Integral of inverse fuctions for noncontinuous distributions
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Let
$ F:[a,b]rightarrow[0,1] $
be an arbitrary (noncontinuous) distribution function. Denote with
$Q(p)=inf{x:p leq F(x)}$
the associated quantile function. I would like to use that
$ int_0^1 Q(p)dp + int_a^b F(x)dx =b $.
If $F$ were strictly monotone (and therefore invertible), this would be a standard application of the rule for integral over inverse functions.
https://en.wikipedia.org/wiki/Integral_of_inverse_functions
A simple picture shows that the result should extend to my case with discontinuities. My feeling is that this must be a standard result, as it is a simple calculation rule for these frequently used objects. Yet, I couldn't find any reference for this. Any suggestion is very much appreciated.
In case of no reference, I must show it hands on. This seems very tedious (cutting the domain in countably many intervals without jumps and no constant parts, doing it piecewise, putting it back together...) Nobody is interested in such a proof. Does anybody have an idea for an elegant short proof?
Looking at the picture, the key intuition is, that integrating "horizontally" and "vertically" is the same, so maybe some smart use of the equivalence between Riemann integration and Lebesgue integration may help. For me this is all very long ago :(
Thanks so much in advance
integration lebesgue-integral distribution-theory inverse-function
asked Nov 15 at 14:15
jonasvw
313
add a comment |
up vote
0
down vote
favorite
Let
$ F:[a,b]rightarrow[0,1] $
be an arbitrary (noncontinuous) distribution function. Denote with
$Q(p)=inf{x:p leq F(x)}$
the associated quantile function. I would like to use that
$ int_0^1 Q(p)dp + int_a^b F(x)dx =b $.
If $F$ were strictly monotone (and therefore invertible), this would be a standard application of the rule for integral over inverse functions.
https://en.wikipedia.org/wiki/Integral_of_inverse_functions
A simple picture shows that the result should extend to my case with discontinuities. My feeling is that this must be a standard result, as it is a simple calculation rule for these frequently used objects. Yet, I couldn't find any reference for this. Any suggestion is very much appreciated.
In case of no reference, I must show it hands on. This seems very tedious (cutting the domain in countably many intervals without jumps and no constant parts, doing it piecewise, putting it back together...) Nobody is interested in such a proof. Does anybody have an idea for an elegant short proof?
Looking at the picture, the key intuition is, that integrating "horizontally" and "vertically" is the same, so maybe some smart use of the equivalence between Riemann integration and Lebesgue integration may help. For me this is all very long ago :(
Thanks so much in advance
integration lebesgue-integral distribution-theory inverse-function
asked Nov 15 at 14:15
jonasvw
313
I think I have a better argument: Any weakly monotone discontinuous function can be approximated in L_1 norm by a continuous strictly increasing one. Apply the standard result to the converging series and use convergence - still, I would so much more prefer a reference.
– jonasvw
Nov 15 at 14:42
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let
$ F:[a,b]rightarrow[0,1] $
be an arbitrary (noncontinuous) distribution function. Denote with
$Q(p)=inf{x:p leq F(x)}$
the associated quantile function. I would like to use that
$ int_0^1 Q(p)dp + int_a^b F(x)dx =b $.
If $F$ were strictly monotone (and therefore invertible), this would be a standard application of the rule for integral over inverse functions.
https://en.wikipedia.org/wiki/Integral_of_inverse_functions
A simple picture shows that the result should extend to my case with discontinuities. My feeling is that this must be a standard result, as it is a simple calculation rule for these frequently used objects. Yet, I couldn't find any reference for this. Any suggestion is very much appreciated.
In case of no reference, I must show it hands on. This seems very tedious (cutting the domain in countably many intervals without jumps and no constant parts, doing it piecewise, putting it back together...) Nobody is interested in such a proof. Does anybody have an idea for an elegant short proof?
Looking at the picture, the key intuition is, that integrating "horizontally" and "vertically" is the same, so maybe some smart use of the equivalence between Riemann integration and Lebesgue integration may help. For me this is all very long ago :(
Thanks so much in advance
integration lebesgue-integral distribution-theory inverse-function
asked Nov 15 at 14:15
jonasvw
313
Let
$ F:[a,b]rightarrow[0,1] $
be an arbitrary (noncontinuous) distribution function. Denote with
$Q(p)=inf{x:p leq F(x)}$
the associated quantile function. I would like to use that
$ int_0^1 Q(p)dp + int_a^b F(x)dx =b $.
If $F$ were strictly monotone (and therefore invertible), this would be a standard application of the rule for integral over inverse functions.
https://en.wikipedia.org/wiki/Integral_of_inverse_functions
A simple picture shows that the result should extend to my case with discontinuities. My feeling is that this must be a standard result, as it is a simple calculation rule for these frequently used objects. Yet, I couldn't find any reference for this. Any suggestion is very much appreciated.
In case of no reference, I must show it hands on. This seems very tedious (cutting the domain in countably many intervals without jumps and no constant parts, doing it piecewise, putting it back together...) Nobody is interested in such a proof. Does anybody have an idea for an elegant short proof?
Looking at the picture, the key intuition is, that integrating "horizontally" and "vertically" is the same, so maybe some smart use of the equivalence between Riemann integration and Lebesgue integration may help. For me this is all very long ago :(
Thanks so much in advance
integration lebesgue-integral distribution-theory inverse-function
integration lebesgue-integral distribution-theory inverse-function
asked Nov 15 at 14:15
jonasvw
313
asked Nov 15 at 14:15
jonasvw
313
edited Nov 15 at 14:20
asked Nov 15 at 14:15
jonasvw
313
asked Nov 15 at 14:15
jonasvw
313
asked Nov 15 at 14:15
jonasvw
313
313
I think I have a better argument: Any weakly monotone discontinuous function can be approximated in L_1 norm by a continuous strictly increasing one. Apply the standard result to the converging series and use convergence - still, I would so much more prefer a reference.
– jonasvw
Nov 15 at 14:42
add a comment |
I think I have a better argument: Any weakly monotone discontinuous function can be approximated in L_1 norm by a continuous strictly increasing one. Apply the standard result to the converging series and use convergence - still, I would so much more prefer a reference.
– jonasvw
Nov 15 at 14:42
I think I have a better argument: Any weakly monotone discontinuous function can be approximated in L_1 norm by a continuous strictly increasing one. Apply the standard result to the converging series and use convergence - still, I would so much more prefer a reference.
– jonasvw
Nov 15 at 14:42
I think I have a better argument: Any weakly monotone discontinuous function can be approximated in L_1 norm by a continuous strictly increasing one. Apply the standard result to the converging series and use convergence - still, I would so much more prefer a reference.
– jonasvw
Nov 15 at 14:42
add a comment |
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I think I have a better argument: Any weakly monotone discontinuous function can be approximated in L_1 norm by a continuous strictly increasing one. Apply the standard result to the converging series and use convergence - still, I would so much more prefer a reference.
– jonasvw
Nov 15 at 14:42