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| Bacle... |
 Posted: Tue Oct 27, 2009 8:50 pm |
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Guest
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Hi, everyone:
I am trying to understand better the statement of the CLT:
If X_1,..,X_n are independent, identically-distributed random variables with mean m and standard deviation s, then the mean of the values of the X_i is normally-distributed, with mean m and std. dev. s.
Please tell me if this application/interpretation of the CLT is correct:
Say I want to estimate the income in a certain area:
So I collect random samples of "large-enough" size (say n>=30) of members of the population, and for each of the samples, I find the average of the incomes of the members. If I take k samples (all of the same size), then, for the i-th sample, the average A_i is a random variable. (An assignment of a random value A_i to the
space of all incomes.)
Then, according to the CLT:
1)The distribution of the A_i is normal N(m,s), with:
2) mean m = the population mean ( i.e., the population of incomes).
3)The standard deviation s is equal to sigma/sqr(n)
where sigma is the population standard deviation, and
n is the sample size. (Maybe this last part is the
actual Law of Large Numbers. )
Is this correct.? |
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| Bacle... |
| Posted: Tue Oct 27, 2009 9:53 pm |
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Guest
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I hope this is not too dumb:
Say we have been able to successfully make use of
the central limit theorem, and so we know what the
sampling distribution of the sample means is :
it has mean=mu; mu is the population mean, and has st. deviation = sigma/sqr(n) , where sigma is the population standard deviation, sqr(n) is the square root of n.
Now: what use is it for us to know the sampling
distribution of the mean?. This information may not
tell us anything useful about the population the
samples came from, if , say, this population is not
symmetric, or has outliers.
So why/how is the knowledge of the sampling mean
useful.?.
Thanks. |
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| Bacle... |
| Posted: Wed Oct 28, 2009 4:12 am |
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Guest
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[quote]On Wed, 28 Oct 2009 02:50:40 EDT, Bacle
bacle at (no spam) yahoo.com> wrote:
Hi, everyone:
I am trying to understand better the statement of
the CLT:
If X_1,..,X_n are independent,
identically-distributed random variables with mean m
and standard deviation s, then the mean of the values
of the X_i is normally-distributed, with mean m and
std. dev. s.
No, CLT does not say that. If you want to understand
it first you
probably should learn what it says.
[/quote]
You cannot conclude that I do not. I did write an incorrect statement, but that does not imply that I don't know what it says.
[quote]
[/quote]
I miswrote: If X_1,..,X_n are a collection of I.I.D
random variables, with finite variance, then their sum
approaches a normally-distributed random variable.
I was mixing up the CLT with the Law of Large
Numbers.
[quote]Please tell me if this application/interpretation of
the CLT is correct:
Say I want to estimate the income in a certain
area:
So I collect random samples of "large-enough" size
(say n>=30) of members of the population, and for
each of the samples, I find the average of the
incomes of the members. If I take k samples (all of
the same size), then, for the i-th sample, the
average A_i is a random variable. (An assignment of a
random value A_i to the
space of all incomes.)
Then, according to the CLT:
1)The distribution of the A_i is normal N(m,s),
with:
2) mean m = the population mean ( i.e., the
population of incomes).
3)The standard deviation s is equal to sigma/sqr(n)
where sigma is the population standard deviation,
and
n is the sample size. (Maybe this last part is the
actual Law of Large Numbers. )
Is this correct.?
David C. Ullrich
"Understanding Godel isn't about following his formal
proof.
That would make a mockery of everything Godel was up
to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)[/quote] |
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| Bacle... |
| Posted: Wed Oct 28, 2009 4:41 am |
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Guest
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O.K: could someone tell me if I fully get the definition.? Not just the literal definition, which is:
" If a sequence of independent, identically distributed random variables, each has finite variance, then, as their number increases, their sum ( or, equivalently, their arithmetic mean) approaches a normally-distributed random variable "
In this case, I am thinking of infinite populations, or sampling with replacement. And n, the sample size,
is larger than 30.
Then , if I take enough samples, take the mean of each sample (each sample size n>30 ), then the sampling distribution of the mean is approximately normal.
Is this correct.?
And I think this follows by the Law of Large Numbers:
Under the conditions necessary for the CLT (If I take a large-enough number of samples, and each sample size n
is larger than 30 , random variables are IID.), it follows that:
The sampling mean tends to the common mean, and the sampling standard deviation tends to sigma/(sqr(n)). |
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| David C. Ullrich... |
| Posted: Wed Oct 28, 2009 5:20 am |
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Guest
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On Wed, 28 Oct 2009 02:50:40 EDT, Bacle <bacle at (no spam) yahoo.com> wrote:
[quote]Hi, everyone:
I am trying to understand better the statement of the CLT:
If X_1,..,X_n are independent, identically-distributed random variables with mean m and standard deviation s, then the mean of the values of the X_i is normally-distributed, with mean m and std. dev. s.
[/quote]
No, CLT does not say that. If you want to understand it first you
probably should learn what it says.
[quote]Please tell me if this application/interpretation of the CLT is correct:
Say I want to estimate the income in a certain area:
So I collect random samples of "large-enough" size (say n>=30) of members of the population, and for each of the samples, I find the average of the incomes of the members. If I take k samples (all of the same size), then, for the i-th sample, the average A_i is a random variable. (An assignment of a random value A_i to the
space of all incomes.)
Then, according to the CLT:
1)The distribution of the A_i is normal N(m,s), with:
2) mean m = the population mean ( i.e., the population of incomes).
3)The standard deviation s is equal to sigma/sqr(n)
where sigma is the population standard deviation, and
n is the sample size. (Maybe this last part is the
actual Law of Large Numbers. )
Is this correct.?
[/quote]
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.) |
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| Herman Rubin... |
| Posted: Wed Oct 28, 2009 11:28 am |
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Guest
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In article <586252478.124237.1256712670167.JavaMail.root at (no spam) gallium.mathforum.org>,
Bacle <bacle at (no spam) yahoo.com> wrote:
[quote]Hi, everyone:
I am trying to understand better the statement of the CLT:
If X_1,..,X_n are independent, identically-distributed random variables with mean m and standard deviation s, then the mean of the values of the X_i is normally-distributed, with mean m and std. dev. s.
[/quote]
Wrong! The mean is not normally distributed UNLESS all
of them are normally distributed.
However, and this is the CLT, the distribution of the
mean approaches the normal distribution. The rate of
approximation is not that great, and there are ways of
estimating the error.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin at (no spam) stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 |
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| Herman Rubin... |
| Posted: Wed Oct 28, 2009 11:33 am |
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Guest
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In article <110378129.124374.1256716420956.JavaMail.root at (no spam) gallium.mathforum.org>,
Bacle <bacle at (no spam) yahoo.com> wrote:
[quote]I hope this is not too dumb:
Say we have been able to successfully make use of
the central limit theorem, and so we know what the
sampling distribution of the sample means is :
it has mean=mu; mu is the population mean, and has st. deviation = sigma/sqr(n) , where sigma is the population standard deviation, sqr(n) is the square root of n.
Now: what use is it for us to know the sampling
distribution of the mean?. This information may not
tell us anything useful about the population the
samples came from, if , say, this population is not
symmetric, or has outliers.
So why/how is the knowledge of the sampling mean
useful.?.
[/quote]
It depends on what you are looking for. The Central
Limit Theorem is a theorem of probability, and you are
asking a question about statistics.
In most cases of interest, the distribution of the
mean of a sample of independent identically distributed
random variables is sufficient to get the distribution
of each, but this is of little practical importance.
One rarely has enough samples of the mean for large
samples, and the computational problem is one with
large errors.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin at (no spam) stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 |
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| jbriggs444... |
| Posted: Thu Oct 29, 2009 7:24 am |
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Guest
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On Oct 28, 1:28 pm, hru... at (no spam) odds.stat.purdue.edu (Herman Rubin) wrote:
[quote]In article <586252478.124237.1256712670167.JavaMail.r... at (no spam) gallium.mathforum.org>,
Bacle <ba... at (no spam) yahoo.com> wrote:
Hi, everyone:
I am trying to understand better the statement of the CLT:
If X_1,..,X_n are independent, identically-distributed random variables with mean m and standard deviation s, then the mean of the values of the X_i is normally-distributed, with mean m and std. dev. s.
Wrong! The mean is not normally distributed UNLESS all
of them are normally distributed.
However, and this is the CLT, the distribution of the
mean approaches the normal distribution. The rate of
approximation is not that great, and there are ways of
estimating the error.
[/quote]
There still seems to be an issue of scaling here. It may approach _a_
normal distribution. But it won't approach _the_ normal distribution
with mean m and standard deviation s.
As a rule of thumb, the standard deviation of the mean of a set of
samples is inversely proportional to the square root of the number of
samples.
This conforms to intuition. If you roll a die once you get a broad
distribution with six uniformly spaced spikes.
If you roll it 100 times and divide the total number of pips by 100
you get a narrower distribution centered at 3.5 with 501 spikes of
varying heights.
If you roll it a million times and take the average you get a
distribution that's narrower yet. Now there are 5,000,001 spikes and
the approximation to a smooth curve is evident.
The tails of the distribution always cut off at 1 and 6. But as the n
increases, the standard deviation decreases. And as the standard
deviation decreases, more and more of them fit into the range between
the tail of the distribution and its mean. |
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| David C. Ullrich... |
| Posted: Thu Oct 29, 2009 7:33 am |
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Guest
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On Wed, 28 Oct 2009 10:12:08 EDT, Bacle <bacle at (no spam) yahoo.com> wrote:
[quote]On Wed, 28 Oct 2009 02:50:40 EDT, Bacle
bacle at (no spam) yahoo.com> wrote:
Hi, everyone:
I am trying to understand better the statement of
the CLT:
If X_1,..,X_n are independent,
identically-distributed random variables with mean m
and standard deviation s, then the mean of the values
of the X_i is normally-distributed, with mean m and
std. dev. s.
No, CLT does not say that. If you want to understand
it first you
probably should learn what it says.
You cannot conclude that I do not. I did write an incorrect statement, but that does not imply that I don't know what it says.
I miswrote: If X_1,..,X_n are a collection of I.I.D
random variables, with finite variance, then their sum
approaches a normally-distributed random variable.
[/quote]
That's not true either. Although it's progress, since "is"
has been changed to "approaches".
A thousand apologies for my "assumption" that you
didn't have CLT quite straight yesterday.
[quote]I was mixing up the CLT with the Law of Large
Numbers.
Please tell me if this application/interpretation of
the CLT is correct:
Say I want to estimate the income in a certain
area:
So I collect random samples of "large-enough" size
(say n>=30) of members of the population, and for
each of the samples, I find the average of the
incomes of the members. If I take k samples (all of
the same size), then, for the i-th sample, the
average A_i is a random variable. (An assignment of a
random value A_i to the
space of all incomes.)
Then, according to the CLT:
1)The distribution of the A_i is normal N(m,s),
with:
2) mean m = the population mean ( i.e., the
population of incomes).
3)The standard deviation s is equal to sigma/sqr(n)
where sigma is the population standard deviation,
and
n is the sample size. (Maybe this last part is the
actual Law of Large Numbers. )
Is this correct.?
David C. Ullrich
"Understanding Godel isn't about following his formal
proof.
That would make a mockery of everything Godel was up
to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
[/quote]
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.) |
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| David C. Ullrich... |
| Posted: Thu Oct 29, 2009 7:40 am |
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Guest
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On Wed, 28 Oct 2009 10:41:15 EDT, Bacle <bacle at (no spam) yahoo.com> wrote:
[quote]O.K: could someone tell me if I fully get the definition.? Not just the literal definition, which is:
" If a sequence of independent, identically distributed random variables, each has finite variance,
then, as their number increases, their sum ( or, equivalently, their arithmetic mean)
approaches a normally-distributed random variable "
[/quote]
Why in the world would you imagine that "sum" is equivalent to "mean"
here?
The sums X_1 + ... X_n do not approach anything in particular. The
means (X_1 + ... + X_n)/n approach something, but that something
is _not_ a normal variable - the means approach a _constant_
(the expected value) in various senses.
I really should apologize _again_ for "assuming" you didn't know
what CLT says.
[quote]In this case, I am thinking of infinite populations, or sampling with replacement. And n, the sample size,
is larger than 30.
Then , if I take enough samples, take the mean of each sample (each sample size n>30 ), then the sampling distribution of the mean is approximately normal.
Is this correct.?
And I think this follows by the Law of Large Numbers:
Under the conditions necessary for the CLT (If I take a large-enough number of samples, and each sample size n
is larger than 30 , random variables are IID.), it follows that:
The sampling mean tends to the common mean, and the sampling standard deviation tends to sigma/(sqr(n)).
[/quote]
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.) |
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| Tim Norfolk... |
| Posted: Thu Oct 29, 2009 4:47 pm |
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Guest
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On Oct 28, 1:33�pm, hru... at (no spam) odds.stat.purdue.edu (Herman Rubin) wrote:
[quote]In article <110378129.124374.1256716420956.JavaMail.r... at (no spam) gallium.mathforum.org>,
Bacle �<ba... at (no spam) yahoo.com> wrote:
I hope this is not too dumb:
� Say we have been able to successfully make use of
the central limit theorem, and so we know what the
sampling distribution of the sample means is :
�it has mean=mu; mu is the population mean, and has st. deviation = �sigma/sqr(n) , where sigma is the population standard deviation, sqr(n) is the square root of n.
� Now: what use is it for us to know the sampling
�distribution of the mean?. This information may not
�tell us anything useful about the population the
�samples came from, if , say, this population is not
�symmetric, or has outliers.
� �So why/how is the knowledge of the sampling mean
�useful.?.
It depends on what you are looking for. �The Central
Limit Theorem is a theorem of probability, and you are
asking a question about statistics.
In most cases of interest, the distribution of the
mean of a sample of independent identically distributed
random variables is sufficient to get the distribution
of each, but this is of little practical importance.
One rarely has enough samples of the mean for large
samples, and the computational problem is one with
large errors.
--
This address is for information only. �I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru... at (no spam) stat.purdue.edu � � � � Phone: (765)494-6054 � FAX: (765)494-0558
[/quote]
Herman;
A related question. Is there a better reference than Feller for the
convergence of the expectations of a certain function in a sequence of
distributions? |
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| jbriggs444... |
| Posted: Fri Oct 30, 2009 3:05 am |
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Guest
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On Oct 29, 1:17 pm, Bacle <ba... at (no spam) yahoo.com> wrote:
[quote]On Wed, 28 Oct 2009 10:41:15 EDT, Bacle
ba... at (no spam) yahoo.com> wrote:
O.K: could someone tell me if I fully get the
definition.? Not just the literal definition, which
is:
" If a sequence of independent, identically
distributed random variables, each has finite
variance,
then, as their number increases, their sum ( or,
equivalently, their arithmetic mean)
approaches a normally-distributed random variable "
Why in the world would you imagine that "sum" is
equivalent to "mean"
here?
The sums X_1 + ... X_n do not approach anything in
particular.
O.K: Let me be more precise. I admit I am being lazy:
" The distribution of the sampling means approaches a normal distribution, as n increases, and if the size of the individual samples is "large-enough" (N>30 is the usual value assumed, AFAIK)"
[/quote]
30? No. I do not think so.
p=.001, x=1000000
p=.999, x=0
Add that distribution to itself 30 times, divide by 30 and see if the
result looks anything like a normal distribution with a mean of 1000.
"In the limit" does not depend on the behavior at 30. Or 1000. Or
1000000. Or 10^100.
Furthermore, distribution of the sampling means approaches a delta
distribution, not a normal distribution. The sequence of
distributions that approaches a normal distribution has some
additional terms in it. |
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| David C. Ullrich... |
| Posted: Fri Oct 30, 2009 5:39 am |
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Guest
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On Thu, 29 Oct 2009 13:17:02 EDT, Bacle <bacle at (no spam) yahoo.com> wrote:
[quote]On Wed, 28 Oct 2009 10:41:15 EDT, Bacle
bacle at (no spam) yahoo.com> wrote:
O.K: could someone tell me if I fully get the
definition.? Not just the literal definition, which
is:
" If a sequence of independent, identically
distributed random variables, each has finite
variance,
then, as their number increases, their sum ( or,
equivalently, their arithmetic mean)
approaches a normally-distributed random variable "
Why in the world would you imagine that "sum" is
equivalent to "mean"
here?
The sums X_1 + ... X_n do not approach anything in
particular.
O.K: Let me be more precise. I admit I am being lazy:
" The distribution of the sampling means approaches a normal distribution, as n increases,
[/quote]
That's more precise. It's still _not_ _true_. The distribution of the
sampling mean approaches a _constant_ distribution.
Yet _again_ I need to apologize for "assuming" that you
didn't have the statement of CLT straight the other day.
I'd tell you what CLT actually says but that would be silly
since you insist you already know. Why not try looking
it up?
[quote]and if the size of the individual samples is "large-enough" (N>30 is the usual value assumed, AFAIK)"
And the law of large numbers says:
The mean mu_m and standard deviation s_m of the sampling
mean m (assuming the conditions of the CLT are satisfied, so that the
mean is normally-distributed) are given ( in the limit) by:
mu_m = mu , mu is the mean of the original population
s_m = sigma/N^1/2 , where N is the size of the sample.
[/quote]
The law of large numbers says nothing at all about normal
distributions or variances.
Saying something is c/N^1/2 "in the limit" is at best
imprecise, unless you mean to say that the limit is 0.
[quote]
The
means (X_1 + ... + X_n)/n approach something, but
that something
is _not_ a normal variable - the means approach a
_constant_
(the expected value) in various senses.
I really should apologize _again_ for "assuming" you
didn't know
what CLT says.
In this case, I am thinking of infinite
populations, or sampling with replacement. And n, the
sample size,
is larger than 30.
Then , if I take enough samples, take the mean of
each sample (each sample size n>30 ), then the
sampling distribution of the mean is approximately
normal.
Is this correct.?
And I think this follows by the Law of Large
Numbers:
Under the conditions necessary for the CLT (If I
take a large-enough number of samples, and each
sample size n
is larger than 30 , random variables are IID.), it
follows that:
The sampling mean tends to the common mean, and the
sampling standard deviation tends to sigma/(sqr(n)).
David C. Ullrich
"Understanding Godel isn't about following his formal
proof.
That would make a mockery of everything Godel was up
to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
[/quote]
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.) |
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| Bacle... |
| Posted: Sat Oct 31, 2009 4:29 pm |
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Guest
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[quote]On Thu, 29 Oct 2009 13:17:02 EDT, Bacle
bacle at (no spam) yahoo.com> wrote:
On Wed, 28 Oct 2009 10:41:15 EDT, Bacle
bacle at (no spam) yahoo.com> wrote:
O.K: could someone tell me if I fully get the
definition.? Not just the literal definition,
which
is:
" If a sequence of independent, identically
distributed random variables, each has finite
variance,
then, as their number increases, their sum ( or,
equivalently, their arithmetic mean)
approaches a normally-distributed random variable.
[/quote]
O.K. I did not know this well at first, true. Your criticism was valid. But while your point is valid, I think you are overdoing the nitpicking :
How do you want me to say it?. The distribution of
the variable 'sampling mean', when considered as a (random)variable, approaches a normal distribution, when the size of each sample is large-enough, and the number of samples drawn is also large-enough.
We random draw samples X_1,..,X_m of size N (use N>30) from the same population ( either from an infinite pop. or sampling is done with replacement), and, for each sample, we find the arithmetic mean of the sampled values. Using this, we define a random variable Y_i :
Y_i =(X_i1+X_i2+...+X_iN)/N
We consider the distribution of the variable Y_i.
As the number of samples grows, i.e., as m->oo, the distribution of the Y_i becomes approximately normal, or the dist. of the Y_i approaches a normal distribution.
The law of large numbers says that the mean of this limiting distribution approaches the mean of the parent population, and the st. deviation. of the distribution
approaches Sigma/N^1/2, where Sigma is the St. Dev. of the parent population.
[quote]"
Why in the world would you imagine that "sum" is
equivalent to "mean"
here?
The sums X_1 + ... X_n do not approach anything in
particular.
O.K: Let me be more precise. I admit I am being
lazy:
" The distribution of the sampling means
approaches a normal distribution, as n increases,
That's more precise. It's still _not_ _true_. The
distribution of the
sampling mean approaches a _constant_ distribution.
Yet _again_ I need to apologize for "assuming" that
you
didn't have the statement of CLT straight the other
day.
I'd tell you what CLT actually says but that would be
silly
since you insist you already know. Why not try
looking
it up?
and if the size of the individual samples is
"large-enough" (N>30 is the usual value assumed,
AFAIK)"
And the law of large numbers says:
The mean mu_m and standard deviation s_m of the
sampling
mean m (assuming the conditions of the CLT are
satisfied, so that the
mean is normally-distributed) are given ( in the
limit) by:
mu_m = mu , mu is the mean of the original
population
s_m = sigma/N^1/2 , where N is the size of the
sample.
The law of large numbers says nothing at all about
normal
distributions or variances.
Saying something is c/N^1/2 "in the limit" is at best
imprecise, unless you mean to say that the limit is
0.
The
means (X_1 + ... + X_n)/n approach something, but
that something
is _not_ a normal variable - the means approach a
_constant_
(the expected value) in various senses.
I really should apologize _again_ for "assuming"
you
didn't know
what CLT says.
In this case, I am thinking of infinite
populations, or sampling with replacement. And n,
the
sample size,
is larger than 30.
Then , if I take enough samples, take the mean
of
each sample (each sample size n>30 ), then the
sampling distribution of the mean is approximately
normal.
Is this correct.?
And I think this follows by the Law of Large
Numbers:
Under the conditions necessary for the CLT (If I
take a large-enough number of samples, and each
sample size n
is larger than 30 , random variables are IID.),
it
follows that:
The sampling mean tends to the common mean, and
the
sampling standard deviation tends to
sigma/(sqr(n)).
David C. Ullrich
"Understanding Godel isn't about following his
formal
proof.
That would make a mockery of everything Godel was
up
to."
(John Jones, "My talk about Godel to the
post-grads."
in sci.logic.)
David C. Ullrich
"Understanding Godel isn't about following his formal
proof.
That would make a mockery of everything Godel was up
to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)[/quote] |
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| Libra/Virgo... |
| Posted: Sun Nov 01, 2009 5:28 am |
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Guest
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http://www.meami.org/On Oct 27, 11:50Â pm, Bacle <ba... at (no spam) yahoo.com>
wrote:
[quote]Hi, everyone:
I am trying to understand better the statement of the CLT:
If X_1,..,X_n are independent, identically-distributed random variables with mean m and standard deviation s, then the mean of the values of the X_i is normally-distributed, with mean m and std. dev. s.
Please tell me if this application/interpretation of the CLT is correct:
  Say I want to estimate the income in a certain area:
 So I collect random samples of "large-enough" size (say n>=30) of members of the population, and for each of the samples, I find the average of the incomes of the members. If I take k samples (all of the same size), then, for the i-th sample, the average A_i is a random variable. (An assignment of a random value A_i to the
space of all incomes.)
 Then, according to the CLT:
1)The distribution of the A_i is normal N(m,s), with:
2) Â mean m = the population mean ( i.e., the population of incomes).
3)The standard deviation s is equal to sigma/sqr(n)
 where sigma is the population standard deviation, and
 n is the sample size. (Maybe this last part is the
 actual Law of Large Numbers. )
 Is this correct.?
[/quote]
Let {X_1,X_2, \cdots, Xn} be a random sample (IID) from a (native)
distribution with well-defined and finite mean μX and variance
\sigma_X^2. Then as n increases, the sampling distributions of the
sample average \overline{X_n}={1\over n}\sum_{i=1}^n{X_i} and the
total sum \overline{T_n}=\sum_{i=1}^n{X_i} approach Normal
distributions with corresponding means and variances:
\mu_{\overline{X_n}}=\mu_X; \sigma_{\overline{X_n}}^2{\sigma_X^2\over n}
\mu_{\overline{T_n}}=n\times\mu_X; \sigma_{\overline{T_n}}^2={n
\times\sigma_X^2}
^[http://wiki.stat.ucla.edu/socr/index.php/
AP_Statistics_Curriculum_2007_Limits_CLT#Symbolic_Statement_of_the_Central_Limit_Theorem]
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