From: hrubin@odds.stat.purdue.edu (Herman Rubin)
Subject: Re: Asymptotic distribution of Shapiro-Wilk
Date: 13 Oct 2000 09:22:50 -0500
Newsgroups: sci.stat.math,sci.math.num-analysis
Summary: [missing]

In article <8s688e$mq7$1@merki.connect.com.au>,
Glen Barnett <glenbarnett@insureware.com> wrote:


			...................

>In it various quantities of interest (including what I need) have asymptotic
>distributions involving a r.v. \zeta, where \zeta is an infinite sum of mean-
>corrected chi-squared(1)'s divided by linearly increasing numbers.

>Specifically, they say that \zeta has the distribution of

>  \Sum_{i=3}^{\infty} (Y_i^2 - 1) / i

>where the Y_i are standard normal.

>(it's been a while, please excuse any silly errors in my LaTeX)

>Now this is very interesting, but the paper gives no clue about
>how to evaluate the percentage points of this creature. I have
>tried asking two of the authors, one was very helpful but couldn't
>answer my question, and the other I haven't heard back from.

>It seems like this particular distribution crops up pretty often in
>related calculations so I am betting some of you have seen it before.

It only requires the use of NSWKF (not sufficiently well
known facts).  The particular one given has some special
features, but the more general one asked for is not too much
harder.  The methods I suggest use numerical integration in
the complex plane involving the moment generating function.
I am assuming that one is interested in the right tail.

So let m be the moment generating function for a cdf F (and
density f).  Then it is fairly well known that the density
is given by

	f(x) = \int exp(-xt) m(t) dt / (2\pi i),

where the integration is from c-i\infty to c+i\infty,
where c is any real number for which the mgf exists,
including the special case c=0 of the characteristic
function.  What is less well known is that if c > 0,
we can divide the integrand by t to obtain the tail
cdf, even if the density does not exist.  So we have
an expression, if the mgf is computable; contour 
integration can be used for both the density and the
cdf, and both should be computed if Newton's method
is used, preferably on the logarithm.

Now to the specific problem.  The mgf of an infinite
linear combination of mean-adjusted squares of normal
random variables with weights w_k (let us avoid i)
is given by 

	m(t)^2  = \prod exp(-2w_k t)/(1 - 2w_k t),

and this exists if and only if \sum w_k^2 is finite.
for the particular case here, the w_k are the reciprocals
of the integers from 3 to infinity.

But we know that

	\Gamma(u) = (\prod exp(u/n)/(1+u/n)) / (u exp(\gamma u)).

A little algebra yields that the product from 3 to infinity
is just \Gamma(u+3)/(2*exp(u*(\gamma - 1.5))).  Substitute
-2t for u in this.

One can get a more complicated, but similar, expression
when the coefficients are of the form 1/h, 1/(h+1), etc.
The expression is that

	m(t)^2 = \Gamma(h-2t)*exp(-2t*\Psi(h))/\Gamma(h).

Also, an initial approximation for the right tail can
be obtained by using the chi-squared distribution for 
the largest positive weights, multiplied by the mgf
at this singularity with the zero terms deleted.
for the case of linear coefficients 
-- 
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, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
hrubin@stat.purdue.edu         Phone: (765)494-6054   FAX: (765)494-0558