Mixing
If $E[|sum_{i < l leq j} xi_l |^gamma] leq (sum_{i < l leq j} u_l)^alpha$ and $sum u_l < infty$ then...
$begingroup$
Suppose $xi_1, xi_2, ldots$ are random variables that satisfy
$$Eleft[left|sum_{i < l leq j} xi_l right|^gammaright] leq left(sum_{i < l leq j} u_lright)^alpha$$
for $gamma geq 0$, $alpha > 1$, $u_l$ non-negative, and $sum u_l < infty$. I want to show $sum xi_l$ converges almost surely.
Let $S_m = sum_{i = 1}^m xi_i$ and $M_m = max_{1 leq i leq m} |S_i|$. This criterion on the expectation can be used to show
$$P(|S_j - S_i| geq lambda) leq frac{1}{lambda^gamma} left(sum_{i < l leq j} u_lright)^alpha$$
Because of this we can use a theorem from Billingsley's book Convergence of Probability Measures (first edition, 1968) to say
$$P(M_m geq lambda) leq frac{K}{lambda^gamma} (u_1 + ldots + u_m)^alpha$$
(Billingsley says to use that theorem to reach the desired result.) $K$ depends only on $gamma$ and $alpha$.
After this I'm not sure how to proceed. Sure, one could take $m to infty$ and then see that the sum is bounded almost surely but I don't see how that can show that the sum is convergent. And I'm not seeing where to judiciously use the Borel-Cantelli lemma.
Basically, I don't see how the bound on $P(M_m geq lambda)$ is useful for proving almost-sure convergence but supposedly it can (and in fact needs to be) used to get the desired result.
So what should I do next?
probability-theory convergence summation random-variables almost-everywhere
asked Jan 7 at 7:28
cgmilcgmil
664316
$endgroup$
add a comment |
$begingroup$
Suppose $xi_1, xi_2, ldots$ are random variables that satisfy
$$Eleft[left|sum_{i < l leq j} xi_l right|^gammaright] leq left(sum_{i < l leq j} u_lright)^alpha$$
for $gamma geq 0$, $alpha > 1$, $u_l$ non-negative, and $sum u_l < infty$. I want to show $sum xi_l$ converges almost surely.
Let $S_m = sum_{i = 1}^m xi_i$ and $M_m = max_{1 leq i leq m} |S_i|$. This criterion on the expectation can be used to show
$$P(|S_j - S_i| geq lambda) leq frac{1}{lambda^gamma} left(sum_{i < l leq j} u_lright)^alpha$$
Because of this we can use a theorem from Billingsley's book Convergence of Probability Measures (first edition, 1968) to say
$$P(M_m geq lambda) leq frac{K}{lambda^gamma} (u_1 + ldots + u_m)^alpha$$
(Billingsley says to use that theorem to reach the desired result.) $K$ depends only on $gamma$ and $alpha$.
After this I'm not sure how to proceed. Sure, one could take $m to infty$ and then see that the sum is bounded almost surely but I don't see how that can show that the sum is convergent. And I'm not seeing where to judiciously use the Borel-Cantelli lemma.
Basically, I don't see how the bound on $P(M_m geq lambda)$ is useful for proving almost-sure convergence but supposedly it can (and in fact needs to be) used to get the desired result.
So what should I do next?
probability-theory convergence summation random-variables almost-everywhere
asked Jan 7 at 7:28
cgmilcgmil
664316
$endgroup$
$begingroup$
Are you sure there is no independence assumption here? The theorems you are quoting are not true without independence.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 7:49
$begingroup$
There is an obvious counterexample with $pm 1$ valued random variables $xi_i$.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 7:50
$begingroup$
@KaviRamaMurthy Billingsley explicitly notes that the variables need not be independent, and the counterexample you mention does not satisfy the assumption mentioned; the tightest $u_l$ you could have would not be summable.
$endgroup$
– cgmil
Jan 10 at 22:44
add a comment |
$begingroup$
Suppose $xi_1, xi_2, ldots$ are random variables that satisfy
$$Eleft[left|sum_{i < l leq j} xi_l right|^gammaright] leq left(sum_{i < l leq j} u_lright)^alpha$$
for $gamma geq 0$, $alpha > 1$, $u_l$ non-negative, and $sum u_l < infty$. I want to show $sum xi_l$ converges almost surely.
Let $S_m = sum_{i = 1}^m xi_i$ and $M_m = max_{1 leq i leq m} |S_i|$. This criterion on the expectation can be used to show
$$P(|S_j - S_i| geq lambda) leq frac{1}{lambda^gamma} left(sum_{i < l leq j} u_lright)^alpha$$
Because of this we can use a theorem from Billingsley's book Convergence of Probability Measures (first edition, 1968) to say
$$P(M_m geq lambda) leq frac{K}{lambda^gamma} (u_1 + ldots + u_m)^alpha$$
(Billingsley says to use that theorem to reach the desired result.) $K$ depends only on $gamma$ and $alpha$.
After this I'm not sure how to proceed. Sure, one could take $m to infty$ and then see that the sum is bounded almost surely but I don't see how that can show that the sum is convergent. And I'm not seeing where to judiciously use the Borel-Cantelli lemma.
Basically, I don't see how the bound on $P(M_m geq lambda)$ is useful for proving almost-sure convergence but supposedly it can (and in fact needs to be) used to get the desired result.
So what should I do next?
probability-theory convergence summation random-variables almost-everywhere
asked Jan 7 at 7:28
cgmilcgmil
664316
$endgroup$
Suppose $xi_1, xi_2, ldots$ are random variables that satisfy
$$Eleft[left|sum_{i < l leq j} xi_l right|^gammaright] leq left(sum_{i < l leq j} u_lright)^alpha$$
for $gamma geq 0$, $alpha > 1$, $u_l$ non-negative, and $sum u_l < infty$. I want to show $sum xi_l$ converges almost surely.
Let $S_m = sum_{i = 1}^m xi_i$ and $M_m = max_{1 leq i leq m} |S_i|$. This criterion on the expectation can be used to show
$$P(|S_j - S_i| geq lambda) leq frac{1}{lambda^gamma} left(sum_{i < l leq j} u_lright)^alpha$$
Because of this we can use a theorem from Billingsley's book Convergence of Probability Measures (first edition, 1968) to say
$$P(M_m geq lambda) leq frac{K}{lambda^gamma} (u_1 + ldots + u_m)^alpha$$
(Billingsley says to use that theorem to reach the desired result.) $K$ depends only on $gamma$ and $alpha$.
After this I'm not sure how to proceed. Sure, one could take $m to infty$ and then see that the sum is bounded almost surely but I don't see how that can show that the sum is convergent. And I'm not seeing where to judiciously use the Borel-Cantelli lemma.
Basically, I don't see how the bound on $P(M_m geq lambda)$ is useful for proving almost-sure convergence but supposedly it can (and in fact needs to be) used to get the desired result.
So what should I do next?
probability-theory convergence summation random-variables almost-everywhere
probability-theory convergence summation random-variables almost-everywhere
asked Jan 7 at 7:28
cgmilcgmil
664316
asked Jan 7 at 7:28
cgmilcgmil
664316
asked Jan 7 at 7:28
cgmilcgmil
664316
asked Jan 7 at 7:28
cgmilcgmil
664316
asked Jan 7 at 7:28
cgmilcgmil
664316
664316
$begingroup$
Are you sure there is no independence assumption here? The theorems you are quoting are not true without independence.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 7:49
$begingroup$
There is an obvious counterexample with $pm 1$ valued random variables $xi_i$.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 7:50
$begingroup$
@KaviRamaMurthy Billingsley explicitly notes that the variables need not be independent, and the counterexample you mention does not satisfy the assumption mentioned; the tightest $u_l$ you could have would not be summable.
$endgroup$
– cgmil
Jan 10 at 22:44
add a comment |
$begingroup$
Are you sure there is no independence assumption here? The theorems you are quoting are not true without independence.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 7:49
$begingroup$
There is an obvious counterexample with $pm 1$ valued random variables $xi_i$.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 7:50
$begingroup$
@KaviRamaMurthy Billingsley explicitly notes that the variables need not be independent, and the counterexample you mention does not satisfy the assumption mentioned; the tightest $u_l$ you could have would not be summable.
$endgroup$
– cgmil
Jan 10 at 22:44
$begingroup$
Are you sure there is no independence assumption here? The theorems you are quoting are not true without independence.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 7:49
$begingroup$
Are you sure there is no independence assumption here? The theorems you are quoting are not true without independence.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 7:49
$begingroup$
There is an obvious counterexample with $pm 1$ valued random variables $xi_i$.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 7:50
$begingroup$
There is an obvious counterexample with $pm 1$ valued random variables $xi_i$.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 7:50
$begingroup$
@KaviRamaMurthy Billingsley explicitly notes that the variables need not be independent, and the counterexample you mention does not satisfy the assumption mentioned; the tightest $u_l$ you could have would not be summable.
$endgroup$
– cgmil
Jan 10 at 22:44
$begingroup$
@KaviRamaMurthy Billingsley explicitly notes that the variables need not be independent, and the counterexample you mention does not satisfy the assumption mentioned; the tightest $u_l$ you could have would not be summable.
$endgroup$
– cgmil
Jan 10 at 22:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I guess we need some kind of weak dependence assumption on $(xi_j)$, but anyway, this is only about how we get the convergence result using the maximal inequality about $M_n$. First, we note that we can bound
$$
P(max_{nle ile N}|S_i-S_n|>lambda)lefrac{K}{lambda^gamma}(u_{n+1} +cdots u_N)^alpha
$$ and hence
$$
P(max_{nle i, jle N}|S_i-S_j|>lambda)lefrac{K'}{lambda^gamma}(u_{n+1} +cdots u_N)^alpha.
$$ (We can regard $xi_j$ as starting from index $j=n+1$.) To show that $S_n$ converges, id est, that $limsup_{ntoinfty} S_n = liminf_{ntoinfty} S_n$, it seems natural for us to control the oscillation defined by
$$
limsup_{i,jto infty}|S_i-S_j| =lim_{ntoinfty}sup_{i,jge n} |S_i-S_j|.
$$ Let $W_{n,N} = sup_{nle i,jle N} |S_i-S_j| $ and $W_n =sup_{ i,jge n} |S_i-S_j|=lim_{Ntoinfty}W_{n,N}$. We have
$$
P(W_n>lambda) =lim_{Ntoinfty}P(W_{n,N}>lambda)
$$by monotonic convergence of $W_{n,N}$. Hence we get the bound
$$
P(W_n>lambda)lefrac{K'}{lambda^gamma}left(sum_{j>n}u_{j} right)^alpha.tag{*}
$$ Finally, we observe
$$
limsup_{ntoinfty} S_n -liminf_{ntoinfty} S_n =lim_{ntoinfty} W_n=:W
$$ and
$$
{S_ntext{ diverges}} = {limsup_{ntoinfty} S_n -liminf_{ntoinfty} S_n>0} =bigcup_{jinmathbb{N}}{Wge 1/j}.
$$ Since it holds that
$$
P(Wge lambda) =lim_{ntoinfty} P(W_nge lambda)= 0
$$ for all $lambda>0$ by the above estimate $(*)$, it follows that ${S_ntext{ diverges}}$ is a countable union of null probability events, and hence has probability $0$.
answered Jan 7 at 8:48
SongSong
10.7k628
$endgroup$
$begingroup$
Thank you for your answer. It was very helpful! I have a follow-up question of a similar nature; perhaps you could answer it too? math.stackexchange.com/q/3069263/360447
$endgroup$
– cgmil
Jan 10 at 22:40
add a comment |
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$begingroup$
I guess we need some kind of weak dependence assumption on $(xi_j)$, but anyway, this is only about how we get the convergence result using the maximal inequality about $M_n$. First, we note that we can bound
$$
P(max_{nle ile N}|S_i-S_n|>lambda)lefrac{K}{lambda^gamma}(u_{n+1} +cdots u_N)^alpha
$$ and hence
$$
P(max_{nle i, jle N}|S_i-S_j|>lambda)lefrac{K'}{lambda^gamma}(u_{n+1} +cdots u_N)^alpha.
$$ (We can regard $xi_j$ as starting from index $j=n+1$.) To show that $S_n$ converges, id est, that $limsup_{ntoinfty} S_n = liminf_{ntoinfty} S_n$, it seems natural for us to control the oscillation defined by
$$
limsup_{i,jto infty}|S_i-S_j| =lim_{ntoinfty}sup_{i,jge n} |S_i-S_j|.
$$ Let $W_{n,N} = sup_{nle i,jle N} |S_i-S_j| $ and $W_n =sup_{ i,jge n} |S_i-S_j|=lim_{Ntoinfty}W_{n,N}$. We have
$$
P(W_n>lambda) =lim_{Ntoinfty}P(W_{n,N}>lambda)
$$by monotonic convergence of $W_{n,N}$. Hence we get the bound
$$
P(W_n>lambda)lefrac{K'}{lambda^gamma}left(sum_{j>n}u_{j} right)^alpha.tag{*}
$$ Finally, we observe
$$
limsup_{ntoinfty} S_n -liminf_{ntoinfty} S_n =lim_{ntoinfty} W_n=:W
$$ and
$$
{S_ntext{ diverges}} = {limsup_{ntoinfty} S_n -liminf_{ntoinfty} S_n>0} =bigcup_{jinmathbb{N}}{Wge 1/j}.
$$ Since it holds that
$$
P(Wge lambda) =lim_{ntoinfty} P(W_nge lambda)= 0
$$ for all $lambda>0$ by the above estimate $(*)$, it follows that ${S_ntext{ diverges}}$ is a countable union of null probability events, and hence has probability $0$.
answered Jan 7 at 8:48
SongSong
10.7k628
$endgroup$
$begingroup$
Thank you for your answer. It was very helpful! I have a follow-up question of a similar nature; perhaps you could answer it too? math.stackexchange.com/q/3069263/360447
$endgroup$
– cgmil
Jan 10 at 22:40
add a comment |
$begingroup$
I guess we need some kind of weak dependence assumption on $(xi_j)$, but anyway, this is only about how we get the convergence result using the maximal inequality about $M_n$. First, we note that we can bound
$$
P(max_{nle ile N}|S_i-S_n|>lambda)lefrac{K}{lambda^gamma}(u_{n+1} +cdots u_N)^alpha
$$ and hence
$$
P(max_{nle i, jle N}|S_i-S_j|>lambda)lefrac{K'}{lambda^gamma}(u_{n+1} +cdots u_N)^alpha.
$$ (We can regard $xi_j$ as starting from index $j=n+1$.) To show that $S_n$ converges, id est, that $limsup_{ntoinfty} S_n = liminf_{ntoinfty} S_n$, it seems natural for us to control the oscillation defined by
$$
limsup_{i,jto infty}|S_i-S_j| =lim_{ntoinfty}sup_{i,jge n} |S_i-S_j|.
$$ Let $W_{n,N} = sup_{nle i,jle N} |S_i-S_j| $ and $W_n =sup_{ i,jge n} |S_i-S_j|=lim_{Ntoinfty}W_{n,N}$. We have
$$
P(W_n>lambda) =lim_{Ntoinfty}P(W_{n,N}>lambda)
$$by monotonic convergence of $W_{n,N}$. Hence we get the bound
$$
P(W_n>lambda)lefrac{K'}{lambda^gamma}left(sum_{j>n}u_{j} right)^alpha.tag{*}
$$ Finally, we observe
$$
limsup_{ntoinfty} S_n -liminf_{ntoinfty} S_n =lim_{ntoinfty} W_n=:W
$$ and
$$
{S_ntext{ diverges}} = {limsup_{ntoinfty} S_n -liminf_{ntoinfty} S_n>0} =bigcup_{jinmathbb{N}}{Wge 1/j}.
$$ Since it holds that
$$
P(Wge lambda) =lim_{ntoinfty} P(W_nge lambda)= 0
$$ for all $lambda>0$ by the above estimate $(*)$, it follows that ${S_ntext{ diverges}}$ is a countable union of null probability events, and hence has probability $0$.
answered Jan 7 at 8:48
SongSong
10.7k628
$endgroup$
$begingroup$
Thank you for your answer. It was very helpful! I have a follow-up question of a similar nature; perhaps you could answer it too? math.stackexchange.com/q/3069263/360447
$endgroup$
– cgmil
Jan 10 at 22:40
add a comment |
$begingroup$
I guess we need some kind of weak dependence assumption on $(xi_j)$, but anyway, this is only about how we get the convergence result using the maximal inequality about $M_n$. First, we note that we can bound
$$
P(max_{nle ile N}|S_i-S_n|>lambda)lefrac{K}{lambda^gamma}(u_{n+1} +cdots u_N)^alpha
$$ and hence
$$
P(max_{nle i, jle N}|S_i-S_j|>lambda)lefrac{K'}{lambda^gamma}(u_{n+1} +cdots u_N)^alpha.
$$ (We can regard $xi_j$ as starting from index $j=n+1$.) To show that $S_n$ converges, id est, that $limsup_{ntoinfty} S_n = liminf_{ntoinfty} S_n$, it seems natural for us to control the oscillation defined by
$$
limsup_{i,jto infty}|S_i-S_j| =lim_{ntoinfty}sup_{i,jge n} |S_i-S_j|.
$$ Let $W_{n,N} = sup_{nle i,jle N} |S_i-S_j| $ and $W_n =sup_{ i,jge n} |S_i-S_j|=lim_{Ntoinfty}W_{n,N}$. We have
$$
P(W_n>lambda) =lim_{Ntoinfty}P(W_{n,N}>lambda)
$$by monotonic convergence of $W_{n,N}$. Hence we get the bound
$$
P(W_n>lambda)lefrac{K'}{lambda^gamma}left(sum_{j>n}u_{j} right)^alpha.tag{*}
$$ Finally, we observe
$$
limsup_{ntoinfty} S_n -liminf_{ntoinfty} S_n =lim_{ntoinfty} W_n=:W
$$ and
$$
{S_ntext{ diverges}} = {limsup_{ntoinfty} S_n -liminf_{ntoinfty} S_n>0} =bigcup_{jinmathbb{N}}{Wge 1/j}.
$$ Since it holds that
$$
P(Wge lambda) =lim_{ntoinfty} P(W_nge lambda)= 0
$$ for all $lambda>0$ by the above estimate $(*)$, it follows that ${S_ntext{ diverges}}$ is a countable union of null probability events, and hence has probability $0$.
answered Jan 7 at 8:48
SongSong
10.7k628
$endgroup$
I guess we need some kind of weak dependence assumption on $(xi_j)$, but anyway, this is only about how we get the convergence result using the maximal inequality about $M_n$. First, we note that we can bound
$$
P(max_{nle ile N}|S_i-S_n|>lambda)lefrac{K}{lambda^gamma}(u_{n+1} +cdots u_N)^alpha
$$ and hence
$$
P(max_{nle i, jle N}|S_i-S_j|>lambda)lefrac{K'}{lambda^gamma}(u_{n+1} +cdots u_N)^alpha.
$$ (We can regard $xi_j$ as starting from index $j=n+1$.) To show that $S_n$ converges, id est, that $limsup_{ntoinfty} S_n = liminf_{ntoinfty} S_n$, it seems natural for us to control the oscillation defined by
$$
limsup_{i,jto infty}|S_i-S_j| =lim_{ntoinfty}sup_{i,jge n} |S_i-S_j|.
$$ Let $W_{n,N} = sup_{nle i,jle N} |S_i-S_j| $ and $W_n =sup_{ i,jge n} |S_i-S_j|=lim_{Ntoinfty}W_{n,N}$. We have
$$
P(W_n>lambda) =lim_{Ntoinfty}P(W_{n,N}>lambda)
$$by monotonic convergence of $W_{n,N}$. Hence we get the bound
$$
P(W_n>lambda)lefrac{K'}{lambda^gamma}left(sum_{j>n}u_{j} right)^alpha.tag{*}
$$ Finally, we observe
$$
limsup_{ntoinfty} S_n -liminf_{ntoinfty} S_n =lim_{ntoinfty} W_n=:W
$$ and
$$
{S_ntext{ diverges}} = {limsup_{ntoinfty} S_n -liminf_{ntoinfty} S_n>0} =bigcup_{jinmathbb{N}}{Wge 1/j}.
$$ Since it holds that
$$
P(Wge lambda) =lim_{ntoinfty} P(W_nge lambda)= 0
$$ for all $lambda>0$ by the above estimate $(*)$, it follows that ${S_ntext{ diverges}}$ is a countable union of null probability events, and hence has probability $0$.
answered Jan 7 at 8:48
SongSong
10.7k628
answered Jan 7 at 8:48
SongSong
10.7k628
answered Jan 7 at 8:48
SongSong
10.7k628
answered Jan 7 at 8:48
SongSong
10.7k628
10.7k628
$begingroup$
Thank you for your answer. It was very helpful! I have a follow-up question of a similar nature; perhaps you could answer it too? math.stackexchange.com/q/3069263/360447
$endgroup$
– cgmil
Jan 10 at 22:40
add a comment |
$begingroup$
Thank you for your answer. It was very helpful! I have a follow-up question of a similar nature; perhaps you could answer it too? math.stackexchange.com/q/3069263/360447
$endgroup$
– cgmil
Jan 10 at 22:40
$begingroup$
Thank you for your answer. It was very helpful! I have a follow-up question of a similar nature; perhaps you could answer it too? math.stackexchange.com/q/3069263/360447
$endgroup$
– cgmil
Jan 10 at 22:40
$begingroup$
Thank you for your answer. It was very helpful! I have a follow-up question of a similar nature; perhaps you could answer it too? math.stackexchange.com/q/3069263/360447
$endgroup$
– cgmil
Jan 10 at 22:40
add a comment |
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Are you sure there is no independence assumption here? The theorems you are quoting are not true without independence.
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– Kavi Rama Murthy
Jan 7 at 7:49
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There is an obvious counterexample with $pm 1$ valued random variables $xi_i$.
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– Kavi Rama Murthy
Jan 7 at 7:50
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@KaviRamaMurthy Billingsley explicitly notes that the variables need not be independent, and the counterexample you mention does not satisfy the assumption mentioned; the tightest $u_l$ you could have would not be summable.
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– cgmil
Jan 10 at 22:44