Separable generating function of a pair of dependent discrete random variables Independence is sufficient but not necessary for the generating function of the sum of two random variables to be the product of their individual generating functions.
I am trying to come up with an example of two dependent rv's whose joint generating function is the product of the individual generating functions.
I have tried U:=X+Y and V:=X-Y where U and V are dependent but X and Y aren't. I ended up with $[\phi(t)]^2$ vs $\phi(t^2)$
Generating function $\phi_{X}(t) := \sum_{k = 0}^\infty Pr\{X=k\} t^k$
 A: This is actually a necessary and sufficient condition (in the case of the joint generating function). Let $X$ and $Y$ have joint generating function
$$\phi_{X,Y}(s,t) = \sum_{i,j=0}^\infty \mathbb P(X=i,Y=j)s^i t^j.$$
The product of the marginal generating functions is
$$
\begin{align*}
\phi_X(s)\phi_Y(t) &= \left(\sum_{i=0}^\infty\mathbb P(X=i)s^i\right)\left(\sum_{j=0}^\infty\mathbb P(Y=j)t^j\right)\\
&= \sum_{i,j=0}^\infty \mathbb P(X=i)\mathbb P(Y=j)s^it^j.
\end{align*}
$$
If we assume these are equal, then $$\mathbb P(X=i,Y=j)=\mathbb P(X=i)\mathbb P(Y=j)$$ for all $i,j\geqslant 0$, so $X$ and $Y$ are independent.
However, it is not the case that
$$ \phi_{X,Y}(s,s) = \phi_X(s)\phi_Y(s)$$ implies that $X$ and $Y$ are independent. This is a subtle distinction, but $\phi_{X,Y}(s,s)$ is the generating function of $X+Y$, not the joint generating function of $(X,Y)$.
For a counterexample, let $X$ and $Y$ each be uniformly distributed over $\{0,1,2\}$, with joint distribution
$$
\mathbb P(X=i,Y=j) = \begin{cases}
\frac19,& (i,j)\in\{(0,0), (1,1), (2,2)\}\\
\frac29,& (i,j)\in\{(0,2), (1,0), (2,1)\}\\
0,&\text{otherwise}
\end{cases}
$$
Then $$\phi_X(s)\phi_Y(s)=\frac13(1+s+s^2)\frac13(1+s+s^2) = \frac1{9}(1+s+s^2)^2$$ and
$$
\begin{align*}
\phi_{X,Y}(s,s) &= \frac19 + \frac29 s + \frac39 s^2 + \frac29 s^3 +\frac19 s^4\\
&= \frac19(1 + 2s + 3s^2 + 2s^3 + s^4)\\
&= \frac19(1+s+s^2)^2\\
&= \phi_X(s)\phi_Y(s).
\end{align*}
 $$
But $X$ and $Y$ are not independent, as for example
$$\mathbb P(X=0)\mathbb P(Y=1)=\frac19\ne0=\mathbb P(X=0,Y=1).$$
