Relation of Law of large numbers and Central limit theorem
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I been reading up on the Law of large numbers and Central limit theorem. While tying to get some intuition I got stuck on the two following claims which bother me. Firstly,
"The Central Limit Theorem is about the journey and the Strong Law of Large Numbers is about the destination"
from,
https://stats.stackexchange.com/questions/49123/central-limit-theorem-and-the-law-of-large-numbers?noredirect=1&lq=1.
I don't understand what he wants to say. Both theorems seam to be talking about the "destination". However of two different sequences of random variables, $sum X_n$ divided by $n$ and $sqrtn$ respectively and under two different modes of convergence.
Another interpretation which I have a hard to to understanding is the idea of thinking of the Central limit theorem in the way outlined here,
https://en.wikipedia.org/wiki/Central_limit_theorem#Independent_sequences
In particular the sentance,
"The classical central limit theorem describes the size and the distributional form of the stochastic fluctuations around the deterministic number µ during this convergence. "
Again it is still a matter of "destintion".
What bothers me is that it is two different kind of
variables converging and under different modes of convergence and this seams not the mentioned at all, I dont understand why people seam to think this is besides to point.
Does anyone know or understand what these guys are saying?
probability-theory
asked May 16 at 11:24
user8321
509
add a comment |Â
up vote
2
down vote
favorite
I been reading up on the Law of large numbers and Central limit theorem. While tying to get some intuition I got stuck on the two following claims which bother me. Firstly,
"The Central Limit Theorem is about the journey and the Strong Law of Large Numbers is about the destination"
from,
https://stats.stackexchange.com/questions/49123/central-limit-theorem-and-the-law-of-large-numbers?noredirect=1&lq=1.
I don't understand what he wants to say. Both theorems seam to be talking about the "destination". However of two different sequences of random variables, $sum X_n$ divided by $n$ and $sqrtn$ respectively and under two different modes of convergence.
Another interpretation which I have a hard to to understanding is the idea of thinking of the Central limit theorem in the way outlined here,
https://en.wikipedia.org/wiki/Central_limit_theorem#Independent_sequences
In particular the sentance,
"The classical central limit theorem describes the size and the distributional form of the stochastic fluctuations around the deterministic number µ during this convergence. "
Again it is still a matter of "destintion".
What bothers me is that it is two different kind of
variables converging and under different modes of convergence and this seams not the mentioned at all, I dont understand why people seam to think this is besides to point.
Does anyone know or understand what these guys are saying?
probability-theory
asked May 16 at 11:24
user8321
509
I think the point they are making with the destination vs journey thing is that CLT tells you the distribution that $S_n$ assumes while converging (hence the journey of $S_n$), while LLN tells you where $S_n$ will end up converging, meaning $mu$, its final destination.
– Dav2357
May 16 at 12:10
@Dav2357 I dont think the CLT does that it tells us where it ends up, just as the LLN. Its also a limit statement.
– user8321
May 16 at 12:13
3
Actually the CLT does not state explicitly $S_n to mu$, instead it says that $sqrtn (S_n - mu) oversetdto mathcalN(0,sigma^2)$, if you rewrite it in the form $S_n oversetdto mathcalN(mu,fracsigma^2n)$ maybe it is more intuitive to see that for $n to infty$ $S_n$ converges to a normal distribution around $mu$ with infinitesimal variance
– Dav2357
May 16 at 12:37
right, it is in distribution and not a.s convergce. The latter way to express it is excellent, it is clear that we are getting closer to something which is constant. It looks kind of werid however, having $n$ on both sides. The limit is the limit and should be something definite, but I guess thats what the acutal statement is i.e $sqrtn(S_n-mu)$.
– user8321
May 16 at 14:55
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I been reading up on the Law of large numbers and Central limit theorem. While tying to get some intuition I got stuck on the two following claims which bother me. Firstly,
"The Central Limit Theorem is about the journey and the Strong Law of Large Numbers is about the destination"
from,
https://stats.stackexchange.com/questions/49123/central-limit-theorem-and-the-law-of-large-numbers?noredirect=1&lq=1.
I don't understand what he wants to say. Both theorems seam to be talking about the "destination". However of two different sequences of random variables, $sum X_n$ divided by $n$ and $sqrtn$ respectively and under two different modes of convergence.
Another interpretation which I have a hard to to understanding is the idea of thinking of the Central limit theorem in the way outlined here,
https://en.wikipedia.org/wiki/Central_limit_theorem#Independent_sequences
In particular the sentance,
"The classical central limit theorem describes the size and the distributional form of the stochastic fluctuations around the deterministic number µ during this convergence. "
Again it is still a matter of "destintion".
What bothers me is that it is two different kind of
variables converging and under different modes of convergence and this seams not the mentioned at all, I dont understand why people seam to think this is besides to point.
Does anyone know or understand what these guys are saying?
probability-theory
asked May 16 at 11:24
user8321
509
I been reading up on the Law of large numbers and Central limit theorem. While tying to get some intuition I got stuck on the two following claims which bother me. Firstly,
"The Central Limit Theorem is about the journey and the Strong Law of Large Numbers is about the destination"
from,
https://stats.stackexchange.com/questions/49123/central-limit-theorem-and-the-law-of-large-numbers?noredirect=1&lq=1.
I don't understand what he wants to say. Both theorems seam to be talking about the "destination". However of two different sequences of random variables, $sum X_n$ divided by $n$ and $sqrtn$ respectively and under two different modes of convergence.
Another interpretation which I have a hard to to understanding is the idea of thinking of the Central limit theorem in the way outlined here,
https://en.wikipedia.org/wiki/Central_limit_theorem#Independent_sequences
In particular the sentance,
"The classical central limit theorem describes the size and the distributional form of the stochastic fluctuations around the deterministic number µ during this convergence. "
Again it is still a matter of "destintion".
What bothers me is that it is two different kind of
variables converging and under different modes of convergence and this seams not the mentioned at all, I dont understand why people seam to think this is besides to point.
Does anyone know or understand what these guys are saying?
probability-theory
probability-theory
asked May 16 at 11:24
user8321
509
asked May 16 at 11:24
user8321
509
edited May 18 at 6:07
asked May 16 at 11:24
user8321
509
asked May 16 at 11:24
user8321
509
asked May 16 at 11:24
user8321
509
509
I think the point they are making with the destination vs journey thing is that CLT tells you the distribution that $S_n$ assumes while converging (hence the journey of $S_n$), while LLN tells you where $S_n$ will end up converging, meaning $mu$, its final destination.
– Dav2357
May 16 at 12:10
@Dav2357 I dont think the CLT does that it tells us where it ends up, just as the LLN. Its also a limit statement.
– user8321
May 16 at 12:13
3
Actually the CLT does not state explicitly $S_n to mu$, instead it says that $sqrtn (S_n - mu) oversetdto mathcalN(0,sigma^2)$, if you rewrite it in the form $S_n oversetdto mathcalN(mu,fracsigma^2n)$ maybe it is more intuitive to see that for $n to infty$ $S_n$ converges to a normal distribution around $mu$ with infinitesimal variance
– Dav2357
May 16 at 12:37
right, it is in distribution and not a.s convergce. The latter way to express it is excellent, it is clear that we are getting closer to something which is constant. It looks kind of werid however, having $n$ on both sides. The limit is the limit and should be something definite, but I guess thats what the acutal statement is i.e $sqrtn(S_n-mu)$.
– user8321
May 16 at 14:55
add a comment |Â
I think the point they are making with the destination vs journey thing is that CLT tells you the distribution that $S_n$ assumes while converging (hence the journey of $S_n$), while LLN tells you where $S_n$ will end up converging, meaning $mu$, its final destination.
– Dav2357
May 16 at 12:10
@Dav2357 I dont think the CLT does that it tells us where it ends up, just as the LLN. Its also a limit statement.
– user8321
May 16 at 12:13
3
Actually the CLT does not state explicitly $S_n to mu$, instead it says that $sqrtn (S_n - mu) oversetdto mathcalN(0,sigma^2)$, if you rewrite it in the form $S_n oversetdto mathcalN(mu,fracsigma^2n)$ maybe it is more intuitive to see that for $n to infty$ $S_n$ converges to a normal distribution around $mu$ with infinitesimal variance
– Dav2357
May 16 at 12:37
right, it is in distribution and not a.s convergce. The latter way to express it is excellent, it is clear that we are getting closer to something which is constant. It looks kind of werid however, having $n$ on both sides. The limit is the limit and should be something definite, but I guess thats what the acutal statement is i.e $sqrtn(S_n-mu)$.
– user8321
May 16 at 14:55
I think the point they are making with the destination vs journey thing is that CLT tells you the distribution that $S_n$ assumes while converging (hence the journey of $S_n$), while LLN tells you where $S_n$ will end up converging, meaning $mu$, its final destination.
– Dav2357
May 16 at 12:10
I think the point they are making with the destination vs journey thing is that CLT tells you the distribution that $S_n$ assumes while converging (hence the journey of $S_n$), while LLN tells you where $S_n$ will end up converging, meaning $mu$, its final destination.
– Dav2357
May 16 at 12:10
@Dav2357 I dont think the CLT does that it tells us where it ends up, just as the LLN. Its also a limit statement.
– user8321
May 16 at 12:13
@Dav2357 I dont think the CLT does that it tells us where it ends up, just as the LLN. Its also a limit statement.
– user8321
May 16 at 12:13
3
3
Actually the CLT does not state explicitly $S_n to mu$, instead it says that $sqrtn (S_n - mu) oversetdto mathcalN(0,sigma^2)$, if you rewrite it in the form $S_n oversetdto mathcalN(mu,fracsigma^2n)$ maybe it is more intuitive to see that for $n to infty$ $S_n$ converges to a normal distribution around $mu$ with infinitesimal variance
– Dav2357
May 16 at 12:37
Actually the CLT does not state explicitly $S_n to mu$, instead it says that $sqrtn (S_n - mu) oversetdto mathcalN(0,sigma^2)$, if you rewrite it in the form $S_n oversetdto mathcalN(mu,fracsigma^2n)$ maybe it is more intuitive to see that for $n to infty$ $S_n$ converges to a normal distribution around $mu$ with infinitesimal variance
– Dav2357
May 16 at 12:37
right, it is in distribution and not a.s convergce. The latter way to express it is excellent, it is clear that we are getting closer to something which is constant. It looks kind of werid however, having $n$ on both sides. The limit is the limit and should be something definite, but I guess thats what the acutal statement is i.e $sqrtn(S_n-mu)$.
– user8321
May 16 at 14:55
right, it is in distribution and not a.s convergce. The latter way to express it is excellent, it is clear that we are getting closer to something which is constant. It looks kind of werid however, having $n$ on both sides. The limit is the limit and should be something definite, but I guess thats what the acutal statement is i.e $sqrtn(S_n-mu)$.
– user8321
May 16 at 14:55
add a comment |Â
1 Answer
1
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oldest
votes
up vote
1
down vote
accepted
Actually the CLT does not state that $S_n to mu$, instead it says that $sqrt n,(S_n−mu) to operatorname N(0,sigma^2)$.
If you rewrite it in the form $S_n to operatorname N(mu,fracsigma^2n)$ maybe it is more intuitive to see that for $n to infty$, $S_n$ converges to a normal distribution around $mu$ with infinitesimal variance.
edited Sep 6 at 17:13
Did
243k23209444
answered Sep 6 at 10:17
user8321
509
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Actually the CLT does not state that $S_n to mu$, instead it says that $sqrt n,(S_n−mu) to operatorname N(0,sigma^2)$.
If you rewrite it in the form $S_n to operatorname N(mu,fracsigma^2n)$ maybe it is more intuitive to see that for $n to infty$, $S_n$ converges to a normal distribution around $mu$ with infinitesimal variance.
edited Sep 6 at 17:13
Did
243k23209444
answered Sep 6 at 10:17
user8321
509
add a comment |Â
up vote
1
down vote
accepted
Actually the CLT does not state that $S_n to mu$, instead it says that $sqrt n,(S_n−mu) to operatorname N(0,sigma^2)$.
If you rewrite it in the form $S_n to operatorname N(mu,fracsigma^2n)$ maybe it is more intuitive to see that for $n to infty$, $S_n$ converges to a normal distribution around $mu$ with infinitesimal variance.
edited Sep 6 at 17:13
Did
243k23209444
answered Sep 6 at 10:17
user8321
509
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Actually the CLT does not state that $S_n to mu$, instead it says that $sqrt n,(S_n−mu) to operatorname N(0,sigma^2)$.
If you rewrite it in the form $S_n to operatorname N(mu,fracsigma^2n)$ maybe it is more intuitive to see that for $n to infty$, $S_n$ converges to a normal distribution around $mu$ with infinitesimal variance.
edited Sep 6 at 17:13
Did
243k23209444
answered Sep 6 at 10:17
user8321
509
Actually the CLT does not state that $S_n to mu$, instead it says that $sqrt n,(S_n−mu) to operatorname N(0,sigma^2)$.
If you rewrite it in the form $S_n to operatorname N(mu,fracsigma^2n)$ maybe it is more intuitive to see that for $n to infty$, $S_n$ converges to a normal distribution around $mu$ with infinitesimal variance.
edited Sep 6 at 17:13
Did
243k23209444
answered Sep 6 at 10:17
user8321
509
edited Sep 6 at 17:13
Did
243k23209444
edited Sep 6 at 17:13
Did
243k23209444
edited Sep 6 at 17:13
Did
243k23209444
243k23209444
answered Sep 6 at 10:17
user8321
509
answered Sep 6 at 10:17
user8321
509
answered Sep 6 at 10:17
user8321
509
509
add a comment |Â
add a comment |Â
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I think the point they are making with the destination vs journey thing is that CLT tells you the distribution that $S_n$ assumes while converging (hence the journey of $S_n$), while LLN tells you where $S_n$ will end up converging, meaning $mu$, its final destination.
– Dav2357
May 16 at 12:10
@Dav2357 I dont think the CLT does that it tells us where it ends up, just as the LLN. Its also a limit statement.
– user8321
May 16 at 12:13
3
Actually the CLT does not state explicitly $S_n to mu$, instead it says that $sqrtn (S_n - mu) oversetdto mathcalN(0,sigma^2)$, if you rewrite it in the form $S_n oversetdto mathcalN(mu,fracsigma^2n)$ maybe it is more intuitive to see that for $n to infty$ $S_n$ converges to a normal distribution around $mu$ with infinitesimal variance
– Dav2357
May 16 at 12:37
right, it is in distribution and not a.s convergce. The latter way to express it is excellent, it is clear that we are getting closer to something which is constant. It looks kind of werid however, having $n$ on both sides. The limit is the limit and should be something definite, but I guess thats what the acutal statement is i.e $sqrtn(S_n-mu)$.
– user8321
May 16 at 14:55