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If I have a joint pdf of multiple random variables, say 3 for simplicity, $f_{X,Y,Z}(x,y,z)$, is it true that the marginal density functions derived from that joint probability distribution ( $f_{X}(x)$, $f_{Y}(y)$, $f_{Z}(z)$ ) are not guaranteed to be valid probability density functions (integrate to 1) in all cases. I can see that the pdf condition holds for the marginals when the random variables are independent but not quite sure what happens when we introduce dependence between them.

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    $\begingroup$ By definition, a marginal distribution describes a subset of the random variables and must be a valid distribution. If you do a calculation that violates that condition, then you have erred. $\endgroup$
    – whuber
    Commented Dec 8, 2023 at 17:15
  • $\begingroup$ I just realized that my claim is equivallent to saying that the volume under the joint distribution is not 1. So dumb... $\endgroup$ Commented Dec 8, 2023 at 17:29
  • $\begingroup$ @Statisticool For closure purposes, you may want to well answer it yourself and accept your own answer :). $\endgroup$
    – Avraham
    Commented Dec 8, 2023 at 17:30

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Using two properties of the joint pdf:
$\int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} f(x, y, z) \, dx \, dy \, dz = 1$, and $f_X(x) = \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} f(x, y, z) \, dy \, dz$
It follows that $\int\limits_{-\infty}^{\infty} f_X(x) \, dx = 1$

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    $\begingroup$ Notice this assumes Fubini's Theorem applies. But you don't need it: the marginal density is that of the variable $X,$ namely, the one obtained by forgetting about all the other variables. It's that simple! $\endgroup$
    – whuber
    Commented Dec 8, 2023 at 18:06

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