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What is the meaning of (sampling) a continuous probability distribution?
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According to theory, a continuous probability function is $f:mathbb Rrightarrowmathbb R$, such that $f$ is continuous and the improper integral of it is $1$. Moreover the function is never negative. The concept makes sense in that these distributions can be thought of as a limit as discrete RVs are summed.
So far so good... But then the words "let $x_1,...,x_n$ be a sample from a contiuous probability distribution" are uttered. This to me sounds a bit mythical, I don't think there is any way we could sample a continuous distribution.
What actually is the appropriate meaning of such sentences?
measure-theory
asked Jul 27 at 15:48
Dole
792514
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up vote
0
down vote
favorite
According to theory, a continuous probability function is $f:mathbb Rrightarrowmathbb R$, such that $f$ is continuous and the improper integral of it is $1$. Moreover the function is never negative. The concept makes sense in that these distributions can be thought of as a limit as discrete RVs are summed.
So far so good... But then the words "let $x_1,...,x_n$ be a sample from a contiuous probability distribution" are uttered. This to me sounds a bit mythical, I don't think there is any way we could sample a continuous distribution.
What actually is the appropriate meaning of such sentences?
measure-theory
asked Jul 27 at 15:48
Dole
792514
1
In math we talk about a lot of things we couldn't actually physically do. Do you object to talking about a number chosen uniformly at random from $[0,1]$, for instance?
– Eric Wofsey
Jul 27 at 15:52
@EricWofsey Given that we can't even define almost all numbers in that interval, yes.
– Dole
Jul 27 at 15:54
OK, then you'll just have to accept that this is a mathematical idealization and not a statement about the real world.
– Eric Wofsey
Jul 27 at 15:55
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
According to theory, a continuous probability function is $f:mathbb Rrightarrowmathbb R$, such that $f$ is continuous and the improper integral of it is $1$. Moreover the function is never negative. The concept makes sense in that these distributions can be thought of as a limit as discrete RVs are summed.
So far so good... But then the words "let $x_1,...,x_n$ be a sample from a contiuous probability distribution" are uttered. This to me sounds a bit mythical, I don't think there is any way we could sample a continuous distribution.
What actually is the appropriate meaning of such sentences?
measure-theory
asked Jul 27 at 15:48
Dole
792514
According to theory, a continuous probability function is $f:mathbb Rrightarrowmathbb R$, such that $f$ is continuous and the improper integral of it is $1$. Moreover the function is never negative. The concept makes sense in that these distributions can be thought of as a limit as discrete RVs are summed.
So far so good... But then the words "let $x_1,...,x_n$ be a sample from a contiuous probability distribution" are uttered. This to me sounds a bit mythical, I don't think there is any way we could sample a continuous distribution.
What actually is the appropriate meaning of such sentences?
measure-theory
asked Jul 27 at 15:48
Dole
792514
edited Jul 27 at 16:27
asked Jul 27 at 15:48
Dole
792514
asked Jul 27 at 15:48
Dole
792514
asked Jul 27 at 15:48
Dole
792514
792514
1
In math we talk about a lot of things we couldn't actually physically do. Do you object to talking about a number chosen uniformly at random from $[0,1]$, for instance?
– Eric Wofsey
Jul 27 at 15:52
@EricWofsey Given that we can't even define almost all numbers in that interval, yes.
– Dole
Jul 27 at 15:54
OK, then you'll just have to accept that this is a mathematical idealization and not a statement about the real world.
– Eric Wofsey
Jul 27 at 15:55
add a comment |Â
1
In math we talk about a lot of things we couldn't actually physically do. Do you object to talking about a number chosen uniformly at random from $[0,1]$, for instance?
– Eric Wofsey
Jul 27 at 15:52
@EricWofsey Given that we can't even define almost all numbers in that interval, yes.
– Dole
Jul 27 at 15:54
OK, then you'll just have to accept that this is a mathematical idealization and not a statement about the real world.
– Eric Wofsey
Jul 27 at 15:55
1
1
In math we talk about a lot of things we couldn't actually physically do. Do you object to talking about a number chosen uniformly at random from $[0,1]$, for instance?
– Eric Wofsey
Jul 27 at 15:52
In math we talk about a lot of things we couldn't actually physically do. Do you object to talking about a number chosen uniformly at random from $[0,1]$, for instance?
– Eric Wofsey
Jul 27 at 15:52
@EricWofsey Given that we can't even define almost all numbers in that interval, yes.
– Dole
Jul 27 at 15:54
@EricWofsey Given that we can't even define almost all numbers in that interval, yes.
– Dole
Jul 27 at 15:54
OK, then you'll just have to accept that this is a mathematical idealization and not a statement about the real world.
– Eric Wofsey
Jul 27 at 15:55
OK, then you'll just have to accept that this is a mathematical idealization and not a statement about the real world.
– Eric Wofsey
Jul 27 at 15:55
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
There is no "appropriate meaning". You are missing nothing. This is not meant to have any exact real-world meaning but is instead a mathematical idealization.
The completely rigorous meaning from a purely mathematical perspective is that $x_1,dots,x_n$ are random variables on some probability space which are independent and have the given continuous distribution function $f$. That is, for any $r_1,dots,r_ninmathbbR$, the probability that $x_i<r_i$ is true for all $i=1,dots,n$ is $prod_i=1^nint_-infty^r_i f(x)dx$.
answered Jul 27 at 15:59
Eric Wofsey
162k12188299
Could the interpretation be along the lines of: "We should be able to get as close as we would like by sampling from the sum of ever increasing amount of discrete variables". I just think that it's a bit weird to have mathematical theorems that does not correspond to stating a fact about any mathematical operations. Sounds like a fantasy novel or something. Although, it could be that I should move over to the Philosophy exchange.
– Dole
Jul 27 at 17:28
I don't know what you mean by "mathematical operations". The meaning in terms of random variables is perfectly "mathematical", by the usual meaning of the term.
– Eric Wofsey
Jul 27 at 17:32
Wofsey Mathematical operations as in $+,-,cdot,textmod$ etc. An example: "Consider a random sample $x_1,x_2$ from gaussian, the expectation of the sum is $E(x_1+x_2)$ ". The literal reading in terms of operations is then: "we can get as close to $E(x_1+x_2)$ as we would like by sampling and summing two samples, from the sum of $n$ discrete RVs, $m$ times and averaging". Since sampling from a continuous distribution is impossible the theorem is otherwise meaningless in terms of mathematical operations.
– Dole
Jul 27 at 17:57
Sampling a discrete distribution is also impossible. How do you propose sampling something which has exactly 1/2 probability, for instance? I really think that your concern here is wholly misplaced.
– Eric Wofsey
Jul 27 at 18:13
The vast majority of mathematics cannot be stated solely in terms of "mathematical operations" in the way you seem to have in mind.
– Eric Wofsey
Jul 27 at 18:21
 |Â
show 1 more comment
up vote
1
down vote
A continuous probability distribution has continuous random variables. The probability density function, pdf, f(x) follows some rules.
$$ f(x) geq 0 $$
$f(x) $ is piece wise continuous and
$$int_-infty^infty f(x) dx = 1 $$
Also if $X$ is a continuous random variable falling in the interval $(a,b)$
$$ Pr(a < X < b) = int_a^b f(x) dx$$
Importantly
$$ Pr(X=c) = int_c^c f(x) dx = 0 $$
answered Jul 27 at 16:13
RHowe
949715
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
There is no "appropriate meaning". You are missing nothing. This is not meant to have any exact real-world meaning but is instead a mathematical idealization.
The completely rigorous meaning from a purely mathematical perspective is that $x_1,dots,x_n$ are random variables on some probability space which are independent and have the given continuous distribution function $f$. That is, for any $r_1,dots,r_ninmathbbR$, the probability that $x_i<r_i$ is true for all $i=1,dots,n$ is $prod_i=1^nint_-infty^r_i f(x)dx$.
answered Jul 27 at 15:59
Eric Wofsey
162k12188299
Could the interpretation be along the lines of: "We should be able to get as close as we would like by sampling from the sum of ever increasing amount of discrete variables". I just think that it's a bit weird to have mathematical theorems that does not correspond to stating a fact about any mathematical operations. Sounds like a fantasy novel or something. Although, it could be that I should move over to the Philosophy exchange.
– Dole
Jul 27 at 17:28
I don't know what you mean by "mathematical operations". The meaning in terms of random variables is perfectly "mathematical", by the usual meaning of the term.
– Eric Wofsey
Jul 27 at 17:32
Wofsey Mathematical operations as in $+,-,cdot,textmod$ etc. An example: "Consider a random sample $x_1,x_2$ from gaussian, the expectation of the sum is $E(x_1+x_2)$ ". The literal reading in terms of operations is then: "we can get as close to $E(x_1+x_2)$ as we would like by sampling and summing two samples, from the sum of $n$ discrete RVs, $m$ times and averaging". Since sampling from a continuous distribution is impossible the theorem is otherwise meaningless in terms of mathematical operations.
– Dole
Jul 27 at 17:57
Sampling a discrete distribution is also impossible. How do you propose sampling something which has exactly 1/2 probability, for instance? I really think that your concern here is wholly misplaced.
– Eric Wofsey
Jul 27 at 18:13
The vast majority of mathematics cannot be stated solely in terms of "mathematical operations" in the way you seem to have in mind.
– Eric Wofsey
Jul 27 at 18:21
 |Â
show 1 more comment
up vote
1
down vote
accepted
There is no "appropriate meaning". You are missing nothing. This is not meant to have any exact real-world meaning but is instead a mathematical idealization.
The completely rigorous meaning from a purely mathematical perspective is that $x_1,dots,x_n$ are random variables on some probability space which are independent and have the given continuous distribution function $f$. That is, for any $r_1,dots,r_ninmathbbR$, the probability that $x_i<r_i$ is true for all $i=1,dots,n$ is $prod_i=1^nint_-infty^r_i f(x)dx$.
answered Jul 27 at 15:59
Eric Wofsey
162k12188299
Could the interpretation be along the lines of: "We should be able to get as close as we would like by sampling from the sum of ever increasing amount of discrete variables". I just think that it's a bit weird to have mathematical theorems that does not correspond to stating a fact about any mathematical operations. Sounds like a fantasy novel or something. Although, it could be that I should move over to the Philosophy exchange.
– Dole
Jul 27 at 17:28
I don't know what you mean by "mathematical operations". The meaning in terms of random variables is perfectly "mathematical", by the usual meaning of the term.
– Eric Wofsey
Jul 27 at 17:32
Wofsey Mathematical operations as in $+,-,cdot,textmod$ etc. An example: "Consider a random sample $x_1,x_2$ from gaussian, the expectation of the sum is $E(x_1+x_2)$ ". The literal reading in terms of operations is then: "we can get as close to $E(x_1+x_2)$ as we would like by sampling and summing two samples, from the sum of $n$ discrete RVs, $m$ times and averaging". Since sampling from a continuous distribution is impossible the theorem is otherwise meaningless in terms of mathematical operations.
– Dole
Jul 27 at 17:57
Sampling a discrete distribution is also impossible. How do you propose sampling something which has exactly 1/2 probability, for instance? I really think that your concern here is wholly misplaced.
– Eric Wofsey
Jul 27 at 18:13
The vast majority of mathematics cannot be stated solely in terms of "mathematical operations" in the way you seem to have in mind.
– Eric Wofsey
Jul 27 at 18:21
 |Â
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
There is no "appropriate meaning". You are missing nothing. This is not meant to have any exact real-world meaning but is instead a mathematical idealization.
The completely rigorous meaning from a purely mathematical perspective is that $x_1,dots,x_n$ are random variables on some probability space which are independent and have the given continuous distribution function $f$. That is, for any $r_1,dots,r_ninmathbbR$, the probability that $x_i<r_i$ is true for all $i=1,dots,n$ is $prod_i=1^nint_-infty^r_i f(x)dx$.
answered Jul 27 at 15:59
Eric Wofsey
162k12188299
There is no "appropriate meaning". You are missing nothing. This is not meant to have any exact real-world meaning but is instead a mathematical idealization.
The completely rigorous meaning from a purely mathematical perspective is that $x_1,dots,x_n$ are random variables on some probability space which are independent and have the given continuous distribution function $f$. That is, for any $r_1,dots,r_ninmathbbR$, the probability that $x_i<r_i$ is true for all $i=1,dots,n$ is $prod_i=1^nint_-infty^r_i f(x)dx$.
answered Jul 27 at 15:59
Eric Wofsey
162k12188299
answered Jul 27 at 15:59
Eric Wofsey
162k12188299
answered Jul 27 at 15:59
Eric Wofsey
162k12188299
answered Jul 27 at 15:59
Eric Wofsey
162k12188299
162k12188299
Could the interpretation be along the lines of: "We should be able to get as close as we would like by sampling from the sum of ever increasing amount of discrete variables". I just think that it's a bit weird to have mathematical theorems that does not correspond to stating a fact about any mathematical operations. Sounds like a fantasy novel or something. Although, it could be that I should move over to the Philosophy exchange.
– Dole
Jul 27 at 17:28
I don't know what you mean by "mathematical operations". The meaning in terms of random variables is perfectly "mathematical", by the usual meaning of the term.
– Eric Wofsey
Jul 27 at 17:32
Wofsey Mathematical operations as in $+,-,cdot,textmod$ etc. An example: "Consider a random sample $x_1,x_2$ from gaussian, the expectation of the sum is $E(x_1+x_2)$ ". The literal reading in terms of operations is then: "we can get as close to $E(x_1+x_2)$ as we would like by sampling and summing two samples, from the sum of $n$ discrete RVs, $m$ times and averaging". Since sampling from a continuous distribution is impossible the theorem is otherwise meaningless in terms of mathematical operations.
– Dole
Jul 27 at 17:57
Sampling a discrete distribution is also impossible. How do you propose sampling something which has exactly 1/2 probability, for instance? I really think that your concern here is wholly misplaced.
– Eric Wofsey
Jul 27 at 18:13
The vast majority of mathematics cannot be stated solely in terms of "mathematical operations" in the way you seem to have in mind.
– Eric Wofsey
Jul 27 at 18:21
 |Â
show 1 more comment
Could the interpretation be along the lines of: "We should be able to get as close as we would like by sampling from the sum of ever increasing amount of discrete variables". I just think that it's a bit weird to have mathematical theorems that does not correspond to stating a fact about any mathematical operations. Sounds like a fantasy novel or something. Although, it could be that I should move over to the Philosophy exchange.
– Dole
Jul 27 at 17:28
I don't know what you mean by "mathematical operations". The meaning in terms of random variables is perfectly "mathematical", by the usual meaning of the term.
– Eric Wofsey
Jul 27 at 17:32
Wofsey Mathematical operations as in $+,-,cdot,textmod$ etc. An example: "Consider a random sample $x_1,x_2$ from gaussian, the expectation of the sum is $E(x_1+x_2)$ ". The literal reading in terms of operations is then: "we can get as close to $E(x_1+x_2)$ as we would like by sampling and summing two samples, from the sum of $n$ discrete RVs, $m$ times and averaging". Since sampling from a continuous distribution is impossible the theorem is otherwise meaningless in terms of mathematical operations.
– Dole
Jul 27 at 17:57
Sampling a discrete distribution is also impossible. How do you propose sampling something which has exactly 1/2 probability, for instance? I really think that your concern here is wholly misplaced.
– Eric Wofsey
Jul 27 at 18:13
The vast majority of mathematics cannot be stated solely in terms of "mathematical operations" in the way you seem to have in mind.
– Eric Wofsey
Jul 27 at 18:21
Could the interpretation be along the lines of: "We should be able to get as close as we would like by sampling from the sum of ever increasing amount of discrete variables". I just think that it's a bit weird to have mathematical theorems that does not correspond to stating a fact about any mathematical operations. Sounds like a fantasy novel or something. Although, it could be that I should move over to the Philosophy exchange.
– Dole
Jul 27 at 17:28
Could the interpretation be along the lines of: "We should be able to get as close as we would like by sampling from the sum of ever increasing amount of discrete variables". I just think that it's a bit weird to have mathematical theorems that does not correspond to stating a fact about any mathematical operations. Sounds like a fantasy novel or something. Although, it could be that I should move over to the Philosophy exchange.
– Dole
Jul 27 at 17:28
I don't know what you mean by "mathematical operations". The meaning in terms of random variables is perfectly "mathematical", by the usual meaning of the term.
– Eric Wofsey
Jul 27 at 17:32
I don't know what you mean by "mathematical operations". The meaning in terms of random variables is perfectly "mathematical", by the usual meaning of the term.
– Eric Wofsey
Jul 27 at 17:32
Wofsey Mathematical operations as in $+,-,cdot,textmod$ etc. An example: "Consider a random sample $x_1,x_2$ from gaussian, the expectation of the sum is $E(x_1+x_2)$ ". The literal reading in terms of operations is then: "we can get as close to $E(x_1+x_2)$ as we would like by sampling and summing two samples, from the sum of $n$ discrete RVs, $m$ times and averaging". Since sampling from a continuous distribution is impossible the theorem is otherwise meaningless in terms of mathematical operations.
– Dole
Jul 27 at 17:57
Wofsey Mathematical operations as in $+,-,cdot,textmod$ etc. An example: "Consider a random sample $x_1,x_2$ from gaussian, the expectation of the sum is $E(x_1+x_2)$ ". The literal reading in terms of operations is then: "we can get as close to $E(x_1+x_2)$ as we would like by sampling and summing two samples, from the sum of $n$ discrete RVs, $m$ times and averaging". Since sampling from a continuous distribution is impossible the theorem is otherwise meaningless in terms of mathematical operations.
– Dole
Jul 27 at 17:57
Sampling a discrete distribution is also impossible. How do you propose sampling something which has exactly 1/2 probability, for instance? I really think that your concern here is wholly misplaced.
– Eric Wofsey
Jul 27 at 18:13
Sampling a discrete distribution is also impossible. How do you propose sampling something which has exactly 1/2 probability, for instance? I really think that your concern here is wholly misplaced.
– Eric Wofsey
Jul 27 at 18:13
The vast majority of mathematics cannot be stated solely in terms of "mathematical operations" in the way you seem to have in mind.
– Eric Wofsey
Jul 27 at 18:21
The vast majority of mathematics cannot be stated solely in terms of "mathematical operations" in the way you seem to have in mind.
– Eric Wofsey
Jul 27 at 18:21
 |Â
show 1 more comment
up vote
1
down vote
A continuous probability distribution has continuous random variables. The probability density function, pdf, f(x) follows some rules.
$$ f(x) geq 0 $$
$f(x) $ is piece wise continuous and
$$int_-infty^infty f(x) dx = 1 $$
Also if $X$ is a continuous random variable falling in the interval $(a,b)$
$$ Pr(a < X < b) = int_a^b f(x) dx$$
Importantly
$$ Pr(X=c) = int_c^c f(x) dx = 0 $$
answered Jul 27 at 16:13
RHowe
949715
add a comment |Â
up vote
1
down vote
A continuous probability distribution has continuous random variables. The probability density function, pdf, f(x) follows some rules.
$$ f(x) geq 0 $$
$f(x) $ is piece wise continuous and
$$int_-infty^infty f(x) dx = 1 $$
Also if $X$ is a continuous random variable falling in the interval $(a,b)$
$$ Pr(a < X < b) = int_a^b f(x) dx$$
Importantly
$$ Pr(X=c) = int_c^c f(x) dx = 0 $$
answered Jul 27 at 16:13
RHowe
949715
add a comment |Â
up vote
1
down vote
up vote
1
down vote
A continuous probability distribution has continuous random variables. The probability density function, pdf, f(x) follows some rules.
$$ f(x) geq 0 $$
$f(x) $ is piece wise continuous and
$$int_-infty^infty f(x) dx = 1 $$
Also if $X$ is a continuous random variable falling in the interval $(a,b)$
$$ Pr(a < X < b) = int_a^b f(x) dx$$
Importantly
$$ Pr(X=c) = int_c^c f(x) dx = 0 $$
answered Jul 27 at 16:13
RHowe
949715
A continuous probability distribution has continuous random variables. The probability density function, pdf, f(x) follows some rules.
$$ f(x) geq 0 $$
$f(x) $ is piece wise continuous and
$$int_-infty^infty f(x) dx = 1 $$
Also if $X$ is a continuous random variable falling in the interval $(a,b)$
$$ Pr(a < X < b) = int_a^b f(x) dx$$
Importantly
$$ Pr(X=c) = int_c^c f(x) dx = 0 $$
answered Jul 27 at 16:13
RHowe
949715
answered Jul 27 at 16:13
RHowe
949715
answered Jul 27 at 16:13
RHowe
949715
answered Jul 27 at 16:13
RHowe
949715
949715
add a comment |Â
add a comment |Â
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1
In math we talk about a lot of things we couldn't actually physically do. Do you object to talking about a number chosen uniformly at random from $[0,1]$, for instance?
– Eric Wofsey
Jul 27 at 15:52
@EricWofsey Given that we can't even define almost all numbers in that interval, yes.
– Dole
Jul 27 at 15:54
OK, then you'll just have to accept that this is a mathematical idealization and not a statement about the real world.
– Eric Wofsey
Jul 27 at 15:55