Mixing
Stieltjes integral of proportion function
$begingroup$
Given a finite set of real numbers $x_1leq x_2leq ... leq x_n$, define the function:
$F(x):= frac{1}{n} underset{k=1}{overset{n}{sum}} chi_{(-infty, x_k]}(x)$
Considering it's induced Stieltjes measure $mu_F$, I would like to verify (or eliminate) a few properties:
(1) Is $mu_Fbig( (-infty,x_1) big)=0$ and $mu_Fbig( (x_n,infty) big)=0$?
(2) Is $mu_F$ supported on ${ x_k }_{k=1}^n$, i.e
$mu_F(mathbb{R})= underset{k=1}{overset{n}{sum}} mu_F({ x_k })$ ?
(3) Is $mu_F big( [x_k,x_{k+1}] big)= frac{1}{n} $ for all $kin [n-1]$?
(4) Is $int_{-infty}^tf(x)dF(x)= frac{1}{n} underset{k=1}{overset{n}{sum}} chi_{(-infty, x_k]}(x)cdot f(x_k)$?
I am trying to understand integration against this Stieljes measure and I am pretty weak on the subject of Stieltjes integral, so I would appreciate pointers on how to relate to this.
measure-theory stieltjes-integral
asked Oct 23 '18 at 14:48
Keen-ameteurKeen-ameteur
1,413316
$endgroup$
add a comment |
$begingroup$
Given a finite set of real numbers $x_1leq x_2leq ... leq x_n$, define the function:
$F(x):= frac{1}{n} underset{k=1}{overset{n}{sum}} chi_{(-infty, x_k]}(x)$
Considering it's induced Stieltjes measure $mu_F$, I would like to verify (or eliminate) a few properties:
(1) Is $mu_Fbig( (-infty,x_1) big)=0$ and $mu_Fbig( (x_n,infty) big)=0$?
(2) Is $mu_F$ supported on ${ x_k }_{k=1}^n$, i.e
$mu_F(mathbb{R})= underset{k=1}{overset{n}{sum}} mu_F({ x_k })$ ?
(3) Is $mu_F big( [x_k,x_{k+1}] big)= frac{1}{n} $ for all $kin [n-1]$?
(4) Is $int_{-infty}^tf(x)dF(x)= frac{1}{n} underset{k=1}{overset{n}{sum}} chi_{(-infty, x_k]}(x)cdot f(x_k)$?
I am trying to understand integration against this Stieljes measure and I am pretty weak on the subject of Stieltjes integral, so I would appreciate pointers on how to relate to this.
measure-theory stieltjes-integral
asked Oct 23 '18 at 14:48
Keen-ameteurKeen-ameteur
1,413316
$endgroup$
add a comment |
$begingroup$
Given a finite set of real numbers $x_1leq x_2leq ... leq x_n$, define the function:
$F(x):= frac{1}{n} underset{k=1}{overset{n}{sum}} chi_{(-infty, x_k]}(x)$
Considering it's induced Stieltjes measure $mu_F$, I would like to verify (or eliminate) a few properties:
(1) Is $mu_Fbig( (-infty,x_1) big)=0$ and $mu_Fbig( (x_n,infty) big)=0$?
(2) Is $mu_F$ supported on ${ x_k }_{k=1}^n$, i.e
$mu_F(mathbb{R})= underset{k=1}{overset{n}{sum}} mu_F({ x_k })$ ?
(3) Is $mu_F big( [x_k,x_{k+1}] big)= frac{1}{n} $ for all $kin [n-1]$?
(4) Is $int_{-infty}^tf(x)dF(x)= frac{1}{n} underset{k=1}{overset{n}{sum}} chi_{(-infty, x_k]}(x)cdot f(x_k)$?
I am trying to understand integration against this Stieljes measure and I am pretty weak on the subject of Stieltjes integral, so I would appreciate pointers on how to relate to this.
measure-theory stieltjes-integral
asked Oct 23 '18 at 14:48
Keen-ameteurKeen-ameteur
1,413316
$endgroup$
Given a finite set of real numbers $x_1leq x_2leq ... leq x_n$, define the function:
$F(x):= frac{1}{n} underset{k=1}{overset{n}{sum}} chi_{(-infty, x_k]}(x)$
Considering it's induced Stieltjes measure $mu_F$, I would like to verify (or eliminate) a few properties:
(1) Is $mu_Fbig( (-infty,x_1) big)=0$ and $mu_Fbig( (x_n,infty) big)=0$?
(2) Is $mu_F$ supported on ${ x_k }_{k=1}^n$, i.e
$mu_F(mathbb{R})= underset{k=1}{overset{n}{sum}} mu_F({ x_k })$ ?
(3) Is $mu_F big( [x_k,x_{k+1}] big)= frac{1}{n} $ for all $kin [n-1]$?
(4) Is $int_{-infty}^tf(x)dF(x)= frac{1}{n} underset{k=1}{overset{n}{sum}} chi_{(-infty, x_k]}(x)cdot f(x_k)$?
I am trying to understand integration against this Stieljes measure and I am pretty weak on the subject of Stieltjes integral, so I would appreciate pointers on how to relate to this.
measure-theory stieltjes-integral
measure-theory stieltjes-integral
asked Oct 23 '18 at 14:48
Keen-ameteurKeen-ameteur
1,413316
asked Oct 23 '18 at 14:48
Keen-ameteurKeen-ameteur
1,413316
edited Oct 24 '18 at 15:43
Keen-ameteur
asked Oct 23 '18 at 14:48
Keen-ameteurKeen-ameteur
1,413316
asked Oct 23 '18 at 14:48
Keen-ameteurKeen-ameteur
1,413316
asked Oct 23 '18 at 14:48
Keen-ameteurKeen-ameteur
1,413316
1,413316
add a comment |
add a comment |
1 Answer
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$begingroup$
You have $mathcal{B}(mathbb{R})= sigma({[c,d):c,dinmathbb{R}, c<d})$. You should make yourself clear, that it holds:
begin{align}
mu_F([c,d))&=F(d-)-F(c-)\
mu_F((c,d])&=F(d+)-F(c+)\
mu_F((c,d))&=F(d-)-F(c+)\
mu_F([c,d])&=F(d+)-F(c-)
end{align}
(1) is true. You should try to solve this on your own.
(2) is also true, actually it holds
$$mu_F=-frac{1}{n}sum_{i=1}^ndelta_{x_k}$$
(3) Is not true, since
$$mu_F([x_k,x_{k+1}])=F(x_{k+1}+)-F(x_k-)=frac{1}{n}big[(n-(k+1))-(n-(k-1))]=-frac{2}{n}$$
(4) Does not make sense. $x$ has to be integrated, so the right hand side should not depend on $x$. Just try to solve the integral, since you know the measure from (2). Note, that
$$int_{(-infty, t]}f(x)dF(x)$$
and
$$int_{(-infty, t)}f(x)dF(x)$$
can have different values. (What happens, if $t=x_k$ for some $kin{1,ldots, n}$?)
Please consider checking my answer. Thanks! :-)
answered Jan 14 at 4:42
user408858user408858
482213
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You have $mathcal{B}(mathbb{R})= sigma({[c,d):c,dinmathbb{R}, c<d})$. You should make yourself clear, that it holds:
begin{align}
mu_F([c,d))&=F(d-)-F(c-)\
mu_F((c,d])&=F(d+)-F(c+)\
mu_F((c,d))&=F(d-)-F(c+)\
mu_F([c,d])&=F(d+)-F(c-)
end{align}
(1) is true. You should try to solve this on your own.
(2) is also true, actually it holds
$$mu_F=-frac{1}{n}sum_{i=1}^ndelta_{x_k}$$
(3) Is not true, since
$$mu_F([x_k,x_{k+1}])=F(x_{k+1}+)-F(x_k-)=frac{1}{n}big[(n-(k+1))-(n-(k-1))]=-frac{2}{n}$$
(4) Does not make sense. $x$ has to be integrated, so the right hand side should not depend on $x$. Just try to solve the integral, since you know the measure from (2). Note, that
$$int_{(-infty, t]}f(x)dF(x)$$
and
$$int_{(-infty, t)}f(x)dF(x)$$
can have different values. (What happens, if $t=x_k$ for some $kin{1,ldots, n}$?)
Please consider checking my answer. Thanks! :-)
answered Jan 14 at 4:42
user408858user408858
482213
$endgroup$
add a comment |
$begingroup$
You have $mathcal{B}(mathbb{R})= sigma({[c,d):c,dinmathbb{R}, c<d})$. You should make yourself clear, that it holds:
begin{align}
mu_F([c,d))&=F(d-)-F(c-)\
mu_F((c,d])&=F(d+)-F(c+)\
mu_F((c,d))&=F(d-)-F(c+)\
mu_F([c,d])&=F(d+)-F(c-)
end{align}
(1) is true. You should try to solve this on your own.
(2) is also true, actually it holds
$$mu_F=-frac{1}{n}sum_{i=1}^ndelta_{x_k}$$
(3) Is not true, since
$$mu_F([x_k,x_{k+1}])=F(x_{k+1}+)-F(x_k-)=frac{1}{n}big[(n-(k+1))-(n-(k-1))]=-frac{2}{n}$$
(4) Does not make sense. $x$ has to be integrated, so the right hand side should not depend on $x$. Just try to solve the integral, since you know the measure from (2). Note, that
$$int_{(-infty, t]}f(x)dF(x)$$
and
$$int_{(-infty, t)}f(x)dF(x)$$
can have different values. (What happens, if $t=x_k$ for some $kin{1,ldots, n}$?)
Please consider checking my answer. Thanks! :-)
answered Jan 14 at 4:42
user408858user408858
482213
$endgroup$
add a comment |
$begingroup$
You have $mathcal{B}(mathbb{R})= sigma({[c,d):c,dinmathbb{R}, c<d})$. You should make yourself clear, that it holds:
begin{align}
mu_F([c,d))&=F(d-)-F(c-)\
mu_F((c,d])&=F(d+)-F(c+)\
mu_F((c,d))&=F(d-)-F(c+)\
mu_F([c,d])&=F(d+)-F(c-)
end{align}
(1) is true. You should try to solve this on your own.
(2) is also true, actually it holds
$$mu_F=-frac{1}{n}sum_{i=1}^ndelta_{x_k}$$
(3) Is not true, since
$$mu_F([x_k,x_{k+1}])=F(x_{k+1}+)-F(x_k-)=frac{1}{n}big[(n-(k+1))-(n-(k-1))]=-frac{2}{n}$$
(4) Does not make sense. $x$ has to be integrated, so the right hand side should not depend on $x$. Just try to solve the integral, since you know the measure from (2). Note, that
$$int_{(-infty, t]}f(x)dF(x)$$
and
$$int_{(-infty, t)}f(x)dF(x)$$
can have different values. (What happens, if $t=x_k$ for some $kin{1,ldots, n}$?)
Please consider checking my answer. Thanks! :-)
answered Jan 14 at 4:42
user408858user408858
482213
$endgroup$
You have $mathcal{B}(mathbb{R})= sigma({[c,d):c,dinmathbb{R}, c<d})$. You should make yourself clear, that it holds:
begin{align}
mu_F([c,d))&=F(d-)-F(c-)\
mu_F((c,d])&=F(d+)-F(c+)\
mu_F((c,d))&=F(d-)-F(c+)\
mu_F([c,d])&=F(d+)-F(c-)
end{align}
(1) is true. You should try to solve this on your own.
(2) is also true, actually it holds
$$mu_F=-frac{1}{n}sum_{i=1}^ndelta_{x_k}$$
(3) Is not true, since
$$mu_F([x_k,x_{k+1}])=F(x_{k+1}+)-F(x_k-)=frac{1}{n}big[(n-(k+1))-(n-(k-1))]=-frac{2}{n}$$
(4) Does not make sense. $x$ has to be integrated, so the right hand side should not depend on $x$. Just try to solve the integral, since you know the measure from (2). Note, that
$$int_{(-infty, t]}f(x)dF(x)$$
and
$$int_{(-infty, t)}f(x)dF(x)$$
can have different values. (What happens, if $t=x_k$ for some $kin{1,ldots, n}$?)
Please consider checking my answer. Thanks! :-)
answered Jan 14 at 4:42
user408858user408858
482213
edited Jan 14 at 5:19
answered Jan 14 at 4:42
user408858user408858
482213
answered Jan 14 at 4:42
user408858user408858
482213
answered Jan 14 at 4:42
user408858user408858
482213
482213
add a comment |
add a comment |
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