Question re: Marginal pmf for Joint Discrete Variables (Textbook Exercise)

Given a discrete joint distribution with pdf $$f(x, y) = p^y (1 - p)^2$$ for $$p \in (0, 1)$$ and $$x, y \in \mathbb{N}, x \leq y$$ (here I include 0 in the natural numbers), find the marginal distributions $$f_y(y), f_x(x)$$.

Finding the marginal distribution for y is trivial: $$f_y (y) = \sum_{i = 0}^y { p^y (1 - p)^2}$$ as x can take on all integers between 0 and y.

Finding the marginal distribution for x, however, is giving me hives; the textbook [1] gives the answer as $$f_x (x)= p^x (1 - p)^2$$, which doesn't make sense to me: x simply establishes a lower bound for y but does not fix it, thus it feels like the marginal pdf of x should be the sum over all possible values of y: $$f_x(x) = \sum_{i = x}^{\infty} {p^i (1 - p)^2}$$.

I'm unable to convince myself I'm either right or wrong, so any guidance is appreciated.

[1] Mathematical Statistics by Rossi

• Are you sure $f_{x}(x)$ has to be $p^{x}(1-p)^{2}$ and not $p^{x}(1-p)$? Commented Nov 21, 2021 at 21:07

$$f_{X}(x) = \sum_{y=x}^{\infty}p^{x}(1-p)^{2}=(1-p)^{2}\sum_{y=x}^{\infty}p^{y}$$
However, for $$\left|p \right|<1, \sum_{y=x}^{\infty}=\frac{p^{x}}{1-p}$$
$$f_{X}(x)= (1-p)^{2}\frac{p^{x}}{1-p}=p^{x}(1-p), \ \ x=0, 1, 2,...$$