Having an old book of exercises without any solution,explaining the concept to me would be helpful.
Two independent variables X,Y are getting values from x∈0,1 and y∈0,1,2.The joint variable (X,Y) has a distribution as P(X=x,Y=y)=K(x+y).
Find the values of X,Y and (X,Y).
My work so far:I have found the marginal values
P(X=0,Y=0)=0
P(X=0,Y=1)=k
P(X=0,Y=2)=2k
P(X=1,Y=0)=k
P(X=1,Y=1)=2k
P(X=1,Y=2)=3k
Their sum must be 1, so each possibility is 0(1/6),k(1/3),2k(1/3),3k(1/6).If assuming i am correct so far then K=1/9 (but what does K mean and what's P(X) and P(Y)?
Yes, $K=1/9$ for the reason you cited. $K$ is just a constant.
Next, by the Law of Total Probability
$$P(X{=}x)~{=~P(X{=}x,Y{=}0)+P(X{=}x,Y{=}1)+P(X{=}x,Y{=}2)\\=~\phantom{K~((x)+(x+1)+(x+2))~\mathbf 1_{x\in\{0,1\}}}\\=~\underline{\phantom{\tfrac13(x+1)}}~\mathbf 1_{x\in\{0,1\}}}$$
And similarly:
$$P(Y{=}y)~{=~{P(X{=}0,Y{=}y)+P(X{=}1,Y{=}y)}\\=~\phantom{K~((y)+(1+y))~\mathbf 1_{y\in\{0,1,2\}}}\\=~\underline{\phantom{\tfrac 19(2y+1)}}~\mathbf 1_{y\in\{0,1,2\}}}$$