Checking the dependencies of two random variables Having a joint probability distribution, is  showing that we can divide Probability distribution into two  sufficient  to prove that the two random variables are independent?
for example:
$$
P_{XY}(n,m)=  P_X(n)P_Y(m) $$
assunig that the separation does not change normalization properties (or proving that it does not).
I'm trying to solve a problem where it says:
for a two random variables with the joing probability distribution.
$$
P_{XY} =C\frac{\alpha^n\beta^m}{\sqrt{n!m!}} $$
for $n,m = 0,1,2,3,...$ check whether the two random variables are independent and find the marginal probability distributions.
 A: Yes, the joint probability mass function factors in that way if and only if the two random variables must be independent.
In general, a standard fundamental theorem in probability is that two random variables $X$ and $Y$ are independent if and only if
$$
P(X\le x\cap Y\le y)=P(X\le x)P(Y\le y)\tag1
$$
holds for all $x,y\in \Bbb R$. You can then show that  $(1)$ is equivalent to $P_{X,Y}(m,n)=P_X(m)P_Y(n)$ for jointly discrete distributions. For one direction, if we assume $P_{X,Y}(m,n)=P_X(m)P_Y(n)$, then
$$
\begin{align}
P(X\le x,Y\le y)
&=\sum_{i\le x}\sum_{j\le y}P_{X,Y}(i,j)
\\&=\sum_{i\le x}\sum_{j\le y}P_{X}(i)P_Y(j)
\\&=\left(\sum_{i\le x}P_X(i)\right)\left(\sum_{j\le y}P_Y(j)\right)
\\&=P(X\le x)P(Y\le y)
\end{align}
$$
On the other hand, if we assume $(1)$, then
\begin{align}
P_{X,Y}(x,y)
&= P(X\le x\cap Y\le y)-P(X\le x-1,Y\le y)
\\ &\phantom{= }-P(X\le x,Y\le y-1)+P(X\le x-1,Y\le y-1)
\\
&= P(X\le x)P(Y\le y)-P(X\le x-1)P(Y\le y)
\\ &\phantom{= }-P(X\le x)P(Y\le y-1)+P(X\le x-1)P(Y\le y-1)
\\
&= \Big[P(X\le x)-P(X\le x-1)\Big]\cdot \Big[P(Y\le y)-P(Y\le y-1)\Big]
\\
&= P_X(x)P_Y(y)
\end{align}
