Convergence of âAAâ Subsequence of sequence strongly convergent of order $alpha$
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Let $(Omega, mathcal F, mathbb P)$ be a probability space and consider a sequence $X = (X_n)_nin mathbb N$ of integrable random variables, let $X_inftyin L_1$ and $alpha > 0$.
Say that $X$ is convergent of order $alpha$ to $X_infty$ if there exists a $Cgeq 0$ s.t.
$$|X_n- X_infty|_1 leq Cfrac1n^alpha, forall n geq 1.$$
In this case I will write $X to_alpha X_infty$. The motivation for this comes from the Euler-Maruyama method for numerically solving SDE's.
One can prove some nice properties, e.g.
- $(X)_nin mathbb N to_alpha X$
- $X to_alpha X_infty$, $n : mathbb N to mathbb N$ monotone $Rightarrow$ $(X_n(k))_k to X_infty$
- $X to_alpha X_infty$ and $X'to_alpha X_infty$ $Rightarrow Xcap X'to_alpha X_infty$
where $(Xcap X')_2n = X_n$ and $(Xcap X')_2n-1 = X'_n$.
What about the following one:
Is is true that $Xto_alpha X_infty$ and ($X in A $ eventually $Leftrightarrow X'in A$ eventually) $Rightarrow X'to_alpha X_infty$?
Here $Xin A$ eventually if $X_nin A$ for large $n$.
It would be nice to have (*), but I don't see how this could be proven.
I was thinking of $A_k := X - X_infty$, then $Xin A_k$ eventually, hence $X'in A_k$ eventually.
In other words
$$forall kin mathbb N : exists i geq 0 : forall j geq i : |X_j' - X_infty|_1 leq C frac1k^alpha$$
But it doesn't seem like that's enough to conclude that $X'to_alpha X_infty$.
(*) In this case this subsequential structure could be induced by a convergence structure.
real-analysis stochastic-processes stochastic-calculus
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up vote
1
down vote
favorite
Let $(Omega, mathcal F, mathbb P)$ be a probability space and consider a sequence $X = (X_n)_nin mathbb N$ of integrable random variables, let $X_inftyin L_1$ and $alpha > 0$.
Say that $X$ is convergent of order $alpha$ to $X_infty$ if there exists a $Cgeq 0$ s.t.
$$|X_n- X_infty|_1 leq Cfrac1n^alpha, forall n geq 1.$$
In this case I will write $X to_alpha X_infty$. The motivation for this comes from the Euler-Maruyama method for numerically solving SDE's.
One can prove some nice properties, e.g.
- $(X)_nin mathbb N to_alpha X$
- $X to_alpha X_infty$, $n : mathbb N to mathbb N$ monotone $Rightarrow$ $(X_n(k))_k to X_infty$
- $X to_alpha X_infty$ and $X'to_alpha X_infty$ $Rightarrow Xcap X'to_alpha X_infty$
where $(Xcap X')_2n = X_n$ and $(Xcap X')_2n-1 = X'_n$.
What about the following one:
Is is true that $Xto_alpha X_infty$ and ($X in A $ eventually $Leftrightarrow X'in A$ eventually) $Rightarrow X'to_alpha X_infty$?
Here $Xin A$ eventually if $X_nin A$ for large $n$.
It would be nice to have (*), but I don't see how this could be proven.
I was thinking of $A_k := X - X_infty$, then $Xin A_k$ eventually, hence $X'in A_k$ eventually.
In other words
$$forall kin mathbb N : exists i geq 0 : forall j geq i : |X_j' - X_infty|_1 leq C frac1k^alpha$$
But it doesn't seem like that's enough to conclude that $X'to_alpha X_infty$.
(*) In this case this subsequential structure could be induced by a convergence structure.
real-analysis stochastic-processes stochastic-calculus
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $(Omega, mathcal F, mathbb P)$ be a probability space and consider a sequence $X = (X_n)_nin mathbb N$ of integrable random variables, let $X_inftyin L_1$ and $alpha > 0$.
Say that $X$ is convergent of order $alpha$ to $X_infty$ if there exists a $Cgeq 0$ s.t.
$$|X_n- X_infty|_1 leq Cfrac1n^alpha, forall n geq 1.$$
In this case I will write $X to_alpha X_infty$. The motivation for this comes from the Euler-Maruyama method for numerically solving SDE's.
One can prove some nice properties, e.g.
- $(X)_nin mathbb N to_alpha X$
- $X to_alpha X_infty$, $n : mathbb N to mathbb N$ monotone $Rightarrow$ $(X_n(k))_k to X_infty$
- $X to_alpha X_infty$ and $X'to_alpha X_infty$ $Rightarrow Xcap X'to_alpha X_infty$
where $(Xcap X')_2n = X_n$ and $(Xcap X')_2n-1 = X'_n$.
What about the following one:
Is is true that $Xto_alpha X_infty$ and ($X in A $ eventually $Leftrightarrow X'in A$ eventually) $Rightarrow X'to_alpha X_infty$?
Here $Xin A$ eventually if $X_nin A$ for large $n$.
It would be nice to have (*), but I don't see how this could be proven.
I was thinking of $A_k := X - X_infty$, then $Xin A_k$ eventually, hence $X'in A_k$ eventually.
In other words
$$forall kin mathbb N : exists i geq 0 : forall j geq i : |X_j' - X_infty|_1 leq C frac1k^alpha$$
But it doesn't seem like that's enough to conclude that $X'to_alpha X_infty$.
(*) In this case this subsequential structure could be induced by a convergence structure.
real-analysis stochastic-processes stochastic-calculus
Let $(Omega, mathcal F, mathbb P)$ be a probability space and consider a sequence $X = (X_n)_nin mathbb N$ of integrable random variables, let $X_inftyin L_1$ and $alpha > 0$.
Say that $X$ is convergent of order $alpha$ to $X_infty$ if there exists a $Cgeq 0$ s.t.
$$|X_n- X_infty|_1 leq Cfrac1n^alpha, forall n geq 1.$$
In this case I will write $X to_alpha X_infty$. The motivation for this comes from the Euler-Maruyama method for numerically solving SDE's.
One can prove some nice properties, e.g.
- $(X)_nin mathbb N to_alpha X$
- $X to_alpha X_infty$, $n : mathbb N to mathbb N$ monotone $Rightarrow$ $(X_n(k))_k to X_infty$
- $X to_alpha X_infty$ and $X'to_alpha X_infty$ $Rightarrow Xcap X'to_alpha X_infty$
where $(Xcap X')_2n = X_n$ and $(Xcap X')_2n-1 = X'_n$.
What about the following one:
Is is true that $Xto_alpha X_infty$ and ($X in A $ eventually $Leftrightarrow X'in A$ eventually) $Rightarrow X'to_alpha X_infty$?
Here $Xin A$ eventually if $X_nin A$ for large $n$.
It would be nice to have (*), but I don't see how this could be proven.
I was thinking of $A_k := X - X_infty$, then $Xin A_k$ eventually, hence $X'in A_k$ eventually.
In other words
$$forall kin mathbb N : exists i geq 0 : forall j geq i : |X_j' - X_infty|_1 leq C frac1k^alpha$$
But it doesn't seem like that's enough to conclude that $X'to_alpha X_infty$.
(*) In this case this subsequential structure could be induced by a convergence structure.
real-analysis stochastic-processes stochastic-calculus
asked Aug 28 at 17:19
Stefan Perko
8,31611641
8,31611641
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1 Answer
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I think the following map be a counterexample. For any sequence $X$, define $X'$ by $X'_n = X_sqrtn$, where the index is rounded down. Certainly $X'$ has the same eventuality set as $X$. However,
$$
|X'_n-X_infty|_1=|X_sqrtn-X_infty|_1le C/(sqrtn)^alpha=C/n^alpha/2
$$
so you can only say $X'to_alpha/2X_infty.$
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I think the following map be a counterexample. For any sequence $X$, define $X'$ by $X'_n = X_sqrtn$, where the index is rounded down. Certainly $X'$ has the same eventuality set as $X$. However,
$$
|X'_n-X_infty|_1=|X_sqrtn-X_infty|_1le C/(sqrtn)^alpha=C/n^alpha/2
$$
so you can only say $X'to_alpha/2X_infty.$
add a comment |Â
up vote
2
down vote
accepted
I think the following map be a counterexample. For any sequence $X$, define $X'$ by $X'_n = X_sqrtn$, where the index is rounded down. Certainly $X'$ has the same eventuality set as $X$. However,
$$
|X'_n-X_infty|_1=|X_sqrtn-X_infty|_1le C/(sqrtn)^alpha=C/n^alpha/2
$$
so you can only say $X'to_alpha/2X_infty.$
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I think the following map be a counterexample. For any sequence $X$, define $X'$ by $X'_n = X_sqrtn$, where the index is rounded down. Certainly $X'$ has the same eventuality set as $X$. However,
$$
|X'_n-X_infty|_1=|X_sqrtn-X_infty|_1le C/(sqrtn)^alpha=C/n^alpha/2
$$
so you can only say $X'to_alpha/2X_infty.$
I think the following map be a counterexample. For any sequence $X$, define $X'$ by $X'_n = X_sqrtn$, where the index is rounded down. Certainly $X'$ has the same eventuality set as $X$. However,
$$
|X'_n-X_infty|_1=|X_sqrtn-X_infty|_1le C/(sqrtn)^alpha=C/n^alpha/2
$$
so you can only say $X'to_alpha/2X_infty.$
answered Aug 28 at 19:19
Mike Earnest
17.5k11749
17.5k11749
add a comment |Â
add a comment |Â
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