Marginal p.m.f. of two random variables with joint p.m.f. $p(x,y) = 2^{-x-y}$

Question

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.

Answer 1

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 $p(y)$.

Answer 2

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 $p(y)$.