What we are given here? We have the joint distribution of $X$ and $Y$. Let's just take the first point, $(0,1)$, and go from there:
Claim. Random variables $X, Y$ are independent.
Disproof. By contradiction; suppose $X, Y$ are independent. Then $P(X = 0 \cap Y=1) = P(X=0)P(Y=1)$. We can then write $\frac{1}{4} = \frac{1}{2}\cdot\frac{1}{4}$, which is clearly false, so $X$ and $Y$ are not independent.
Hope this helps!
A: You don't need to determine whether the random variables are independent or not to answer the questions in the text of your question. 
For discrete random variables $X$ and $Y$ taking on values in the respective sets $\{u_i\}$ and $\{v_j\}$, $$E[XY] = \sum_i\sum_j u_iv_jp_{X,Y}(u_i, v_j)\tag{1}$$ where $p_{X,Y}(u_i, v_j)$ is the joint probability mass function of the random variables. For your problem, $\{u_i\} = \{v_j\} = \{-1,0,+1\}$ and it is easily seen that in the nine terms in $(1)$, either $u_iv_j = 0$ or $p_{X,Y}(u_i, v_j)=0$ and so $E[XY] = E[X]E[Y]$ just as would happen if $X$ and $Y$ were independent (they are not independent in this case).
Turning to the matter of the variances, $X+Y$ is a random variable taking on values $-1$ and $+1$ with respective probabilities 
\begin{align}P(X+Y=-1) &= p_{X,Y}(0, -1) + p_{X,Y}(-1, 0) = \frac 12\\
P(X+Y=+1) &= p_{X,Y}(0, +1) + p_{X,Y}(+1, 0) = \frac 12
\end{align}
where the claim that both probabilities equal $\frac 12$ is justified by the fact that the means of $X$ and $Y$ are given to be zero and so it must be that $p_{X,Y}(-1, 0) = p_{X,Y}(+1, 0)$ and $p_{X,Y}(0, -1)= p_{X,Y}(0, +1)$. Consequently, $\operatorname{var}(X+Y) = 1$. I leave it to you to determine what the variances of $X$ and $Y$ are and whether $\operatorname{var}(X+Y)$ equals the sum of $\operatorname{var}(X)$ and $\operatorname{var}(Y)$ or not. You will definitely be surprised since I know that it is valid only if X and Y are independent is not correct. It is not what you don't know that will kill you; it is what you know that just a'int so.
