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Convergence of integrals under weak convergence of measure and compact convergence
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I'm trying to solve Problem 2.4.12 on page 64 of Karatzas-Shreve's book "Brownian motion and stochastic calculus":
My attempt is to use triangle inequality (denoting $Omega=C[0,infty)$)
$$|int_Omega f_ndP_n-int_Omega fdP|leq |int_Omega (f_n-f)dP_n|+|int_Omega fdP_n-int_Omega fdP|quadstar,$$
and estimate the first term with dominated convergence, the second using weak convergence of the measures.
For $epsilon>0$, since $(P_n)_n$ is tight (by Prohorov thm) I can choose a compact $Ksubset Omega$ such that $P_n(K)geq 1-epsilon$.
Fix $mgeq1$, then by dominated convergence for $n$ large enough I have
$$ |int_Omega (f_n-f)dP_m|=|int_K (f_n-f)dP_m|+|int_K^c (f_n-f)dP_m|
leq epsilon+epsilonsup_ngeq1,omegain K^c|f_n(omega)-f(omega)|.$$
To conclude I need $f$ to be bounded. Does this follow from the fact that $f$ is the compact-limit of uniformly bounded functions? (Note that I need boundedness also to use weak convergence on the second term of $star$)
Is it correct that if $forall mgeq1$ $|int_Omega (f_n-f)dP_m|to_ntoinfty0$, then
$ |int_Omega (f_n-f)dP_n|to_ntoinfty0?$
To be fair I think it's true only if the first convergence is uniform in $m$.
Thanks in advance :)
functional-analysis weak-convergence
asked Jul 28 at 13:54
Demetrio Masciurett
285
add a comment |Â
up vote
1
down vote
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I'm trying to solve Problem 2.4.12 on page 64 of Karatzas-Shreve's book "Brownian motion and stochastic calculus":
My attempt is to use triangle inequality (denoting $Omega=C[0,infty)$)
$$|int_Omega f_ndP_n-int_Omega fdP|leq |int_Omega (f_n-f)dP_n|+|int_Omega fdP_n-int_Omega fdP|quadstar,$$
and estimate the first term with dominated convergence, the second using weak convergence of the measures.
For $epsilon>0$, since $(P_n)_n$ is tight (by Prohorov thm) I can choose a compact $Ksubset Omega$ such that $P_n(K)geq 1-epsilon$.
Fix $mgeq1$, then by dominated convergence for $n$ large enough I have
$$ |int_Omega (f_n-f)dP_m|=|int_K (f_n-f)dP_m|+|int_K^c (f_n-f)dP_m|
leq epsilon+epsilonsup_ngeq1,omegain K^c|f_n(omega)-f(omega)|.$$
To conclude I need $f$ to be bounded. Does this follow from the fact that $f$ is the compact-limit of uniformly bounded functions? (Note that I need boundedness also to use weak convergence on the second term of $star$)
Is it correct that if $forall mgeq1$ $|int_Omega (f_n-f)dP_m|to_ntoinfty0$, then
$ |int_Omega (f_n-f)dP_n|to_ntoinfty0?$
To be fair I think it's true only if the first convergence is uniform in $m$.
Thanks in advance :)
functional-analysis weak-convergence
asked Jul 28 at 13:54
Demetrio Masciurett
285
You know that the sequence is uniformly bounded, so the same bound holds also for the pointwise limit
– Bob
Jul 28 at 21:23
Re your 2nd question: What's wrong about doing the estimate for $m=n$?
– saz
Jul 29 at 5:39
@saz: by fixing $mgeq1,epsilon>0$ I found $N(epsilon,m)geq1$ such that $forall ngeq N(epsilon,m),$ $int_Omega(f_n-f)dP_m<epsilon$. My problem is that $N(epsilon,m)$ depends on $m$. How do I know that $sup_m N(epsilon,m)$ is finite?
– Demetrio Masciurett
Jul 29 at 8:58
@Bob: thanks, since I have point-wise convergence and boundedness I even don't need to separate the integral in $K$ and $K^c$: I can apply dom convergence directly on $Omega$, right?
– Demetrio Masciurett
Jul 29 at 9:45
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to solve Problem 2.4.12 on page 64 of Karatzas-Shreve's book "Brownian motion and stochastic calculus":
My attempt is to use triangle inequality (denoting $Omega=C[0,infty)$)
$$|int_Omega f_ndP_n-int_Omega fdP|leq |int_Omega (f_n-f)dP_n|+|int_Omega fdP_n-int_Omega fdP|quadstar,$$
and estimate the first term with dominated convergence, the second using weak convergence of the measures.
For $epsilon>0$, since $(P_n)_n$ is tight (by Prohorov thm) I can choose a compact $Ksubset Omega$ such that $P_n(K)geq 1-epsilon$.
Fix $mgeq1$, then by dominated convergence for $n$ large enough I have
$$ |int_Omega (f_n-f)dP_m|=|int_K (f_n-f)dP_m|+|int_K^c (f_n-f)dP_m|
leq epsilon+epsilonsup_ngeq1,omegain K^c|f_n(omega)-f(omega)|.$$
To conclude I need $f$ to be bounded. Does this follow from the fact that $f$ is the compact-limit of uniformly bounded functions? (Note that I need boundedness also to use weak convergence on the second term of $star$)
Is it correct that if $forall mgeq1$ $|int_Omega (f_n-f)dP_m|to_ntoinfty0$, then
$ |int_Omega (f_n-f)dP_n|to_ntoinfty0?$
To be fair I think it's true only if the first convergence is uniform in $m$.
Thanks in advance :)
functional-analysis weak-convergence
asked Jul 28 at 13:54
Demetrio Masciurett
285
I'm trying to solve Problem 2.4.12 on page 64 of Karatzas-Shreve's book "Brownian motion and stochastic calculus":
My attempt is to use triangle inequality (denoting $Omega=C[0,infty)$)
$$|int_Omega f_ndP_n-int_Omega fdP|leq |int_Omega (f_n-f)dP_n|+|int_Omega fdP_n-int_Omega fdP|quadstar,$$
and estimate the first term with dominated convergence, the second using weak convergence of the measures.
For $epsilon>0$, since $(P_n)_n$ is tight (by Prohorov thm) I can choose a compact $Ksubset Omega$ such that $P_n(K)geq 1-epsilon$.
Fix $mgeq1$, then by dominated convergence for $n$ large enough I have
$$ |int_Omega (f_n-f)dP_m|=|int_K (f_n-f)dP_m|+|int_K^c (f_n-f)dP_m|
leq epsilon+epsilonsup_ngeq1,omegain K^c|f_n(omega)-f(omega)|.$$
To conclude I need $f$ to be bounded. Does this follow from the fact that $f$ is the compact-limit of uniformly bounded functions? (Note that I need boundedness also to use weak convergence on the second term of $star$)
Is it correct that if $forall mgeq1$ $|int_Omega (f_n-f)dP_m|to_ntoinfty0$, then
$ |int_Omega (f_n-f)dP_n|to_ntoinfty0?$
To be fair I think it's true only if the first convergence is uniform in $m$.
Thanks in advance :)
functional-analysis weak-convergence
asked Jul 28 at 13:54
Demetrio Masciurett
285
asked Jul 28 at 13:54
Demetrio Masciurett
285
asked Jul 28 at 13:54
Demetrio Masciurett
285
asked Jul 28 at 13:54
Demetrio Masciurett
285
285
You know that the sequence is uniformly bounded, so the same bound holds also for the pointwise limit
– Bob
Jul 28 at 21:23
Re your 2nd question: What's wrong about doing the estimate for $m=n$?
– saz
Jul 29 at 5:39
@saz: by fixing $mgeq1,epsilon>0$ I found $N(epsilon,m)geq1$ such that $forall ngeq N(epsilon,m),$ $int_Omega(f_n-f)dP_m<epsilon$. My problem is that $N(epsilon,m)$ depends on $m$. How do I know that $sup_m N(epsilon,m)$ is finite?
– Demetrio Masciurett
Jul 29 at 8:58
@Bob: thanks, since I have point-wise convergence and boundedness I even don't need to separate the integral in $K$ and $K^c$: I can apply dom convergence directly on $Omega$, right?
– Demetrio Masciurett
Jul 29 at 9:45
add a comment |Â
You know that the sequence is uniformly bounded, so the same bound holds also for the pointwise limit
– Bob
Jul 28 at 21:23
Re your 2nd question: What's wrong about doing the estimate for $m=n$?
– saz
Jul 29 at 5:39
@saz: by fixing $mgeq1,epsilon>0$ I found $N(epsilon,m)geq1$ such that $forall ngeq N(epsilon,m),$ $int_Omega(f_n-f)dP_m<epsilon$. My problem is that $N(epsilon,m)$ depends on $m$. How do I know that $sup_m N(epsilon,m)$ is finite?
– Demetrio Masciurett
Jul 29 at 8:58
@Bob: thanks, since I have point-wise convergence and boundedness I even don't need to separate the integral in $K$ and $K^c$: I can apply dom convergence directly on $Omega$, right?
– Demetrio Masciurett
Jul 29 at 9:45
You know that the sequence is uniformly bounded, so the same bound holds also for the pointwise limit
– Bob
Jul 28 at 21:23
You know that the sequence is uniformly bounded, so the same bound holds also for the pointwise limit
– Bob
Jul 28 at 21:23
Re your 2nd question: What's wrong about doing the estimate for $m=n$?
– saz
Jul 29 at 5:39
Re your 2nd question: What's wrong about doing the estimate for $m=n$?
– saz
Jul 29 at 5:39
@saz: by fixing $mgeq1,epsilon>0$ I found $N(epsilon,m)geq1$ such that $forall ngeq N(epsilon,m),$ $int_Omega(f_n-f)dP_m<epsilon$. My problem is that $N(epsilon,m)$ depends on $m$. How do I know that $sup_m N(epsilon,m)$ is finite?
– Demetrio Masciurett
Jul 29 at 8:58
@saz: by fixing $mgeq1,epsilon>0$ I found $N(epsilon,m)geq1$ such that $forall ngeq N(epsilon,m),$ $int_Omega(f_n-f)dP_m<epsilon$. My problem is that $N(epsilon,m)$ depends on $m$. How do I know that $sup_m N(epsilon,m)$ is finite?
– Demetrio Masciurett
Jul 29 at 8:58
@Bob: thanks, since I have point-wise convergence and boundedness I even don't need to separate the integral in $K$ and $K^c$: I can apply dom convergence directly on $Omega$, right?
– Demetrio Masciurett
Jul 29 at 9:45
@Bob: thanks, since I have point-wise convergence and boundedness I even don't need to separate the integral in $K$ and $K^c$: I can apply dom convergence directly on $Omega$, right?
– Demetrio Masciurett
Jul 29 at 9:45
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Finally got it:
$|int_Omega (f_n-f)dP_n|leq int_K_epsilon |f_n-f|dP_n +int_K^c_epsilon |f_n-f|dP_n
leq sup_omegain K_epsilon|f_n(omega)-f(omega)|+epsilon||f_n-f||_infty.$
Since $epsilon$ is arbitrary, letting $nto infty$ the last term goes to zero by uniform compact convergence of $(f_n)_ngeq 1$ to $f$.
answered Aug 4 at 11:59
Demetrio Masciurett
285
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Finally got it:
$|int_Omega (f_n-f)dP_n|leq int_K_epsilon |f_n-f|dP_n +int_K^c_epsilon |f_n-f|dP_n
leq sup_omegain K_epsilon|f_n(omega)-f(omega)|+epsilon||f_n-f||_infty.$
Since $epsilon$ is arbitrary, letting $nto infty$ the last term goes to zero by uniform compact convergence of $(f_n)_ngeq 1$ to $f$.
answered Aug 4 at 11:59
Demetrio Masciurett
285
add a comment |Â
up vote
1
down vote
accepted
Finally got it:
$|int_Omega (f_n-f)dP_n|leq int_K_epsilon |f_n-f|dP_n +int_K^c_epsilon |f_n-f|dP_n
leq sup_omegain K_epsilon|f_n(omega)-f(omega)|+epsilon||f_n-f||_infty.$
Since $epsilon$ is arbitrary, letting $nto infty$ the last term goes to zero by uniform compact convergence of $(f_n)_ngeq 1$ to $f$.
answered Aug 4 at 11:59
Demetrio Masciurett
285
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Finally got it:
$|int_Omega (f_n-f)dP_n|leq int_K_epsilon |f_n-f|dP_n +int_K^c_epsilon |f_n-f|dP_n
leq sup_omegain K_epsilon|f_n(omega)-f(omega)|+epsilon||f_n-f||_infty.$
Since $epsilon$ is arbitrary, letting $nto infty$ the last term goes to zero by uniform compact convergence of $(f_n)_ngeq 1$ to $f$.
answered Aug 4 at 11:59
Demetrio Masciurett
285
Finally got it:
$|int_Omega (f_n-f)dP_n|leq int_K_epsilon |f_n-f|dP_n +int_K^c_epsilon |f_n-f|dP_n
leq sup_omegain K_epsilon|f_n(omega)-f(omega)|+epsilon||f_n-f||_infty.$
Since $epsilon$ is arbitrary, letting $nto infty$ the last term goes to zero by uniform compact convergence of $(f_n)_ngeq 1$ to $f$.
answered Aug 4 at 11:59
Demetrio Masciurett
285
answered Aug 4 at 11:59
Demetrio Masciurett
285
answered Aug 4 at 11:59
Demetrio Masciurett
285
answered Aug 4 at 11:59
Demetrio Masciurett
285
285
add a comment |Â
add a comment |Â
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You know that the sequence is uniformly bounded, so the same bound holds also for the pointwise limit
– Bob
Jul 28 at 21:23
Re your 2nd question: What's wrong about doing the estimate for $m=n$?
– saz
Jul 29 at 5:39
@saz: by fixing $mgeq1,epsilon>0$ I found $N(epsilon,m)geq1$ such that $forall ngeq N(epsilon,m),$ $int_Omega(f_n-f)dP_m<epsilon$. My problem is that $N(epsilon,m)$ depends on $m$. How do I know that $sup_m N(epsilon,m)$ is finite?
– Demetrio Masciurett
Jul 29 at 8:58
@Bob: thanks, since I have point-wise convergence and boundedness I even don't need to separate the integral in $K$ and $K^c$: I can apply dom convergence directly on $Omega$, right?
– Demetrio Masciurett
Jul 29 at 9:45