Once again 2 parts.
1) Show that t distribution with infinity degrees of freedom is the
same as normal distribution from 0 to 1 or tsub infinity=N(0,1)
2) show that 
square root of n (Uhat-U)/ S sub n = t distributed with n-1 degrees of
freedom.

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Request for Question Clarification by rbnn-ga on 28 Oct 2002 11:11 PST

In question 2: 

 i. Can you delineate the formula more precisely somehow? I am not
sure what the square root is of. For example, parenthesize all terms.

 ii. What is U, Uhat, and S_n ?

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Request for Question Clarification by calebu2-ga on 28 Oct 2002 11:21 PST

rbnn: I'm guessing sqrt(n)*(mu-hat - mu)/S_n

where mu-hat is the sample mean, mu is the true mean and S_n is the
sample standard deviation.

Anyway, fire away.

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Request for Question Clarification by rbnn-ga on 28 Oct 2002 14:59 PST

In order for me to answer 

1), I would need to know the precise definition of "t distribution
with infinity degrees of freedom" that is being used. If the
definition involves taking the pointwise limit of a sequence of
probability distributions, then 1) may be a lot of work, because there
is some subtlety in showing that the pointwise limit of a sequence of
density functions converges. (This would probably be more work than
the price for me, although researchers more familiar with the field
might do it quicker.).

For 2) again; the statement is a standard result in the literature and
can be used as tantamount to a definition. It might be desirable to
know the definition of t distribution used in order to give a
non-tautological answer.

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Clarification of Question by brkyrhrt99_00-ga on 28 Oct 2002 19:10 PST

rbnn calebu 2 is correct.  Here is the definition
Tn=(square rot of N*Z)/square root of xsubn is said to have a t
distribution with n degrees of freedom
for part 1) lim as x aproaches infinity P(Tn < a) = P(Z < a) that's
the hint he gave.  He also said we could use the fact that sample mean
is the approxiamtion of actual mean
for part 2) if x is normally distributed then
           1) muhat and Ssubnsquared are independent
           2) Ssubnsquared*(n-1)/Theta squared is chi distributed with
n-1 degrees of freedom
    These are the clues for part 2

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Clarification of Question by brkyrhrt99_00-ga on 28 Oct 2002 19:12 PST

the clarification I posted earlier for part 2 he said those are the
steps to solving and proving this.   THanks

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Request for Question Clarification by rbnn-ga on 29 Oct 2002 03:55 PST

You still are 
 1) Not defining the characters in your definition of t distribution.
     a) By Tn do you mean T_n, that is, T subscript n ? Is this the T
distribution with n degrees of freedom?

     b) What is N? is it different from n?

     c) What is Z (usually it's the unit normal)

     d) What is xsubn, or x_n, that is x with subscript n ?

     e) Most important, you still have not defined the "t distribution
with infinity degrees of freedom."

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Clarification of Question by brkyrhrt99_00-ga on 29 Oct 2002 09:03 PST

Yes T_n is t with n degrees of freedom
z is the unit normal
t with infinity degrees of freedom is defined to be normal and that's
what I have to prove.  Basically I need to derive the definition of
t_infinity.

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Clarification of Question by brkyrhrt99_00-ga on 30 Oct 2002 11:03 PST

I have solved the first part I just need the second part.
it consists of 3 steps
1) proving muhat and Ssubnsquared are independent 
2) proving Ssubnsquared*(n-1)/Theta squared is chi distributed with
n-1 degrees of freedom
3) proving square root of (n)*(muhat-mu)/ S_n = t_n-1 

Anyone's help is greatly appreciated.

brkyrhrt,

Most of my knowledge of probability and statistics comes from the
notes of one professor who taught me most of what I know at this
level. Luckily he still has his course notes available online. I'm
going to give an answer to part (2) of your question as a combination
of my own knowledge and the formal proof given in his notes. Then
again, given that my knowledge is a subset of these notes, I should
probably give Prof. Weber full credit for this answer :)

Prof Weber's Statistics course homepage at Cambridge University :
http://www.statslab.cam.ac.uk/~rrw1/stats/index00.html

The lecture notes corresponding to your question (in postscript
format) :
http://www.statslab.cam.ac.uk/~rrw1/stats/S10.ps

A pdf version of those notes (this link will probably expire - I
created it by uploading the ps file to www.ps2pdf.com) :
http://www.ps2pdf.com/tgetfile/S10.pdf?key=1036087512&name=S10.pdf

Here is a paraphrase of Lemma 10.4 (which answers part 2) :

Let X1, ..., Xn be IID N(mu, sig^2)

Define Xbar = sum(Xi)/n

Define Sxx = sum((Xi - Xbar)^2)

Then (there are 6 parts, I will prove each part as we go along) :

i) Xbar ~ N(mu, sig^2/n) and n(Xbar - mu)^2 ~ sig^2 * ChiSq(1)

Proof:
The sum of n IID normal variables has mean n*mu and variance n*sig^2.
Hence sum(Xi) ~ N(n*mu, n*sig^2).

When X ~ N(mu, sig^2), aX ~ N(a*mu, a^2 * sig^2). So (1/n)sum(Xi) ~
N(n*mu/n, n*sig^2/(n^2) ~ N(mu, sig^2/n)

the distribution of n(Xbar - mu)^2 follows from the definition of
ChiSq(1) as the square of a N(0,1) variable.

ii) Xi - mu ~ N(0, sig^2), so sum((Xi - mu)^2) ~ sig^2 * ChiSq(n)

Proof:
Follows directly from the definition of ChiSq(n) = sum of n
independent ChiSq(1)s.

iii) sum((Xi - mu)^2) = Sxx + n(Xbar - mu)^2

Proof:
sum((Xi - mu)^2) = sum(Xi^2 - 2*mu*Xi + mu^2)
= sum(Xi^2 - 2*Xi*Xbar + Xbar^2 + 2*Xi*Xbar - Xbar^2 - 2*mu*Xi + mu^2)
= sum((Xi - Xbar)^2) + sum(2*Xi*Xbar - Xbar^2 - 2*mu*Xi + mu^2)
= sum((Xi - Xbar)^2) + 2*sum(Xi)*Xbar - n*Xbar^2 - 2*mu*Sum(Xi) +
n*mu^2
= sum((Xi - Xbar)^2) + 2*n*Xbar^2 - n*Xbar^2 - 2*mu*n*xbar + n*mu^2
= sum((Xi - Xbar)^2) + n*(Xbar - mu)^2

iv) Not relevant for what you wanted

v) Xbar and Sxx are independent

Proof:

Let Y = A(X - mu*1) be a linear transformation of X
(1 here is a n-vector of ones)

Define A so that Y1 = sqrt(n)*(Xbar - mu) (ie. the first row of A is
(1/sqrt(n),... , 1/sqrt(n))) and so that the remaining rows of A are
orthogonal to the first row (This is possible if A is n by n)

[i'm going to use the notation ^t to mean transpose]

A orthogonal means that A^tA = I

Now sum(Yi) = Y^tY = (X - mu*1)^t A^tA (X - mu*1) = sum((Xi - mu)^2)

Then sum(Y2^2, ..., Yn^2) = sum((Xi - mu)^2) - Y1^2
= sum((Xi - mu)^2) - n(Xbar - mu)^2
= Sxx (by part iii)

but sqrt(n)*cov(Sxx, Xbar) = cov(sum(Y2^2, ..., Yn^2), Y1^2) = 0 as Y1
and Y2, ..., Yn are independent by the orthogonality of A.

vi) Sxx ~ sig^2ChiSq(n-1)

Proof:
Yi ~ N(0,sig^2)

So Sxx = sum(Y2^2, ..., Yn^2) = sum of n-1 squared normal variables.

SO FINALLY :

sqrt(n) * (Xbar - mu)/sig ~ N(0,1)

sqrt(Sxx / sig^2(n-1)) ~ ChiSq(n-1)

So :

sqrt(n) * (Xbar - mu)         N(0,1)
---------------------   ~   ---------- = t(n-1)
sqrt(Sxx / (n-1))           Chisq(n-1)

This is assuming that you treat the definition of a t-distribution
with n degrees of freedom to be the ratio of an independent Normal
distribution and the squareroot of a ChiSq(n) distribution scaled by
1/n).

Note that this definition of a t-distribution helps you with part (1)
of your question, since you can ask what happens to the distribution
of :

sqrt(Chisq(n)/n) as n goes to infinity.

Because a Chisq(n) distribution has a mean of n and variance of 2n the
distribution of Chisq(n)/n -> 1. It's been a while since I've done
that formally - so I'm not going to dare write something that is
mathematically weak.

Hope that helps.

Regards

calebu2-ga

Information from Mathworld on the t-distribution
http://mathworld.wolfram.com/Studentst-Distribution.html

Information from Mathworld on the Chi-squared distribution
http://mathworld.wolfram.com/Chi-SquaredDistribution.html

thanks