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Lebesgue integral w.r.t law and Stieltjes integral w.r.t distribution function
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Consider a generic probability space $(Omega,mathcalF,mathbbP)$ and a measurable space
$(mathbbR,mathcalB)$ where $mathcalB$ is meant to denote the Borel $sigma$-algebra. A measurable function
$$X,:,OmegarightarrowmathbbR$$
such that $sigma(X)subseteqmathcalF$ is then referred to as random variable and allows for the definition of a law (or distribution)
$$mu:mathcalBrightarrow[0,1]quadtextsuch thatquad BmapstomathbbP(Xin B)quadtextfor allquad BinmathcalB.$$
Since all intervals of the type $(-infty,x]$ are elements of $mathcalB$, it follows that one can define a distribution function
$$F(x)=mu(-infty,x].$$
Due to the properties of $mu$, $F$ is easily shown to be monotonically increasing, right-continuous, and normalized to take values in $[0,1]$.
Building on these results most texts in probability theory (at least those that I have encountered so far) simply go on asserting that - provided certain conditions on $f$ are met - we can define the Stieltjes integral w.r.t. distribution function $F$
$$int_a^bf(x),textdF(x)$$ to be equal to Lebesgue integral w.r.t. the law of $X$
$$int_(a,b]f,textdmu,$$
such that $int_a^bf(x),textdF(x)$ and $int_(a,b],f,textdmu$ are essentially just two different ways of writing exactly the same thing.
As $F$ and $mu$ are related by the fact that $F(x)=mu(-infty,x]$, I am not really comfortable to accept the above as a matter of definition. Therefore, I was wondering whether there is any way to see that the statement above is actually true?
Thank you very much.
Best wishes,
Jon
real-analysis probability-theory measure-theory
asked Jul 15 at 13:06
J.Beck
184
add a comment |Â
up vote
0
down vote
favorite
Consider a generic probability space $(Omega,mathcalF,mathbbP)$ and a measurable space
$(mathbbR,mathcalB)$ where $mathcalB$ is meant to denote the Borel $sigma$-algebra. A measurable function
$$X,:,OmegarightarrowmathbbR$$
such that $sigma(X)subseteqmathcalF$ is then referred to as random variable and allows for the definition of a law (or distribution)
$$mu:mathcalBrightarrow[0,1]quadtextsuch thatquad BmapstomathbbP(Xin B)quadtextfor allquad BinmathcalB.$$
Since all intervals of the type $(-infty,x]$ are elements of $mathcalB$, it follows that one can define a distribution function
$$F(x)=mu(-infty,x].$$
Due to the properties of $mu$, $F$ is easily shown to be monotonically increasing, right-continuous, and normalized to take values in $[0,1]$.
Building on these results most texts in probability theory (at least those that I have encountered so far) simply go on asserting that - provided certain conditions on $f$ are met - we can define the Stieltjes integral w.r.t. distribution function $F$
$$int_a^bf(x),textdF(x)$$ to be equal to Lebesgue integral w.r.t. the law of $X$
$$int_(a,b]f,textdmu,$$
such that $int_a^bf(x),textdF(x)$ and $int_(a,b],f,textdmu$ are essentially just two different ways of writing exactly the same thing.
As $F$ and $mu$ are related by the fact that $F(x)=mu(-infty,x]$, I am not really comfortable to accept the above as a matter of definition. Therefore, I was wondering whether there is any way to see that the statement above is actually true?
Thank you very much.
Best wishes,
Jon
real-analysis probability-theory measure-theory
asked Jul 15 at 13:06
J.Beck
184
Essentially $F(x_2)-F(x_1)=mu (x_2-x_1)$. Then define $dF(x)$ and $dmu$.
– herb steinberg
Jul 15 at 15:03
Thank you very much for your comment. I am afraid, though, that I do not follow. Since $mu:mathcalBrightarrow[0,1]$, I don't see how $(x_1-x_2)$ could be an argument of $mu$. As the Lebesgue-Stieltjes measure is defined such that $F(x_2)-F(x_1)=mu(x_1,x_2]$, is it legitimate to argue that $textdF=F(x+textdx)-F(x)=mu(x,x+textdx]=textdmu$? I have to admit, that I have a hard time believing that $mu(x,x+textdx]=textdmu$, i.e. that the differential of $mu$ is the measure assigned to the interval $(x,x+textdx]$.
– J.Beck
Jul 15 at 16:41
Your comment is correct. I should have written $mu(x_1,x_2)$. As for $dmu$, I don't see any other way to define it other than $dmu=mu(x,x+dx)$.
– herb steinberg
Jul 15 at 19:02
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider a generic probability space $(Omega,mathcalF,mathbbP)$ and a measurable space
$(mathbbR,mathcalB)$ where $mathcalB$ is meant to denote the Borel $sigma$-algebra. A measurable function
$$X,:,OmegarightarrowmathbbR$$
such that $sigma(X)subseteqmathcalF$ is then referred to as random variable and allows for the definition of a law (or distribution)
$$mu:mathcalBrightarrow[0,1]quadtextsuch thatquad BmapstomathbbP(Xin B)quadtextfor allquad BinmathcalB.$$
Since all intervals of the type $(-infty,x]$ are elements of $mathcalB$, it follows that one can define a distribution function
$$F(x)=mu(-infty,x].$$
Due to the properties of $mu$, $F$ is easily shown to be monotonically increasing, right-continuous, and normalized to take values in $[0,1]$.
Building on these results most texts in probability theory (at least those that I have encountered so far) simply go on asserting that - provided certain conditions on $f$ are met - we can define the Stieltjes integral w.r.t. distribution function $F$
$$int_a^bf(x),textdF(x)$$ to be equal to Lebesgue integral w.r.t. the law of $X$
$$int_(a,b]f,textdmu,$$
such that $int_a^bf(x),textdF(x)$ and $int_(a,b],f,textdmu$ are essentially just two different ways of writing exactly the same thing.
As $F$ and $mu$ are related by the fact that $F(x)=mu(-infty,x]$, I am not really comfortable to accept the above as a matter of definition. Therefore, I was wondering whether there is any way to see that the statement above is actually true?
Thank you very much.
Best wishes,
Jon
real-analysis probability-theory measure-theory
asked Jul 15 at 13:06
J.Beck
184
Consider a generic probability space $(Omega,mathcalF,mathbbP)$ and a measurable space
$(mathbbR,mathcalB)$ where $mathcalB$ is meant to denote the Borel $sigma$-algebra. A measurable function
$$X,:,OmegarightarrowmathbbR$$
such that $sigma(X)subseteqmathcalF$ is then referred to as random variable and allows for the definition of a law (or distribution)
$$mu:mathcalBrightarrow[0,1]quadtextsuch thatquad BmapstomathbbP(Xin B)quadtextfor allquad BinmathcalB.$$
Since all intervals of the type $(-infty,x]$ are elements of $mathcalB$, it follows that one can define a distribution function
$$F(x)=mu(-infty,x].$$
Due to the properties of $mu$, $F$ is easily shown to be monotonically increasing, right-continuous, and normalized to take values in $[0,1]$.
Building on these results most texts in probability theory (at least those that I have encountered so far) simply go on asserting that - provided certain conditions on $f$ are met - we can define the Stieltjes integral w.r.t. distribution function $F$
$$int_a^bf(x),textdF(x)$$ to be equal to Lebesgue integral w.r.t. the law of $X$
$$int_(a,b]f,textdmu,$$
such that $int_a^bf(x),textdF(x)$ and $int_(a,b],f,textdmu$ are essentially just two different ways of writing exactly the same thing.
As $F$ and $mu$ are related by the fact that $F(x)=mu(-infty,x]$, I am not really comfortable to accept the above as a matter of definition. Therefore, I was wondering whether there is any way to see that the statement above is actually true?
Thank you very much.
Best wishes,
Jon
real-analysis probability-theory measure-theory
asked Jul 15 at 13:06
J.Beck
184
asked Jul 15 at 13:06
J.Beck
184
asked Jul 15 at 13:06
J.Beck
184
asked Jul 15 at 13:06
J.Beck
184
184
Essentially $F(x_2)-F(x_1)=mu (x_2-x_1)$. Then define $dF(x)$ and $dmu$.
– herb steinberg
Jul 15 at 15:03
Thank you very much for your comment. I am afraid, though, that I do not follow. Since $mu:mathcalBrightarrow[0,1]$, I don't see how $(x_1-x_2)$ could be an argument of $mu$. As the Lebesgue-Stieltjes measure is defined such that $F(x_2)-F(x_1)=mu(x_1,x_2]$, is it legitimate to argue that $textdF=F(x+textdx)-F(x)=mu(x,x+textdx]=textdmu$? I have to admit, that I have a hard time believing that $mu(x,x+textdx]=textdmu$, i.e. that the differential of $mu$ is the measure assigned to the interval $(x,x+textdx]$.
– J.Beck
Jul 15 at 16:41
Your comment is correct. I should have written $mu(x_1,x_2)$. As for $dmu$, I don't see any other way to define it other than $dmu=mu(x,x+dx)$.
– herb steinberg
Jul 15 at 19:02
add a comment |Â
Essentially $F(x_2)-F(x_1)=mu (x_2-x_1)$. Then define $dF(x)$ and $dmu$.
– herb steinberg
Jul 15 at 15:03
Thank you very much for your comment. I am afraid, though, that I do not follow. Since $mu:mathcalBrightarrow[0,1]$, I don't see how $(x_1-x_2)$ could be an argument of $mu$. As the Lebesgue-Stieltjes measure is defined such that $F(x_2)-F(x_1)=mu(x_1,x_2]$, is it legitimate to argue that $textdF=F(x+textdx)-F(x)=mu(x,x+textdx]=textdmu$? I have to admit, that I have a hard time believing that $mu(x,x+textdx]=textdmu$, i.e. that the differential of $mu$ is the measure assigned to the interval $(x,x+textdx]$.
– J.Beck
Jul 15 at 16:41
Your comment is correct. I should have written $mu(x_1,x_2)$. As for $dmu$, I don't see any other way to define it other than $dmu=mu(x,x+dx)$.
– herb steinberg
Jul 15 at 19:02
Essentially $F(x_2)-F(x_1)=mu (x_2-x_1)$. Then define $dF(x)$ and $dmu$.
– herb steinberg
Jul 15 at 15:03
Essentially $F(x_2)-F(x_1)=mu (x_2-x_1)$. Then define $dF(x)$ and $dmu$.
– herb steinberg
Jul 15 at 15:03
Thank you very much for your comment. I am afraid, though, that I do not follow. Since $mu:mathcalBrightarrow[0,1]$, I don't see how $(x_1-x_2)$ could be an argument of $mu$. As the Lebesgue-Stieltjes measure is defined such that $F(x_2)-F(x_1)=mu(x_1,x_2]$, is it legitimate to argue that $textdF=F(x+textdx)-F(x)=mu(x,x+textdx]=textdmu$? I have to admit, that I have a hard time believing that $mu(x,x+textdx]=textdmu$, i.e. that the differential of $mu$ is the measure assigned to the interval $(x,x+textdx]$.
– J.Beck
Jul 15 at 16:41
Thank you very much for your comment. I am afraid, though, that I do not follow. Since $mu:mathcalBrightarrow[0,1]$, I don't see how $(x_1-x_2)$ could be an argument of $mu$. As the Lebesgue-Stieltjes measure is defined such that $F(x_2)-F(x_1)=mu(x_1,x_2]$, is it legitimate to argue that $textdF=F(x+textdx)-F(x)=mu(x,x+textdx]=textdmu$? I have to admit, that I have a hard time believing that $mu(x,x+textdx]=textdmu$, i.e. that the differential of $mu$ is the measure assigned to the interval $(x,x+textdx]$.
– J.Beck
Jul 15 at 16:41
Your comment is correct. I should have written $mu(x_1,x_2)$. As for $dmu$, I don't see any other way to define it other than $dmu=mu(x,x+dx)$.
– herb steinberg
Jul 15 at 19:02
Your comment is correct. I should have written $mu(x_1,x_2)$. As for $dmu$, I don't see any other way to define it other than $dmu=mu(x,x+dx)$.
– herb steinberg
Jul 15 at 19:02
add a comment |Â
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Essentially $F(x_2)-F(x_1)=mu (x_2-x_1)$. Then define $dF(x)$ and $dmu$.
– herb steinberg
Jul 15 at 15:03
Thank you very much for your comment. I am afraid, though, that I do not follow. Since $mu:mathcalBrightarrow[0,1]$, I don't see how $(x_1-x_2)$ could be an argument of $mu$. As the Lebesgue-Stieltjes measure is defined such that $F(x_2)-F(x_1)=mu(x_1,x_2]$, is it legitimate to argue that $textdF=F(x+textdx)-F(x)=mu(x,x+textdx]=textdmu$? I have to admit, that I have a hard time believing that $mu(x,x+textdx]=textdmu$, i.e. that the differential of $mu$ is the measure assigned to the interval $(x,x+textdx]$.
– J.Beck
Jul 15 at 16:41
Your comment is correct. I should have written $mu(x_1,x_2)$. As for $dmu$, I don't see any other way to define it other than $dmu=mu(x,x+dx)$.
– herb steinberg
Jul 15 at 19:02