 |
|
| Science Forum Index » Mathematics Forum » Central Limit Theorem. Is This Correct.?... |
|
Page 2 of 2 Goto page Previous 1, 2 |
|
| Author |
Message |
| David C. Ullrich... |
 Posted: Sun Nov 01, 2009 10:19 am |
|
|
|
Guest
|
On Sat, 31 Oct 2009 22:29:33 EDT, Bacle <bacle at (no spam) yahoo.com> wrote:
[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.
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?.
[/quote]
Correctly, umambiguously and precisely.
[quote]The distribution of
the variable 'sampling mean', when considered as a (random)variable, approaches a normal distribution,
[/quote]
No it doesn't. Perhaps what you meant by that is true.
It's true that the distribution of the mean is "approximately normal",
in some sense. But one has to say _exactly_ what is meant by that.
It is simply not true that the distributions of the mean approach
a normal distribution. You can repeat it as many times as you
want, that's not going to make it true.
The easiest way to say _exactly_ what we mean when we say the
distribution of the mean is "approximately normal" involves
the distribution of something _otther_ than the mean.
[quote]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,
[/quote]
More or less, although as noted above that's too vague to be
actually _true_.
[quote]or the dist. of the Y_i approaches a normal distribution.
[/quote]
No. The two are not the same.
[quote]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,
[/quote]
So we have these distributions with standard deviation sigma/sqrt(n),
great.
Now exactly what is the std dev of the normal distribution that you
insist these are approaching? It seems to be a normal distribtion
with std dev 0. There's no such thing.
[quote]where Sigma is the St. Dev. of the parent population.
"
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]
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.) |
|
|
| Back to top |
|
|
|
| Bacle... |
| Posted: Sun Nov 01, 2009 5:24 pm |
|
|
|
Guest
|
[quote]On Oct 29, 1:17 pm, Bacle <ba... at (no spam) yahoo.com> wrote:
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)"
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.
[/quote]
Could you please tell me.?. I am not a statistician.
I am teaching an intro class where I need to present
the material to non-math majors , most being freshmen.
I have gone through the formal statement, and I stated it as well as I could.
I would appreciate your help. |
|
|
| Back to top |
|
|
|
| adamk... |
| Posted: Sun Nov 01, 2009 6:00 pm |
|
|
|
Guest
|
[quote]On Sat, 31 Oct 2009 22:29:33 EDT, Bacle
bacle at (no spam) yahoo.com> wrote:
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.
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?.
Correctly, umambiguously and precisely.
The distribution of
the variable 'sampling mean', when considered as
a (random)variable, approaches a normal distribution,
No it doesn't. Perhaps what you meant by that is
true.
[/quote]
Would you say what is wrong with my statement.?.
I am not considering the issues of finite variance,
nor of degenerate parent distributions.
[quote]
It's true that the distribution of the mean is
"approximately normal",
in some sense. But one has to say _exactly_ what is
meant by that.
[/quote]
I am teaching an undergraduate class for non-math-major freshmen. I am looking for a statement that they can understand. Maybe I should have included that.
I want to tell them that they are justified in using the normal distribution with mean the population mean
and standard deviation equal to the parent pop's standard deviation divided by the square root of the sample size, when working with the variable sampling mean, i.e., when the variable being studied is obtained by averaging out the values of individual samples of fixed size N (as a rule of thumb use N>30), and if we
take a large-enough number of samples.
( I know that just stating the numerical value of the sampling mean approaches the numerical value of the population mean, and this is not precise-enough mathematically, but I cannot afford to develop all the context to explain this more precisely.)
[quote]It is simply not true that the distributions of the
mean approach
a normal distribution. You can repeat it as many
times as you
want, that's not going to make it true.
[/quote]
I meant the sampling mean, i.e., the variable
obtained by averaging out the values obtained in
any given sample.
I am ignoring issues like having finite variance, and
ignoring possible degenerate parent distributions.
[quote]
The easiest way to say _exactly_ what we mean when we
say the
distribution of the mean is "approximately normal"
involves
the distribution of something _otther_ than the mean.
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,
More or less, although as noted above that's too
vague to be
actually _true_.
or the dist. of the Y_i approaches a normal
distribution.
No. The two are not the same.
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,
So we have these distributions with standard
deviation sigma/sqrt(n),
great.
Now exactly what is the std dev of the normal
distribution that you
insist these are approaching? It seems to be a normal
distribtion
with std dev 0. There's no such thing.
[/quote]
Not so here. n here is the size of the individual samples, not the number of samples.
[quote]
where Sigma is the St. Dev. of the parent
population.
"
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.)
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] |
|
|
| Back to top |
|
|
|
| Bacle... |
| Posted: Sun Nov 01, 2009 6:26 pm |
|
|
|
Guest
|
[quote]On Sat, 31 Oct 2009 22:29:33 EDT, Bacle
bacle at (no spam) yahoo.com> wrote:
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.
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?.
Correctly, umambiguously and precisely.
The distribution of
the variable 'sampling mean', when considered
as
a (random)variable, approaches a normal
distribution,
No it doesn't. Perhaps what you meant by that is
true.
Would you say what is wrong with my statement.?.
I am not considering the issues of finite variance,
nor of degenerate parent distributions.
[/quote]
Please use your own post or at least use your own statements, and not mine.
[quote]
It's true that the distribution of the mean is
"approximately normal",
in some sense. But one has to say _exactly_ what
is
meant by that.
I am teaching an undergraduate class for
for non-math-major freshmen. I am looking for a
statement that they can understand. Maybe I should
have included that.
I want to tell them that they are justified in
in using the normal distribution with mean the
population mean
and standard deviation equal to the parent pop's
standard deviation divided by the square root of the
sample size, when working with the variable sampling
mean, i.e., when the variable being studied is
obtained by averaging out the values of individual
samples of fixed size N (as a rule of thumb use
N>30), and if we
take a large-enough number of samples.
[/quote]
This sounds reasonable, and it is used in this precise way.
[quote]
( I know that just stating the numerical value of the
sampling mean approaches the numerical value of the
population mean, and this is not precise-enough
mathematically, but I cannot afford to develop all
the context to explain this more precisely.)
It is simply not true that the distributions of
the
mean approach
a normal distribution. You can repeat it as many
times as you
want, that's not going to make it true.
I meant the sampling mean, i.e., the variable
obtained by averaging out the values obtained in
any given sample.
I am ignoring issues like having finite variance,
ce, and
ignoring possible degenerate parent distributions.
The easiest way to say _exactly_ what we mean when
we
say the
distribution of the mean is "approximately normal"
involves
the distribution of something _otther_ than the
mean.
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,
More or less, although as noted above that's too
vague to be
actually _true_.
or the dist. of the Y_i approaches a normal
distribution.
No. The two are not the same.
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,
So we have these distributions with standard
deviation sigma/sqrt(n),
great.
Now exactly what is the std dev of the normal
distribution that you
insist these are approaching? It seems to be a
normal
distribtion
with std dev 0. There's no such thing.
Not so here. n here is the size of the individual
al samples, not the number of samples.
where Sigma is the St. Dev. of the parent
population.
"
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.)
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] |
|
|
| Back to top |
|
|
|
| David C. Ullrich... |
| Posted: Mon Nov 02, 2009 8:51 am |
|
|
|
Guest
|
On Sun, 01 Nov 2009 23:00:23 EST, adamk <adamk at (no spam) adamk.net> wrote:
[quote]On Sat, 31 Oct 2009 22:29:33 EDT, Bacle
bacle at (no spam) yahoo.com> wrote:
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.
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?.
Correctly, umambiguously and precisely.
The distribution of
the variable 'sampling mean', when considered as
a (random)variable, approaches a normal distribution,
No it doesn't. Perhaps what you meant by that is
true.
Would you say what is wrong with my statement.?.
[/quote]
I did, in the very next sentence.
[quote]I am not considering the issues of finite variance,
nor of degenerate parent distributions.
It's true that the distribution of the mean is
"approximately normal",
in some sense. But one has to say _exactly_ what is
meant by that.
I am teaching an undergraduate class for non-math-major freshmen. I am looking for a statement that they can understand. Maybe I should have included that.
I want to tell them that they are justified in using the normal distribution with mean the population mean
and standard deviation equal to the parent pop's standard deviation divided by the square root of the sample size, when working with the variable sampling mean, i.e., when the variable being studied is obtained by averaging out the values of individual samples of fixed size N (as a rule of thumb use N>30), and if we
take a large-enough number of samples.
[/quote]
Then tell them that. I thought the topic was math, sorry.
[quote]( I know that just stating the numerical value of the sampling mean approaches the numerical value of the population mean, and this is not precise-enough mathematically, but I cannot afford to develop all the context to explain this more precisely.)
It is simply not true that the distributions of the
mean approach
a normal distribution. You can repeat it as many
times as you
want, that's not going to make it true.
I meant the sampling mean, i.e., the variable
obtained by averaging out the values obtained in
any given sample.
[/quote]
This is ridiculous. You keep repeating what you meant,
as though the fact I say it's wrong _must_ be because
I didn't understand you.
Read my previous post. All of it. Carefully.
[quote]I am ignoring issues like having finite variance, and
ignoring possible degenerate parent distributions.
The easiest way to say _exactly_ what we mean when we
say the
distribution of the mean is "approximately normal"
involves
the distribution of something _otther_ than the mean.
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,
More or less, although as noted above that's too
vague to be
actually _true_.
or the dist. of the Y_i approaches a normal
distribution.
No. The two are not the same.
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,
So we have these distributions with standard
deviation sigma/sqrt(n),
great.
Now exactly what is the std dev of the normal
distribution that you
insist these are approaching? It seems to be a normal
distribtion
with std dev 0. There's no such thing.
Not so here. n here is the size of the individual samples, not the number of samples.
where Sigma is the St. Dev. of the parent
population.
"
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.)
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.) |
|
|
| Back to top |
|
|
|
|
|
All times are GMT - 5 Hours
The time now is Mon Dec 14, 2009 3:58 pm
|
|