Given the PDF of a continuous joint distribution:$$f(x,y)$$ Why isn't it true that taking a cross sectional area of the distribution parallel to the y axis represents the following: $$\int_{-\infty}^{\infty}f(x,y)dy = P(X=x)$$

Surely integrating for all possible values of y with respect to y, is akin to summing all probabilities P(x, y), where x is fixed at a given value. If that is the case, that would be calculating the probability that the random variable X is equal to a given value x, as you have exhausted all instances where X = x. Why is it that the cross sectional area still yields a probability density as opposed to an absolute probability?

Thank you in advance.

  • 1
    $\begingroup$ Welcome to Math.SE! What you are looking for is precisely the marginal distribution, please, check the following (en.wikipedia.org/wiki/Marginal_distribution) $\endgroup$ – hyperkahler Dec 25 '18 at 1:22
  • $\begingroup$ @Arteom Thank you for the link! On the wiki page, it states that for continuous random variables, integrating over y gives the marginal PDF of x. I'm still confused as to why the result is still a density function. If my understanding is correct you are essentially summing the products of all infinitesimal intervals of y by their respective probability densities. Doesn't that yield an "absolute" probability in the same way that it does for a PDF where there is only one random variable? $\endgroup$ – z-Dust Dec 25 '18 at 1:45
  • $\begingroup$ In short, because $P(X=x)=0$. $\endgroup$ – kludg Dec 25 '18 at 10:28
  • $\begingroup$ Yes, but that doesn't explain why infinitesimal interval * probability density = probability density. Why doesn't a cross section represent a probability in the same way that an area under a PDF of one random variable does when the axis in both cases represent the same things? Obviously I'm overlooking something but I can't tell what. $\endgroup$ – z-Dust Dec 25 '18 at 13:26

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