Hi everyone...

   Last week I posted a rather theoretical question on
the list. I received many responses, thanks to everyone who
contributed.
Anyway, below I write some selected responses (at
least the ones which I found useful) plus some short comments.

   For those who may have missed the question I repeat
it below (hope the others won't mind):

   " We all know that if X is a normally distributed
random variable then
Y=f(X)=a*X+b (a, b are real, Y is, of course a random
variable) is also
normally distributed. My question is does there exist
any NON-LINEAR
CONTINUOUS function that has this property:

g(.) is  a non-linear continuous function such that if
X is a normally distributed random variable then the random
variable Y=g(X) is also normally distributed "

   In fact the problem is still an open, the results below assume some 
additional conditions other than mere continuity. 
                            
First is a reply from Ted Harding 
(from University of Manchester I presume):

================ Ted Harding ===============================

It is not difficult to show that only linear
transformations
preserve normality, at any rate under the condition
that the
transformation is differentiable.

Suppose X is normally distributed, say N(0,1). Let dP
denote
the probability that X is in the small interval dx:

  dP = C.exp(-½ x^2) dx

(where C = 1/sqrt(2pi)).

Now suppose x = g(y), and let dx corrspond to dy, so
that
dP, the probability that X is in dx, is also the
probability
that Y is in dy. If p(y) is the density function for
Y, then

  dP = p(y)dy = C.exp(-½ x^2) dx
     = C.exp(-½ g(y)^2) |g'(y)| dy

where g' is the derivative of g, since dx = |g'(y)|dy 
  

Now Y is to be Normally distributed, say N(m,D^2)
for some values of m and D, so

  C.exp(-½ g(y)^2) |g'(y)| = (C/D).exp(½(y - m)^2 /
D^2)

or

  dg/dy = +/- (1/D).exp(½(y - m)^2 / D^2 - ½ g^2)

which (according to sign +/-) gives two possible
differential
equations for g(y). Take the "+" case, and suppose D
is given.

Clearly one solution is   g = (y - m)/D   and it is
then possible
to choose m so that the solution g(y) goes through any
given
point (y0,g0) for g(0)>0.

Now the above differential equation belongs to a class
such
that there is only one solution through a given point
(y0,g0)
(essentially, it satisfies Liouville's conditions).
Hence 
the _only_ solution is of this form, i.e. is linear.

There is a similar argument for the "-" case, again
leading
to a linear transformation g = -(y - m)/D.

======================================================================

No comments here.

The next one is from Emmanuel S. Aziz


===========Emmanuel S. Azis==========================================

I guess that you are interested in a class g(X) of
non-linear continuous
functions of a normal random variable X. In addition
if  g(.) is
differentiable and all its derivatives exist at X=a,  
then use Taylor
expansion of g(.) with/without remainder to write g(.)
as a polynomial in X.
Clearly the distribution of this NOT normal.

===========================================================================

  Interesting idea here. But we should be careful
about asserting that if all 
derivatives 
exists at a point then a function has a Taylor
expansion there. If you know
distribution theory (most people studying systems and
control theory usually have at 
least heard about it) then the so-called test function
is infinitely differentiable at its center
(it is a symmetric function) yet has NO Taylor series
expansion at that point. Weird functions with
this property are usually called non-analytic.
  Secondly it may be possible that the sum of a
Gaussian + non-Gaussian rand. vars. add up to make a 
Gaussian rand. var. Conversely, it is also poosible
for two correlated Gaussian r.vs to add up to 
non-Gaussian rand. var. I think more work is needed
for this argument. 

Finally, below is a response from Terence Iles of
Cardiff Univ.
He proposes that there may be non-linear
transformations
which are approximately Gaussian under some
conditions.

==========Terence Iles====================================

A research student of mine, Gareth James, has just
completed a
thesis in which he studied your question (in the
context of
regression models). We have a theoretical result, not
yet in print,
along the lines that if the coefficient of variation
of a normal
distribution is small, the mean is positive, and the
transformation is
monotonic and 'not too curved', then the transformed
random
variable is approximately normal. For example if you
have a N(9,1)
and square root it, the result is approximately
Normal.
We would be very interested to know of the context of
your
question, and of any other responses that you get.

Gareth looked at a range of power
transformations X**p with p ranging from -1 to -.25
then +.25 to 2,
and simulations showed that with a CV less than 1/4 or
so the
transformed data are pretty closely approximated by
the normal
distribution. For a CV less than 1/10 you would need a
very large
sample to be able to tell the difference. He found the
same for log.
We don't know for sure whether the result is true for
other non-
linear transfromations, but for exp(x) something
different certainly
applies.    

You could use Gareth's PhD for a reference, we won't
have a paper
in print within your 4 month time-scale. 'Non-Linear
Errors-in-
Variables Regression' G.G. James PhD thesis,
University of Cardiff
2002.                 

======================================================================


Well this summary does not mean that the topic has to end.
If anyone has more suggestions please feel free to send me an
e-mail.


Best wishes,


Hendra Ishwara Nurdin