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

  • $\begingroup$ Are you sure $f_{x}(x)$ has to be $p^{x}(1-p)^{2}$ and not $p^{x}(1-p)$? $\endgroup$
    – Fiodor1234
    Nov 21, 2021 at 21:07

1 Answer 1


$f_{X}(x) = \sum_{y=x}^{\infty}p^{x}(1-p)^{2}=(1-p)^{2}\sum_{y=x}^{\infty}p^{y}$

However, for $\left|p \right|<1, \sum_{y=x}^{\infty}=\frac{p^{x}}{1-p} $

Hence, a plug-in that to the first equation we get

$f_{X}(x)= (1-p)^{2}\frac{p^{x}}{1-p}=p^{x}(1-p), \ \ x=0, 1, 2,...$

  • $\begingroup$ Well, that's embarrassingly simple. Thanks. $\endgroup$
    – tbert
    Nov 22, 2021 at 7:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.