How is Riemann–Stieltjes Integration insufficient for developing modern probability theory?
If we consider Riemann–Stieltjes integration then it can perfectly account for mixed probability distribution (a continuous R.V with some point mass). So why would we still need Lebesgue Integration theory?
Is it because the Riemann integrable class is not large enough, or is it because under Riemann integration interchanging limits and integration is too hard(usually requiring uniform convergence)?
probability integration lebesgue-integral
asked Nov 25 '14 at 22:13
user2804929
834
add a comment |
If we consider Riemann–Stieltjes integration then it can perfectly account for mixed probability distribution (a continuous R.V with some point mass). So why would we still need Lebesgue Integration theory?
Is it because the Riemann integrable class is not large enough, or is it because under Riemann integration interchanging limits and integration is too hard(usually requiring uniform convergence)?
probability integration lebesgue-integral
asked Nov 25 '14 at 22:13
user2804929
834
add a comment |
If we consider Riemann–Stieltjes integration then it can perfectly account for mixed probability distribution (a continuous R.V with some point mass). So why would we still need Lebesgue Integration theory?
Is it because the Riemann integrable class is not large enough, or is it because under Riemann integration interchanging limits and integration is too hard(usually requiring uniform convergence)?
probability integration lebesgue-integral
asked Nov 25 '14 at 22:13
user2804929
834
If we consider Riemann–Stieltjes integration then it can perfectly account for mixed probability distribution (a continuous R.V with some point mass). So why would we still need Lebesgue Integration theory?
Is it because the Riemann integrable class is not large enough, or is it because under Riemann integration interchanging limits and integration is too hard(usually requiring uniform convergence)?
probability integration lebesgue-integral
probability integration lebesgue-integral
asked Nov 25 '14 at 22:13
user2804929
834
asked Nov 25 '14 at 22:13
user2804929
834
asked Nov 25 '14 at 22:13
user2804929
834
asked Nov 25 '14 at 22:13
user2804929
834
asked Nov 25 '14 at 22:13
user2804929
834
834
add a comment |
add a comment |
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The points that you mention are already good reasons. There really would be no advantage in not using Lebesgue integration.
Another reason is that we want to be able to measure sets like ${X geq 0 }$, which directly leads us to Lebesgue measure theory. So it is very natural to then also use Lebesgue integration.
Another important reason: Riemann/Stieltjes integration might be enough to calculate the expected value or variance for a single random variable, but we want to be able to define random variables on the same space, for example to be able to look at the sum of random variables, or to have a certain dependency of the random variables. For this, we need the notion of a general probability space $(Omega, mathcal{A},P)$. This also leads us to the Lebesgue measure/integration theory.
answered Nov 28 at 14:29
Tki Deneb
26710
add a comment |
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1 Answer
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The points that you mention are already good reasons. There really would be no advantage in not using Lebesgue integration.
Another reason is that we want to be able to measure sets like ${X geq 0 }$, which directly leads us to Lebesgue measure theory. So it is very natural to then also use Lebesgue integration.
Another important reason: Riemann/Stieltjes integration might be enough to calculate the expected value or variance for a single random variable, but we want to be able to define random variables on the same space, for example to be able to look at the sum of random variables, or to have a certain dependency of the random variables. For this, we need the notion of a general probability space $(Omega, mathcal{A},P)$. This also leads us to the Lebesgue measure/integration theory.
answered Nov 28 at 14:29
Tki Deneb
26710
add a comment |
The points that you mention are already good reasons. There really would be no advantage in not using Lebesgue integration.
Another reason is that we want to be able to measure sets like ${X geq 0 }$, which directly leads us to Lebesgue measure theory. So it is very natural to then also use Lebesgue integration.
Another important reason: Riemann/Stieltjes integration might be enough to calculate the expected value or variance for a single random variable, but we want to be able to define random variables on the same space, for example to be able to look at the sum of random variables, or to have a certain dependency of the random variables. For this, we need the notion of a general probability space $(Omega, mathcal{A},P)$. This also leads us to the Lebesgue measure/integration theory.
answered Nov 28 at 14:29
Tki Deneb
26710
add a comment |
The points that you mention are already good reasons. There really would be no advantage in not using Lebesgue integration.
Another reason is that we want to be able to measure sets like ${X geq 0 }$, which directly leads us to Lebesgue measure theory. So it is very natural to then also use Lebesgue integration.
Another important reason: Riemann/Stieltjes integration might be enough to calculate the expected value or variance for a single random variable, but we want to be able to define random variables on the same space, for example to be able to look at the sum of random variables, or to have a certain dependency of the random variables. For this, we need the notion of a general probability space $(Omega, mathcal{A},P)$. This also leads us to the Lebesgue measure/integration theory.
answered Nov 28 at 14:29
Tki Deneb
26710
The points that you mention are already good reasons. There really would be no advantage in not using Lebesgue integration.
Another reason is that we want to be able to measure sets like ${X geq 0 }$, which directly leads us to Lebesgue measure theory. So it is very natural to then also use Lebesgue integration.
Another important reason: Riemann/Stieltjes integration might be enough to calculate the expected value or variance for a single random variable, but we want to be able to define random variables on the same space, for example to be able to look at the sum of random variables, or to have a certain dependency of the random variables. For this, we need the notion of a general probability space $(Omega, mathcal{A},P)$. This also leads us to the Lebesgue measure/integration theory.
answered Nov 28 at 14:29
Tki Deneb
26710
answered Nov 28 at 14:29
Tki Deneb
26710
answered Nov 28 at 14:29
Tki Deneb
26710
answered Nov 28 at 14:29
Tki Deneb
26710
26710
add a comment |
add a comment |
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