Pointwise convergence for predictable processes implies ucp convergence
my task is to proof the following:
Let $H^n$ be a sequence of uniformly bounded predictable processes such that
$$lim_{ntoinfty}H^n_t= 0$$ almost surely holds for all $tgeq 0$.
Then we have $$H^nxrightarrow{ucp} 0,$$which means $sup_{sin [0,T]} H^n_sto 0$ for all $Tgeq 0$.
The problem is, I am not even sure whether this statement is true.
Assume the deterministic process
$$ H^n=begin{cases} t^n&text{for } t<1\
0&text{for } tgeq 1.end{cases}$$
Then we have $$H^n_tto 0 text{ for all } tgeq 0$$ but $sup_{sin[0,1]} H^n_s = 1$ and hence $H^nnotxrightarrow{ucp} 0$.
My ansatz for a proof would be to show the statement for continuous processes (which seems fairly easy) and then apply the monotone class theorem.
Can anyone tell me whether this statement is actually true and in case it is, what is wrong with my "counterexample"?
Thank you very much.
convergence stochastic-processes stochastic-calculus
asked 2 days ago
Agnetha TimaraAgnetha Timara
1086
add a comment |
my task is to proof the following:
Let $H^n$ be a sequence of uniformly bounded predictable processes such that
$$lim_{ntoinfty}H^n_t= 0$$ almost surely holds for all $tgeq 0$.
Then we have $$H^nxrightarrow{ucp} 0,$$which means $sup_{sin [0,T]} H^n_sto 0$ for all $Tgeq 0$.
The problem is, I am not even sure whether this statement is true.
Assume the deterministic process
$$ H^n=begin{cases} t^n&text{for } t<1\
0&text{for } tgeq 1.end{cases}$$
Then we have $$H^n_tto 0 text{ for all } tgeq 0$$ but $sup_{sin[0,1]} H^n_s = 1$ and hence $H^nnotxrightarrow{ucp} 0$.
My ansatz for a proof would be to show the statement for continuous processes (which seems fairly easy) and then apply the monotone class theorem.
Can anyone tell me whether this statement is actually true and in case it is, what is wrong with my "counterexample"?
Thank you very much.
convergence stochastic-processes stochastic-calculus
asked 2 days ago
Agnetha TimaraAgnetha Timara
1086
The fact that $H^n_tto 0 text{ for all } tgeq 0$ is wrong for $t=1$, so your counterexample doesn't prove that the statement that you are trying to prove is false. The almost surely part doesn't refer to the $([0,1], lambda)$ space ( where $lambda$ is Lebesgue measure) but to $mathbb{P}$. In you example the $H^n $ are determistic so the convergence is everywhere and not almost everywhere to the indicator function $1_{{1}}(t)$ over $tin [0,1]$.
– TheBridge
yesterday
Thanks for your comment. The process $H$ is defined to be zero at 1. Hence it is a right continuous process. And for each $t, epsilon>0$ you can find a $Nin mathbb N$ such that $H^n_t<epsilon$ for all $ngeq N$. But there is no $epsilon<1$ such that $H^n_t < epsilon$ for all $tgeq 0$.
– Agnetha Timara
yesterday
Ok with this specifications (which I missed in my comment) your counterexample is correct and the statement is false ! Your example is the same as the wiki example to show that pointwise convergence doesn't entails uniform convergence on compacts (en.wikipedia.org/wiki/Uniform_convergence).
– TheBridge
yesterday
add a comment |
my task is to proof the following:
Let $H^n$ be a sequence of uniformly bounded predictable processes such that
$$lim_{ntoinfty}H^n_t= 0$$ almost surely holds for all $tgeq 0$.
Then we have $$H^nxrightarrow{ucp} 0,$$which means $sup_{sin [0,T]} H^n_sto 0$ for all $Tgeq 0$.
The problem is, I am not even sure whether this statement is true.
Assume the deterministic process
$$ H^n=begin{cases} t^n&text{for } t<1\
0&text{for } tgeq 1.end{cases}$$
Then we have $$H^n_tto 0 text{ for all } tgeq 0$$ but $sup_{sin[0,1]} H^n_s = 1$ and hence $H^nnotxrightarrow{ucp} 0$.
My ansatz for a proof would be to show the statement for continuous processes (which seems fairly easy) and then apply the monotone class theorem.
Can anyone tell me whether this statement is actually true and in case it is, what is wrong with my "counterexample"?
Thank you very much.
convergence stochastic-processes stochastic-calculus
asked 2 days ago
Agnetha TimaraAgnetha Timara
1086
my task is to proof the following:
Let $H^n$ be a sequence of uniformly bounded predictable processes such that
$$lim_{ntoinfty}H^n_t= 0$$ almost surely holds for all $tgeq 0$.
Then we have $$H^nxrightarrow{ucp} 0,$$which means $sup_{sin [0,T]} H^n_sto 0$ for all $Tgeq 0$.
The problem is, I am not even sure whether this statement is true.
Assume the deterministic process
$$ H^n=begin{cases} t^n&text{for } t<1\
0&text{for } tgeq 1.end{cases}$$
Then we have $$H^n_tto 0 text{ for all } tgeq 0$$ but $sup_{sin[0,1]} H^n_s = 1$ and hence $H^nnotxrightarrow{ucp} 0$.
My ansatz for a proof would be to show the statement for continuous processes (which seems fairly easy) and then apply the monotone class theorem.
Can anyone tell me whether this statement is actually true and in case it is, what is wrong with my "counterexample"?
Thank you very much.
convergence stochastic-processes stochastic-calculus
convergence stochastic-processes stochastic-calculus
asked 2 days ago
Agnetha TimaraAgnetha Timara
1086
asked 2 days ago
Agnetha TimaraAgnetha Timara
1086
asked 2 days ago
Agnetha TimaraAgnetha Timara
1086
asked 2 days ago
Agnetha TimaraAgnetha Timara
1086
asked 2 days ago
Agnetha TimaraAgnetha Timara
1086
1086
The fact that $H^n_tto 0 text{ for all } tgeq 0$ is wrong for $t=1$, so your counterexample doesn't prove that the statement that you are trying to prove is false. The almost surely part doesn't refer to the $([0,1], lambda)$ space ( where $lambda$ is Lebesgue measure) but to $mathbb{P}$. In you example the $H^n $ are determistic so the convergence is everywhere and not almost everywhere to the indicator function $1_{{1}}(t)$ over $tin [0,1]$.
– TheBridge
yesterday
Thanks for your comment. The process $H$ is defined to be zero at 1. Hence it is a right continuous process. And for each $t, epsilon>0$ you can find a $Nin mathbb N$ such that $H^n_t<epsilon$ for all $ngeq N$. But there is no $epsilon<1$ such that $H^n_t < epsilon$ for all $tgeq 0$.
– Agnetha Timara
yesterday
Ok with this specifications (which I missed in my comment) your counterexample is correct and the statement is false ! Your example is the same as the wiki example to show that pointwise convergence doesn't entails uniform convergence on compacts (en.wikipedia.org/wiki/Uniform_convergence).
– TheBridge
yesterday
add a comment |
The fact that $H^n_tto 0 text{ for all } tgeq 0$ is wrong for $t=1$, so your counterexample doesn't prove that the statement that you are trying to prove is false. The almost surely part doesn't refer to the $([0,1], lambda)$ space ( where $lambda$ is Lebesgue measure) but to $mathbb{P}$. In you example the $H^n $ are determistic so the convergence is everywhere and not almost everywhere to the indicator function $1_{{1}}(t)$ over $tin [0,1]$.
– TheBridge
yesterday
Thanks for your comment. The process $H$ is defined to be zero at 1. Hence it is a right continuous process. And for each $t, epsilon>0$ you can find a $Nin mathbb N$ such that $H^n_t<epsilon$ for all $ngeq N$. But there is no $epsilon<1$ such that $H^n_t < epsilon$ for all $tgeq 0$.
– Agnetha Timara
yesterday
Ok with this specifications (which I missed in my comment) your counterexample is correct and the statement is false ! Your example is the same as the wiki example to show that pointwise convergence doesn't entails uniform convergence on compacts (en.wikipedia.org/wiki/Uniform_convergence).
– TheBridge
yesterday
The fact that $H^n_tto 0 text{ for all } tgeq 0$ is wrong for $t=1$, so your counterexample doesn't prove that the statement that you are trying to prove is false. The almost surely part doesn't refer to the $([0,1], lambda)$ space ( where $lambda$ is Lebesgue measure) but to $mathbb{P}$. In you example the $H^n $ are determistic so the convergence is everywhere and not almost everywhere to the indicator function $1_{{1}}(t)$ over $tin [0,1]$.
– TheBridge
yesterday
The fact that $H^n_tto 0 text{ for all } tgeq 0$ is wrong for $t=1$, so your counterexample doesn't prove that the statement that you are trying to prove is false. The almost surely part doesn't refer to the $([0,1], lambda)$ space ( where $lambda$ is Lebesgue measure) but to $mathbb{P}$. In you example the $H^n $ are determistic so the convergence is everywhere and not almost everywhere to the indicator function $1_{{1}}(t)$ over $tin [0,1]$.
– TheBridge
yesterday
Thanks for your comment. The process $H$ is defined to be zero at 1. Hence it is a right continuous process. And for each $t, epsilon>0$ you can find a $Nin mathbb N$ such that $H^n_t<epsilon$ for all $ngeq N$. But there is no $epsilon<1$ such that $H^n_t < epsilon$ for all $tgeq 0$.
– Agnetha Timara
yesterday
Thanks for your comment. The process $H$ is defined to be zero at 1. Hence it is a right continuous process. And for each $t, epsilon>0$ you can find a $Nin mathbb N$ such that $H^n_t<epsilon$ for all $ngeq N$. But there is no $epsilon<1$ such that $H^n_t < epsilon$ for all $tgeq 0$.
– Agnetha Timara
yesterday
Ok with this specifications (which I missed in my comment) your counterexample is correct and the statement is false ! Your example is the same as the wiki example to show that pointwise convergence doesn't entails uniform convergence on compacts (en.wikipedia.org/wiki/Uniform_convergence).
– TheBridge
yesterday
Ok with this specifications (which I missed in my comment) your counterexample is correct and the statement is false ! Your example is the same as the wiki example to show that pointwise convergence doesn't entails uniform convergence on compacts (en.wikipedia.org/wiki/Uniform_convergence).
– TheBridge
yesterday
add a comment |
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The fact that $H^n_tto 0 text{ for all } tgeq 0$ is wrong for $t=1$, so your counterexample doesn't prove that the statement that you are trying to prove is false. The almost surely part doesn't refer to the $([0,1], lambda)$ space ( where $lambda$ is Lebesgue measure) but to $mathbb{P}$. In you example the $H^n $ are determistic so the convergence is everywhere and not almost everywhere to the indicator function $1_{{1}}(t)$ over $tin [0,1]$.
– TheBridge
yesterday
Thanks for your comment. The process $H$ is defined to be zero at 1. Hence it is a right continuous process. And for each $t, epsilon>0$ you can find a $Nin mathbb N$ such that $H^n_t<epsilon$ for all $ngeq N$. But there is no $epsilon<1$ such that $H^n_t < epsilon$ for all $tgeq 0$.
– Agnetha Timara
yesterday
Ok with this specifications (which I missed in my comment) your counterexample is correct and the statement is false ! Your example is the same as the wiki example to show that pointwise convergence doesn't entails uniform convergence on compacts (en.wikipedia.org/wiki/Uniform_convergence).
– TheBridge
yesterday