Joint PDF cross sectional area - what does it represent?
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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.
probability statistics probability-distributions
asked Dec 25 '18 at 1:14
z-Dustz-Dust
11
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add a comment |
$begingroup$
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.
probability statistics probability-distributions
asked Dec 25 '18 at 1:14
z-Dustz-Dust
11
$endgroup$
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
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@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
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In short, because $P(X=x)=0$.
$endgroup$
– kludg
Dec 25 '18 at 10:28
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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
add a comment |
$begingroup$
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.
probability statistics probability-distributions
asked Dec 25 '18 at 1:14
z-Dustz-Dust
11
$endgroup$
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.
probability statistics probability-distributions
probability statistics probability-distributions
asked Dec 25 '18 at 1:14
z-Dustz-Dust
11
asked Dec 25 '18 at 1:14
z-Dustz-Dust
11
edited Dec 25 '18 at 2:00
z-Dust
asked Dec 25 '18 at 1:14
z-Dustz-Dust
11
asked Dec 25 '18 at 1:14
z-Dustz-Dust
11
asked Dec 25 '18 at 1:14
z-Dustz-Dust
11
11
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
add a comment |
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
1
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
$begingroup$
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
$begingroup$
@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
$begingroup$
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
$begingroup$
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
add a comment |
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$begingroup$
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