For random variables $(X_k)_{k=1}^infty$ in $mathbb{R}$, find $(c_k)_{k=1}^infty$ such that $P(lim...
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I'm stuck with the following problem:
Let $(X_k)_{k=1}^infty$ be a sequence of real-valued random variables. Show that there is a real sequence $(c_k)_{k=1}^infty$ such that
$$Pleft(lim_{krightarrowinfty}frac{X_k}{c_k}=0right)=1.$$
With the typical approach for proving almost sure convergence through the Borel-Cantelli-lemma you arrive at
$$Pleft(lim_{krightarrowinfty}frac{X_k}{c_k}neq 0right)leq sum_{n=1}^infty Pleft(limsup_{krightarrowinfty}|X_k|>frac{c_k}{n}right)$$
and it would be sufficient to find $(c_k)_{k=1}^infty$ such that
$$sum_{k=1}^infty Pleft(|X_k|>frac{c_k}{n}right)<infty.$$
If the $X_k$ were identically distributed, I think a construction of the form $c_k=kcdot q_{1-1/(k^2)}$ could work, where $q_alpha$ is an $alpha$-quantile of $|X_1|$, but I can't do this in the general case.
How can I solve the general case (no further assumption about the $X_k$)? Is Borel-Cantelli even a good approach here?
probability convergence borel-cantelli-lemmas
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I'm stuck with the following problem:
Let $(X_k)_{k=1}^infty$ be a sequence of real-valued random variables. Show that there is a real sequence $(c_k)_{k=1}^infty$ such that
$$Pleft(lim_{krightarrowinfty}frac{X_k}{c_k}=0right)=1.$$
With the typical approach for proving almost sure convergence through the Borel-Cantelli-lemma you arrive at
$$Pleft(lim_{krightarrowinfty}frac{X_k}{c_k}neq 0right)leq sum_{n=1}^infty Pleft(limsup_{krightarrowinfty}|X_k|>frac{c_k}{n}right)$$
and it would be sufficient to find $(c_k)_{k=1}^infty$ such that
$$sum_{k=1}^infty Pleft(|X_k|>frac{c_k}{n}right)<infty.$$
If the $X_k$ were identically distributed, I think a construction of the form $c_k=kcdot q_{1-1/(k^2)}$ could work, where $q_alpha$ is an $alpha$-quantile of $|X_1|$, but I can't do this in the general case.
How can I solve the general case (no further assumption about the $X_k$)? Is Borel-Cantelli even a good approach here?
probability convergence borel-cantelli-lemmas
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm stuck with the following problem:
Let $(X_k)_{k=1}^infty$ be a sequence of real-valued random variables. Show that there is a real sequence $(c_k)_{k=1}^infty$ such that
$$Pleft(lim_{krightarrowinfty}frac{X_k}{c_k}=0right)=1.$$
With the typical approach for proving almost sure convergence through the Borel-Cantelli-lemma you arrive at
$$Pleft(lim_{krightarrowinfty}frac{X_k}{c_k}neq 0right)leq sum_{n=1}^infty Pleft(limsup_{krightarrowinfty}|X_k|>frac{c_k}{n}right)$$
and it would be sufficient to find $(c_k)_{k=1}^infty$ such that
$$sum_{k=1}^infty Pleft(|X_k|>frac{c_k}{n}right)<infty.$$
If the $X_k$ were identically distributed, I think a construction of the form $c_k=kcdot q_{1-1/(k^2)}$ could work, where $q_alpha$ is an $alpha$-quantile of $|X_1|$, but I can't do this in the general case.
How can I solve the general case (no further assumption about the $X_k$)? Is Borel-Cantelli even a good approach here?
probability convergence borel-cantelli-lemmas
I'm stuck with the following problem:
Let $(X_k)_{k=1}^infty$ be a sequence of real-valued random variables. Show that there is a real sequence $(c_k)_{k=1}^infty$ such that
$$Pleft(lim_{krightarrowinfty}frac{X_k}{c_k}=0right)=1.$$
With the typical approach for proving almost sure convergence through the Borel-Cantelli-lemma you arrive at
$$Pleft(lim_{krightarrowinfty}frac{X_k}{c_k}neq 0right)leq sum_{n=1}^infty Pleft(limsup_{krightarrowinfty}|X_k|>frac{c_k}{n}right)$$
and it would be sufficient to find $(c_k)_{k=1}^infty$ such that
$$sum_{k=1}^infty Pleft(|X_k|>frac{c_k}{n}right)<infty.$$
If the $X_k$ were identically distributed, I think a construction of the form $c_k=kcdot q_{1-1/(k^2)}$ could work, where $q_alpha$ is an $alpha$-quantile of $|X_1|$, but I can't do this in the general case.
How can I solve the general case (no further assumption about the $X_k$)? Is Borel-Cantelli even a good approach here?
probability convergence borel-cantelli-lemmas
probability convergence borel-cantelli-lemmas
asked Nov 15 at 8:59
Mau314
36418
36418
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If $P{|X_k| >frac {c_k} {2^{k}}} <frac 1 {2^{k}}$ then $sum_k P{|X_k| >frac 1 {2^{k}}c_k} <infty$ and Borel Cantelli Lemma tells you that $frac {X_k} {c_k} to 0$ almost surely. You can always find $c_k$'s satisfying this condition, right? [All you need is the fact that $P{|X_k| >t} to 0$ as $t to infty$].
Yes, that's in spirit what I wanted to indicate with this quantile-construction (just that I compensate $n$ with $k$ rather than $2^k$) and now I don't know why I thought there's a problem with that. Thanks.
– Mau314
Nov 15 at 9:21
I think you have a small typo: it should be $c_k/2^k$ in the first set
– mathworker21
Nov 15 at 10:51
@mathworker21 Thanks. You are absolutely right.
– Kavi Rama Murthy
Nov 15 at 11:49
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
If $P{|X_k| >frac {c_k} {2^{k}}} <frac 1 {2^{k}}$ then $sum_k P{|X_k| >frac 1 {2^{k}}c_k} <infty$ and Borel Cantelli Lemma tells you that $frac {X_k} {c_k} to 0$ almost surely. You can always find $c_k$'s satisfying this condition, right? [All you need is the fact that $P{|X_k| >t} to 0$ as $t to infty$].
Yes, that's in spirit what I wanted to indicate with this quantile-construction (just that I compensate $n$ with $k$ rather than $2^k$) and now I don't know why I thought there's a problem with that. Thanks.
– Mau314
Nov 15 at 9:21
I think you have a small typo: it should be $c_k/2^k$ in the first set
– mathworker21
Nov 15 at 10:51
@mathworker21 Thanks. You are absolutely right.
– Kavi Rama Murthy
Nov 15 at 11:49
add a comment |
up vote
2
down vote
If $P{|X_k| >frac {c_k} {2^{k}}} <frac 1 {2^{k}}$ then $sum_k P{|X_k| >frac 1 {2^{k}}c_k} <infty$ and Borel Cantelli Lemma tells you that $frac {X_k} {c_k} to 0$ almost surely. You can always find $c_k$'s satisfying this condition, right? [All you need is the fact that $P{|X_k| >t} to 0$ as $t to infty$].
Yes, that's in spirit what I wanted to indicate with this quantile-construction (just that I compensate $n$ with $k$ rather than $2^k$) and now I don't know why I thought there's a problem with that. Thanks.
– Mau314
Nov 15 at 9:21
I think you have a small typo: it should be $c_k/2^k$ in the first set
– mathworker21
Nov 15 at 10:51
@mathworker21 Thanks. You are absolutely right.
– Kavi Rama Murthy
Nov 15 at 11:49
add a comment |
up vote
2
down vote
up vote
2
down vote
If $P{|X_k| >frac {c_k} {2^{k}}} <frac 1 {2^{k}}$ then $sum_k P{|X_k| >frac 1 {2^{k}}c_k} <infty$ and Borel Cantelli Lemma tells you that $frac {X_k} {c_k} to 0$ almost surely. You can always find $c_k$'s satisfying this condition, right? [All you need is the fact that $P{|X_k| >t} to 0$ as $t to infty$].
If $P{|X_k| >frac {c_k} {2^{k}}} <frac 1 {2^{k}}$ then $sum_k P{|X_k| >frac 1 {2^{k}}c_k} <infty$ and Borel Cantelli Lemma tells you that $frac {X_k} {c_k} to 0$ almost surely. You can always find $c_k$'s satisfying this condition, right? [All you need is the fact that $P{|X_k| >t} to 0$ as $t to infty$].
edited Nov 15 at 11:49
answered Nov 15 at 9:04
Kavi Rama Murthy
42.1k31751
42.1k31751
Yes, that's in spirit what I wanted to indicate with this quantile-construction (just that I compensate $n$ with $k$ rather than $2^k$) and now I don't know why I thought there's a problem with that. Thanks.
– Mau314
Nov 15 at 9:21
I think you have a small typo: it should be $c_k/2^k$ in the first set
– mathworker21
Nov 15 at 10:51
@mathworker21 Thanks. You are absolutely right.
– Kavi Rama Murthy
Nov 15 at 11:49
add a comment |
Yes, that's in spirit what I wanted to indicate with this quantile-construction (just that I compensate $n$ with $k$ rather than $2^k$) and now I don't know why I thought there's a problem with that. Thanks.
– Mau314
Nov 15 at 9:21
I think you have a small typo: it should be $c_k/2^k$ in the first set
– mathworker21
Nov 15 at 10:51
@mathworker21 Thanks. You are absolutely right.
– Kavi Rama Murthy
Nov 15 at 11:49
Yes, that's in spirit what I wanted to indicate with this quantile-construction (just that I compensate $n$ with $k$ rather than $2^k$) and now I don't know why I thought there's a problem with that. Thanks.
– Mau314
Nov 15 at 9:21
Yes, that's in spirit what I wanted to indicate with this quantile-construction (just that I compensate $n$ with $k$ rather than $2^k$) and now I don't know why I thought there's a problem with that. Thanks.
– Mau314
Nov 15 at 9:21
I think you have a small typo: it should be $c_k/2^k$ in the first set
– mathworker21
Nov 15 at 10:51
I think you have a small typo: it should be $c_k/2^k$ in the first set
– mathworker21
Nov 15 at 10:51
@mathworker21 Thanks. You are absolutely right.
– Kavi Rama Murthy
Nov 15 at 11:49
@mathworker21 Thanks. You are absolutely right.
– Kavi Rama Murthy
Nov 15 at 11:49
add a comment |
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