Mixing
Proving that CLT doesn't hold for a given sequence.
$begingroup$
During examination of compound Poisson process, with log-normal distribution I came across to the following problem.
I have examined the following form $$L=sum_{i=1}^{N}X_{i}$$
And $X_{i}sim LogN(mu,sigma^{2})$.
And I have checked the conditions of Lyapunov version of the CLT for finite sum of conditional representation of $L$. And I found the conditions are not satisfied for the case $delta=2k$.
Moreover I can derive form conditional representation of CPP, that the moment increase in order by $approx e^{e}$. (I mean initial moments not central ones).
Though all finite order moments exist and are computable, but increasing. (Like with Log-normal distribution itself but exponentially faster).
So here is my question.
$textbf{Having disproved Lyapunov condition, can I state that my sequence}$
$textbf{doesn't satisfy CLT?}$
As it is more likely to converge to heavy-tailed distibution.
And another question.
$textbf{Are their any conditions, which can let me tell that the sequence doesn't}$
$textbf{satisfy CLT?}$
Any ideas on the problem can be helpful.
Thank you very much.
central-limit-theorem
asked Jan 16 at 18:55
kolobokishkolobokish
41139
$endgroup$
add a comment |
$begingroup$
During examination of compound Poisson process, with log-normal distribution I came across to the following problem.
I have examined the following form $$L=sum_{i=1}^{N}X_{i}$$
And $X_{i}sim LogN(mu,sigma^{2})$.
And I have checked the conditions of Lyapunov version of the CLT for finite sum of conditional representation of $L$. And I found the conditions are not satisfied for the case $delta=2k$.
Moreover I can derive form conditional representation of CPP, that the moment increase in order by $approx e^{e}$. (I mean initial moments not central ones).
Though all finite order moments exist and are computable, but increasing. (Like with Log-normal distribution itself but exponentially faster).
So here is my question.
$textbf{Having disproved Lyapunov condition, can I state that my sequence}$
$textbf{doesn't satisfy CLT?}$
As it is more likely to converge to heavy-tailed distibution.
And another question.
$textbf{Are their any conditions, which can let me tell that the sequence doesn't}$
$textbf{satisfy CLT?}$
Any ideas on the problem can be helpful.
Thank you very much.
central-limit-theorem
asked Jan 16 at 18:55
kolobokishkolobokish
41139
$endgroup$
add a comment |
$begingroup$
During examination of compound Poisson process, with log-normal distribution I came across to the following problem.
I have examined the following form $$L=sum_{i=1}^{N}X_{i}$$
And $X_{i}sim LogN(mu,sigma^{2})$.
And I have checked the conditions of Lyapunov version of the CLT for finite sum of conditional representation of $L$. And I found the conditions are not satisfied for the case $delta=2k$.
Moreover I can derive form conditional representation of CPP, that the moment increase in order by $approx e^{e}$. (I mean initial moments not central ones).
Though all finite order moments exist and are computable, but increasing. (Like with Log-normal distribution itself but exponentially faster).
So here is my question.
$textbf{Having disproved Lyapunov condition, can I state that my sequence}$
$textbf{doesn't satisfy CLT?}$
As it is more likely to converge to heavy-tailed distibution.
And another question.
$textbf{Are their any conditions, which can let me tell that the sequence doesn't}$
$textbf{satisfy CLT?}$
Any ideas on the problem can be helpful.
Thank you very much.
central-limit-theorem
asked Jan 16 at 18:55
kolobokishkolobokish
41139
$endgroup$
During examination of compound Poisson process, with log-normal distribution I came across to the following problem.
I have examined the following form $$L=sum_{i=1}^{N}X_{i}$$
And $X_{i}sim LogN(mu,sigma^{2})$.
And I have checked the conditions of Lyapunov version of the CLT for finite sum of conditional representation of $L$. And I found the conditions are not satisfied for the case $delta=2k$.
Moreover I can derive form conditional representation of CPP, that the moment increase in order by $approx e^{e}$. (I mean initial moments not central ones).
Though all finite order moments exist and are computable, but increasing. (Like with Log-normal distribution itself but exponentially faster).
So here is my question.
$textbf{Having disproved Lyapunov condition, can I state that my sequence}$
$textbf{doesn't satisfy CLT?}$
As it is more likely to converge to heavy-tailed distibution.
And another question.
$textbf{Are their any conditions, which can let me tell that the sequence doesn't}$
$textbf{satisfy CLT?}$
Any ideas on the problem can be helpful.
Thank you very much.
central-limit-theorem
central-limit-theorem
asked Jan 16 at 18:55
kolobokishkolobokish
41139
asked Jan 16 at 18:55
kolobokishkolobokish
41139
asked Jan 16 at 18:55
kolobokishkolobokish
41139
asked Jan 16 at 18:55
kolobokishkolobokish
41139
asked Jan 16 at 18:55
kolobokishkolobokish
41139
41139
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
You may want to look at Lindeberg's condition as well, which, as far as I know, is the most general version of the CLT that provides necessary and sufficient conditions for a sequence of random variables to converge in distribution to the standard normal.
answered Jan 16 at 18:59
OldGodzillaOldGodzilla
58227
$endgroup$
$begingroup$
Thank you very much. However I can't check the Lindeberg's condition that easily, as I have no closed form for truncated integral in my case.(But even if I could, would i be able to deduce that sequence doesn't satisfy CLT?)
$endgroup$
– kolobokish
Jan 16 at 19:02
1
$begingroup$
If you can lower bound the truncated integral, that may be all you need. If you can show that it fails the condition, then it can't follow the CLT (but may still converge to a different distribution).
$endgroup$
– OldGodzilla
Jan 16 at 19:55
$begingroup$
Thank you very much. Could you give any reference for that? I mean that negation of Lindeberg leads to negation of CLT.
$endgroup$
– kolobokish
Jan 16 at 20:09
1
$begingroup$
It's not true in general, only under a certain condition about the variance described in the wikipedia en.wikipedia.org/wiki/Lindeberg%27s_condition#Statement
$endgroup$
– OldGodzilla
Jan 16 at 20:15
$begingroup$
Thank you very much, again. I'll try to use this.
$endgroup$
– kolobokish
Jan 16 at 20:20
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You may want to look at Lindeberg's condition as well, which, as far as I know, is the most general version of the CLT that provides necessary and sufficient conditions for a sequence of random variables to converge in distribution to the standard normal.
answered Jan 16 at 18:59
OldGodzillaOldGodzilla
58227
$endgroup$
$begingroup$
Thank you very much. However I can't check the Lindeberg's condition that easily, as I have no closed form for truncated integral in my case.(But even if I could, would i be able to deduce that sequence doesn't satisfy CLT?)
$endgroup$
– kolobokish
Jan 16 at 19:02
1
$begingroup$
If you can lower bound the truncated integral, that may be all you need. If you can show that it fails the condition, then it can't follow the CLT (but may still converge to a different distribution).
$endgroup$
– OldGodzilla
Jan 16 at 19:55
$begingroup$
Thank you very much. Could you give any reference for that? I mean that negation of Lindeberg leads to negation of CLT.
$endgroup$
– kolobokish
Jan 16 at 20:09
1
$begingroup$
It's not true in general, only under a certain condition about the variance described in the wikipedia en.wikipedia.org/wiki/Lindeberg%27s_condition#Statement
$endgroup$
– OldGodzilla
Jan 16 at 20:15
$begingroup$
Thank you very much, again. I'll try to use this.
$endgroup$
– kolobokish
Jan 16 at 20:20
add a comment |
$begingroup$
You may want to look at Lindeberg's condition as well, which, as far as I know, is the most general version of the CLT that provides necessary and sufficient conditions for a sequence of random variables to converge in distribution to the standard normal.
answered Jan 16 at 18:59
OldGodzillaOldGodzilla
58227
$endgroup$
$begingroup$
Thank you very much. However I can't check the Lindeberg's condition that easily, as I have no closed form for truncated integral in my case.(But even if I could, would i be able to deduce that sequence doesn't satisfy CLT?)
$endgroup$
– kolobokish
Jan 16 at 19:02
1
$begingroup$
If you can lower bound the truncated integral, that may be all you need. If you can show that it fails the condition, then it can't follow the CLT (but may still converge to a different distribution).
$endgroup$
– OldGodzilla
Jan 16 at 19:55
$begingroup$
Thank you very much. Could you give any reference for that? I mean that negation of Lindeberg leads to negation of CLT.
$endgroup$
– kolobokish
Jan 16 at 20:09
1
$begingroup$
It's not true in general, only under a certain condition about the variance described in the wikipedia en.wikipedia.org/wiki/Lindeberg%27s_condition#Statement
$endgroup$
– OldGodzilla
Jan 16 at 20:15
$begingroup$
Thank you very much, again. I'll try to use this.
$endgroup$
– kolobokish
Jan 16 at 20:20
add a comment |
$begingroup$
You may want to look at Lindeberg's condition as well, which, as far as I know, is the most general version of the CLT that provides necessary and sufficient conditions for a sequence of random variables to converge in distribution to the standard normal.
answered Jan 16 at 18:59
OldGodzillaOldGodzilla
58227
$endgroup$
You may want to look at Lindeberg's condition as well, which, as far as I know, is the most general version of the CLT that provides necessary and sufficient conditions for a sequence of random variables to converge in distribution to the standard normal.
answered Jan 16 at 18:59
OldGodzillaOldGodzilla
58227
answered Jan 16 at 18:59
OldGodzillaOldGodzilla
58227
answered Jan 16 at 18:59
OldGodzillaOldGodzilla
58227
answered Jan 16 at 18:59
OldGodzillaOldGodzilla
58227
58227
$begingroup$
Thank you very much. However I can't check the Lindeberg's condition that easily, as I have no closed form for truncated integral in my case.(But even if I could, would i be able to deduce that sequence doesn't satisfy CLT?)
$endgroup$
– kolobokish
Jan 16 at 19:02
1
$begingroup$
If you can lower bound the truncated integral, that may be all you need. If you can show that it fails the condition, then it can't follow the CLT (but may still converge to a different distribution).
$endgroup$
– OldGodzilla
Jan 16 at 19:55
$begingroup$
Thank you very much. Could you give any reference for that? I mean that negation of Lindeberg leads to negation of CLT.
$endgroup$
– kolobokish
Jan 16 at 20:09
1
$begingroup$
It's not true in general, only under a certain condition about the variance described in the wikipedia en.wikipedia.org/wiki/Lindeberg%27s_condition#Statement
$endgroup$
– OldGodzilla
Jan 16 at 20:15
$begingroup$
Thank you very much, again. I'll try to use this.
$endgroup$
– kolobokish
Jan 16 at 20:20
add a comment |
$begingroup$
Thank you very much. However I can't check the Lindeberg's condition that easily, as I have no closed form for truncated integral in my case.(But even if I could, would i be able to deduce that sequence doesn't satisfy CLT?)
$endgroup$
– kolobokish
Jan 16 at 19:02
1
$begingroup$
If you can lower bound the truncated integral, that may be all you need. If you can show that it fails the condition, then it can't follow the CLT (but may still converge to a different distribution).
$endgroup$
– OldGodzilla
Jan 16 at 19:55
$begingroup$
Thank you very much. Could you give any reference for that? I mean that negation of Lindeberg leads to negation of CLT.
$endgroup$
– kolobokish
Jan 16 at 20:09
1
$begingroup$
It's not true in general, only under a certain condition about the variance described in the wikipedia en.wikipedia.org/wiki/Lindeberg%27s_condition#Statement
$endgroup$
– OldGodzilla
Jan 16 at 20:15
$begingroup$
Thank you very much, again. I'll try to use this.
$endgroup$
– kolobokish
Jan 16 at 20:20
$begingroup$
Thank you very much. However I can't check the Lindeberg's condition that easily, as I have no closed form for truncated integral in my case.(But even if I could, would i be able to deduce that sequence doesn't satisfy CLT?)
$endgroup$
– kolobokish
Jan 16 at 19:02
$begingroup$
Thank you very much. However I can't check the Lindeberg's condition that easily, as I have no closed form for truncated integral in my case.(But even if I could, would i be able to deduce that sequence doesn't satisfy CLT?)
$endgroup$
– kolobokish
Jan 16 at 19:02
1
1
$begingroup$
If you can lower bound the truncated integral, that may be all you need. If you can show that it fails the condition, then it can't follow the CLT (but may still converge to a different distribution).
$endgroup$
– OldGodzilla
Jan 16 at 19:55
$begingroup$
If you can lower bound the truncated integral, that may be all you need. If you can show that it fails the condition, then it can't follow the CLT (but may still converge to a different distribution).
$endgroup$
– OldGodzilla
Jan 16 at 19:55
$begingroup$
Thank you very much. Could you give any reference for that? I mean that negation of Lindeberg leads to negation of CLT.
$endgroup$
– kolobokish
Jan 16 at 20:09
$begingroup$
Thank you very much. Could you give any reference for that? I mean that negation of Lindeberg leads to negation of CLT.
$endgroup$
– kolobokish
Jan 16 at 20:09
1
1
$begingroup$
It's not true in general, only under a certain condition about the variance described in the wikipedia en.wikipedia.org/wiki/Lindeberg%27s_condition#Statement
$endgroup$
– OldGodzilla
Jan 16 at 20:15
$begingroup$
It's not true in general, only under a certain condition about the variance described in the wikipedia en.wikipedia.org/wiki/Lindeberg%27s_condition#Statement
$endgroup$
– OldGodzilla
Jan 16 at 20:15
$begingroup$
Thank you very much, again. I'll try to use this.
$endgroup$
– kolobokish
Jan 16 at 20:20
$begingroup$
Thank you very much, again. I'll try to use this.
$endgroup$
– kolobokish
Jan 16 at 20:20
add a comment |
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