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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

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  • $\begingroup$ 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

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

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

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

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

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