Marginal p.m.f. of two random variables with joint p.m.f. $p(x,y) = 2^{-x-y}$ Two discrete random variables $X$ and $Y$, whose values are positive integers, have the joint probability mass function:
$$p(x,y) = 2^{-x-y}$$
I need to determine the marginal probability mass functions, which I believe to be defined as $p(x) = \sum p(x,y)$ for $y$ and $p(y) = \sum p(x,y)$ for $x$.
But there is no interval of definition, so how am I suppose to do the table or use the formula to find the marginal probability mass functions?
I think I am missing a formula but I just can't find it.
 A: You are right for the formula of the marginal. 
Since the joint probability is $$ p(x,y)= 2^{-x-y}$$
if you are not sure about the support of $(X,Y)$ (which is include in $\mathbb{N}\times\mathbb{N}$ from your hypothesis), use that you must have 
$$\sum_{(x,y)\in A\times B} 2^{-x-y}=P((X,Y)\in \mathbb{N}\times\mathbb{N})= 1$$
to determine the support of it.
Since $2^{-x-y}=2^{-x}2^{-y}$ are positive elements, we can use Fubini, so 
\begin{align*}
\sum_{(x,y)\in A\times B} 2^{-x-y}&=\sum_{x\in A}\sum_{y\in B} 2^{-x-y}\\
& = \sum_{x\in A}\sum_{y\in B}2^{-x}2^{-y} \\
& = \sum_{x\in A} 2^{-x}\sum_{y\in B}2^{-y} 
\end{align*}
Since 
$$ \sum_{n = 1}^\infty 2^{-n} = 1  $$
It implies $A=B=\{1,2,\dots \}$
Now, using the formula you gave :
\begin{align*}
 p(x)&=\sum_{y=1}^{\infty}p(x,y)\\
& =\sum_{y=1}^{\infty}2^{-x-y} \\
& =\sum_{y=1}^{\infty}2^{-x}2^{-y} \\
& =2^{-x}\sum_{y=1}^{\infty}2^{-y}=2^{-x}
\end{align*}
same method for $p(y)$.
A: Guide:
$X$ and $Y$ takes positive integers.
Hence 
$$p(x) = \sum_{y=1}^\infty p(x,y)$$
Geometric sum might help you to solve the problem.
Similarly for $p(y)$.
