Q:

Finding the probability mass function for a pair of random variables $(X,Y)$

A pair of random variables $(X,Y)$ which take only integer values have the following pmf: 
$$ \begin{equation*}
 p_{x,y}=P\{X=m,Y=n\}=\begin{cases} 
K, \forall  \;\; m = 0, 1, 2\;\;; \mbox{ and } n=0,...,m;\\
0, \mbox{ otherwise. } 
\end{cases}
\end{equation*}
$$

I need to find K. 
I need to determine the mean and variance of $Y$.

I've created a table to determine the probability of each value of $X$ (0,1,2) and $Y$ (0,1,2). Using this, I think $K$ is $1/9$, but I'm not entirely sure, and it's just by using logic that I came to this. 
However, I don't think that this is correct, because in the next step, I'm getting  $E[Y] = 2$ and $Var[Y] = - 7/3$ based on the previous values. 
Since variance can't be negative, I know this isn't correct. Can someone help point me in the right direction? I'm sure it's a pretty simple solution, but I'm looking for the exact steps to find this.
Thanks! 

A:

Your table should take into account the fact that $p_{x,y}=0$ if $y>x$ (if this is what you mean with : $p_{x,y}=K$ for $m=0,1,2$ and $n=0,\dots,m$.
There are $6$ couples $(x,y)$ in $\{0,1,2\}^2$ with $0\le y\le x\le 2$. So $K$ should be $\frac16$.
For $0\le k\le 2$, ${\rm P}(Y=k)=\sum_{j=0}^2 p_{j,k}=\sum_{j=0}^k \frac16 = \frac{k+1}{6}$.
So 
$${\rm E}(Y)=\sum_{k=0}^2 k\frac{k+1}{6} = \frac43$$
and
$${\rm Var}(Y)={\rm E}(Y^2)-{\rm E}(Y)^2=\sum_{k=0}^2 k^2\frac{k+1}{6}-\frac{16}{9} = \frac73-\frac{16}{9}=\frac59$$

