Science > Physics > what conditon should a probability characteristic function satisfy?
| Topic: |
Science > Physics |
| User: |
"Luna Moon" |
| Date: |
18 Aug 2006 02:01:17 AM |
| Object: |
what conditon should a probability characteristic function satisfy? |
Hi all,
I am wondering is there a condtion for a function to be probability
characteristic function, i.e., it has a valid probability distribution
function associated with it? I remembered there is such a theorem
existing, and I just could not get it back into my memory...
I am also looking for methods to a method to directly fit data to its
characteristic function, instead of probability distribution. Could
anybody please give me some pointers?
Thx!!!
Luna
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| User: "David Jones" |
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| Title: Re: what conditon should a probability characteristic function satisfy? |
18 Aug 2006 04:54:49 AM |
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Luna Moon wrote:
Hi all,
I am wondering is there a condtion for a function to be probability
characteristic function, i.e., it has a valid probability
distribution
function associated with it? I remembered there is such a theorem
existing, and I just could not get it back into my memory...
As far as I know, there is nothing simple apart from, ("phi(t)"
represents the characteristic function at t)
(i) phi(0)=1
(ii) log(phi(t)) can't be exactly a finite polynomial in t unless the
dergree of the polynomial is less than or equal to 2.
(iii) the inverse transformation (related to the inverse Fourier
transform but with a particular scaling) gives a "a valid probability
distribution function", but this is the questuion you are asking, so
it is of no help.
Hower you should see the book: "Characteristic Functions" by E.Lukacs,
Griffin 1970 (2nd Ed).
I am also looking for methods to a method to directly fit data to
its
characteristic function, instead of probability distribution. Could
anybody please give me some pointers?
There has been previous work on this: I think it was often in the
context of fitting stable distributions where the characteristic
function has a simple formula but there is no usaeable formula for the
density for use in maximum likelihood. So you might look in the
literature for stable distributions. Alternatively, you might search
for the keyword "empirical characteristic function".
You might also be interested in the follwing as a extension from just
model fitting:
Epps TW & Pulley LB (1983) A test for normality based on the empirical
characteristic function. Biometrika, 70 (3) 723-726.
David Jones
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| User: "Herman Rubin" |
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| Title: Re: what conditon should a probability characteristic function satisfy? |
18 Aug 2006 09:00:13 AM |
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In article <44e58e6b$1@news.nwl.ac.uk>, David Jones <dajxxx@ceh.ac.uk> wrote:
Luna Moon wrote:
Hi all,
I am wondering is there a condtion for a function to be probability
characteristic function, i.e., it has a valid probability
distribution
function associated with it? I remembered there is such a theorem
existing, and I just could not get it back into my memory...
As far as I know, there is nothing simple apart from, ("phi(t)"
represents the characteristic function at t)
(i) phi(0)=1
(ii) log(phi(t)) can't be exactly a finite polynomial in t unless the
dergree of the polynomial is less than or equal to 2.
(iii) the inverse transformation (related to the inverse Fourier
transform but with a particular scaling) gives a "a valid probability
distribution function", but this is the questuion you are asking, so
it is of no help.
A LITTLE more can be said, but not much. There are
situations in which more can be invoked, but not much.
Hower you should see the book: "Characteristic Functions" by E.Lukacs,
Griffin 1970 (2nd Ed).
An excellent idea.
I am also looking for methods to a method to directly fit data to
its
characteristic function, instead of probability distribution. Could
anybody please give me some pointers?
The sample characteristic function is an unbiased
estimate of the population characteristic function,
and thus can be used for fitting. As it is the
average of bounded complex random variables, its
joint distribution is asymptotically normal.
But beware of using it for large values; these need
substantial discounting. Convergence is uniform in
BOUNDED regions, but not in unbounded regions except
in rare cases.
There has been previous work on this: I think it was often in the
context of fitting stable distributions where the characteristic
function has a simple formula but there is no usaeable formula for the
density for use in maximum likelihood. So you might look in the
literature for stable distributions. Alternatively, you might search
for the keyword "empirical characteristic function".
You might also be interested in the follwing as a extension from just
model fitting:
Epps TW & Pulley LB (1983) A test for normality based on the empirical
characteristic function. Biometrika, 70 (3) 723-726.
David Jones
--
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@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
.
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| User: "David Jones" |
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| Title: Re: what conditon should a probability characteristic function satisfy? |
18 Aug 2006 09:38:56 AM |
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Herman Rubin wrote:
In article <44e58e6b$1@news.nwl.ac.uk>, David Jones
<dajxxx@ceh.ac.uk> wrote:
Luna Moon wrote:
Hi all,
<snip>
I am also looking for methods to a method to directly fit data to
its
characteristic function, instead of probability distribution.
Could
anybody please give me some pointers?
The sample characteristic function is an unbiased
estimate of the population characteristic function,
and thus can be used for fitting.
.... and, more importantly, it is a consistent estimate
As it is the
average of bounded complex random variables, its
joint distribution is asymptotically normal.
But beware of using it for large values; these need
substantial discounting. Convergence is uniform in
BOUNDED regions, but not in unbounded regions except
in rare cases.
I recall seeing a paper that looked at the question of choosing which
values of "t" should be used
for a finite set of instances within an objective function so as to
get the best estimates of model parameters ... these were rather close
to zero I think. However things have moved on it seems to consider
integrals over t in forming an objective function. A recent reference
I just found is at www.mysmu.edu/faculty/yujun/YuER.pdf
David Jones
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| User: "Luna Moon" |
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| Title: Re: what conditon should a probability characteristic function satisfy? |
18 Aug 2006 10:59:16 PM |
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Thanks so much guys for your pointers! I really appreciate your
effective help!
Have a nice weekend!
Luna.
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| User: "Luna Moon" |
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| Title: Re: what conditon should a probability characteristic function satisfy? |
19 Aug 2006 01:50:45 AM |
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Hi guys,
I've just found that this "positive definite function" requirement for
a function to be a probability characteristic function is not a
constructive and operational one and it is hard to use it to construct
characteristic functions... Suppose I want to create some function that
is characteristic function, what guidelines should I follow to make the
function "postive definite"?
Thanks again!
Luna
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| User: "ArtflDodgr" |
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| Title: Re: what conditon should a probability characteristic function satisfy? |
19 Aug 2006 11:47:48 AM |
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In article <1155970245.104912.16680@m79g2000cwm.googlegroups.com>,
"Luna Moon" <lunamoonmoon@gmail.com> wrote:
Hi guys,
I've just found that this "positive definite function" requirement for
a function to be a probability characteristic function is not a
constructive and operational one and it is hard to use it to construct
characteristic functions... Suppose I want to create some function that
is characteristic function, what guidelines should I follow to make the
function "postive definite"?
There is one simple sufficient condition ("polya's theorem") for a
function to be a characteristic function, which leads to many examples.
If f:[0,infinity) --> R is continuous and convex, with f(0) = 1, then
g(t) := f(|t|) is a characteristic function.
--
A.
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| User: "Herman Rubin" |
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| Title: Re: what conditon should a probability characteristic function satisfy? |
19 Aug 2006 12:18:12 PM |
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In article <artfldodgr-442C88.09474819082006@news.west.cox.net>,
ArtflDodgr <artfldodgr@aol.com> wrote:
In article <1155970245.104912.16680@m79g2000cwm.googlegroups.com>,
"Luna Moon" <lunamoonmoon@gmail.com> wrote:
Hi guys,
I've just found that this "positive definite function" requirement for
a function to be a probability characteristic function is not a
constructive and operational one and it is hard to use it to construct
characteristic functions... Suppose I want to create some function that
is characteristic function, what guidelines should I follow to make the
function "postive definite"?
There is one simple sufficient condition ("polya's theorem") for a
function to be a characteristic function, which leads to many examples.
If f:[0,infinity) --> R is continuous and convex, with f(0) = 1, then
g(t) := f(|t|) is a characteristic function.
There are many other ways. The above produces no distributions
which have a first moment.
--
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@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
.
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| User: "G=EMC^2 Glazier" |
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| Title: Re: what conditon should a probability characteristic functionsati... |
19 Aug 2006 05:13:54 PM |
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Spin up or spin down for an electron Bert
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| User: "David Jones" |
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| Title: Re: what conditon should a probability characteristic function satisfy? |
06 Sep 2006 11:34:30 AM |
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Luna Moon wrote:
Hi guys,
I've just found that this "positive definite function" requirement
for
a function to be a probability characteristic function is not a
constructive and operational one and it is hard to use it to
construct
characteristic functions... Suppose I want to create some function
that is characteristic function, what guidelines should I follow to
make the function "postive definite"?
Thanks again!
Luna
If this question is still of interest, I have come across some results
in the book below that deal with creating new characteristic functions
out of known characteristic functions. See Section 6.10 (p 275) of
Moran PAP.(1968) Introduction to Probabilitry theory. Clarendon Press,
Oxford.
David Jones
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| User: "David Jones" |
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| Title: Re: what conditon should a probability characteristic function satisfy? |
21 Aug 2006 04:39:54 AM |
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Luna Moon wrote:
Hi guys,
I've just found that this "positive definite function" requirement
for
a function to be a probability characteristic function is not a
constructive and operational one and it is hard to use it to
construct
characteristic functions... Suppose I want to create some function
that is characteristic function, what guidelines should I follow to
make the function "postive definite"?
Thanks again!
Luna
You could start from the methods commonly used for constructing new
distributions when working from the "probability distribution" side of
the question and see if any of these carry over in a useful way. Two
obvious possibilities are:
(i) construction by addition ... the distribution function of the sum
of two independent random variables is given by the convolution of the
pair of distribution functions. This is quite complicated from the
"probability distribution" side, but rather simpler for characteristic
functions, since you just need the product of the characteristic
functions. This can be extended to the product of more than two and,
for appropriate conditions (weightings of the underlying rv's) to
infinite products.
(ii) construction by compounding .. this is an extension of simple
finite mixtures, which I assume are an obvious possibility (giving a
weighted sum of the characteristic functions). Write the distribution
function of a rv in terms of a number of parameters and let some or
all of these parameters be selected at random from given probability
distributions. Then the distribution function or characteristic
function of the final random variable is the integral of the original
distribution function or characteristic function (as a function of the
parameters) over the distribution of the parameters.
For example, the above possiblities lead to different ways of
combining an "ordinary" distribution such as the Normal, with an
unusual one such as a stable distribution.
You could also try consulting a table or listing of known
characteristic functions to see if there is a family that suits your
needs. You can find some in books which are a collection of tables,
but one separate table is available as:
Oberhettinger F (1973) Fourier Transforms of Distributions and Their
Inverses: a collection of Tables. Academic Press, NY.
Before you go too far with this, you should consider whether you need
to consider including the multivariate case in what you are trying to
do, as this may infulence some of your decisions.
.
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| User: "Luna Moon" |
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| Title: Re: what conditon should a probability characteristic function satisfy? |
22 Aug 2006 06:58:46 PM |
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David Jones wrote:
Luna Moon wrote:
Hi guys,
I've just found that this "positive definite function" requirement
for
a function to be a probability characteristic function is not a
constructive and operational one and it is hard to use it to
construct
characteristic functions... Suppose I want to create some function
that is characteristic function, what guidelines should I follow to
make the function "postive definite"?
Thanks again!
Luna
You could start from the methods commonly used for constructing new
distributions when working from the "probability distribution" side of
the question and see if any of these carry over in a useful way. Two
obvious possibilities are:
(i) construction by addition ... the distribution function of the sum
of two independent random variables is given by the convolution of the
pair of distribution functions. This is quite complicated from the
"probability distribution" side, but rather simpler for characteristic
functions, since you just need the product of the characteristic
functions. This can be extended to the product of more than two and,
for appropriate conditions (weightings of the underlying rv's) to
infinite products.
(ii) construction by compounding .. this is an extension of simple
finite mixtures, which I assume are an obvious possibility (giving a
weighted sum of the characteristic functions). Write the distribution
function of a rv in terms of a number of parameters and let some or
all of these parameters be selected at random from given probability
distributions. Then the distribution function or characteristic
function of the final random variable is the integral of the original
distribution function or characteristic function (as a function of the
parameters) over the distribution of the parameters.
For example, the above possiblities lead to different ways of
combining an "ordinary" distribution such as the Normal, with an
unusual one such as a stable distribution.
You could also try consulting a table or listing of known
characteristic functions to see if there is a family that suits your
needs. You can find some in books which are a collection of tables,
but one separate table is available as:
Oberhettinger F (1973) Fourier Transforms of Distributions and Their
Inverses: a collection of Tables. Academic Press, NY.
Before you go too far with this, you should consider whether you need
to consider including the multivariate case in what you are trying to
do, as this may infulence some of your decisions.
Thanks a lot David!
Compouding and addition make a lot sense. But they tend to produce
characteristic functions that are very complicated.
Since I am using characteristic function in data fiting(fitting data
directly to the empirical characteristic function), I am looking for a
characteristic function model that is simple, and flexible(with some
parameters) which can capture quite a number of probability models.
For example, in probability distribution domain, the Weibull
distribution and many others are with 2-3 parameters are quite flexible
and good for data fitting. Gaussian is a good one, but I am looking for
positive valued probability distributions.
I am looking for a characteristic function model, with 2-3 parameters,
and then it can be fit to data flexibly...
Any more comments? Thanks again!
.
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| User: "Herman Rubin" |
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| Title: Re: what conditon should a probability characteristic function satisfy? |
23 Aug 2006 01:06:33 PM |
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In article <1156291125.980142.162130@i3g2000cwc.googlegroups.com>,
Luna Moon <lunamoonmoon@gmail.com> wrote:
David Jones wrote:
Luna Moon wrote:
..................
For example, the above possiblities lead to different ways of
combining an "ordinary" distribution such as the Normal, with an
unusual one such as a stable distribution.
You could also try consulting a table or listing of known
characteristic functions to see if there is a family that suits your
needs. You can find some in books which are a collection of tables,
but one separate table is available as:
Oberhettinger F (1973) Fourier Transforms of Distributions and Their
Inverses: a collection of Tables. Academic Press, NY.
Before you go too far with this, you should consider whether you need
to consider including the multivariate case in what you are trying to
do, as this may infulence some of your decisions.
Thanks a lot David!
Compouding and addition make a lot sense. But they tend to produce
characteristic functions that are very complicated.
Since I am using characteristic function in data fiting(fitting data
directly to the empirical characteristic function), I am looking for a
characteristic function model that is simple, and flexible(with some
parameters) which can capture quite a number of probability models.
For example, in probability distribution domain, the Weibull
distribution and many others are with 2-3 parameters are quite flexible
and good for data fitting. Gaussian is a good one, but I am looking for
positive valued probability distributions.
I am looking for a characteristic function model, with 2-3 parameters,
and then it can be fit to data flexibly...
Any more comments? Thanks again!
Other than such things as compound Poisson, stable laws,
and a few other categories, I do not know of any such.
It is relatively simple to produce probability models,
some of which lead to simple calculations, and some of
which do not. Even here there are limitations caused
by the requirement that probabilities are positive.
There have been many attempts to get a wide class of
probability distributions with a few parameters. The
Pearson family was a successful fit on 4 moments, but
the Gram-Charlier had a substantial chance of producing
negative probabilities. Similarly, attempts which are
not highly restricted in producing characteristic
functions are likely to produce functions which do
not satisfy the conditions.
A similar example is the compounding distribution of
a compound Poisson distribution. The probability of
getting a value k on a given trial is
\int exp(-t)*t^k/k! dM(t).
Now one could take k! times the proportion of results
which are k, and use these to "fit" the k-th moment of
the measure dQ(t) = exp(-t) dM(t), but the sample
proportions, if anything other than 0 occurs, will
not satisfy the conditions for moments. The fitting
is poor for the measure M, but not poor for the true
probabilities of the outcomes.
--
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@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
.
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| User: "David Jones" |
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| Title: Re: what conditon should a probability characteristic function satisfy? |
23 Aug 2006 04:41:20 AM |
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Luna Moon wrote:
Compouding and addition make a lot sense. But they tend to produce
characteristic functions that are very complicated.
Since I am using characteristic function in data fiting(fitting data
directly to the empirical characteristic function), I am looking for
a
characteristic function model that is simple, and flexible(with some
parameters) which can capture quite a number of probability models.
For example, in probability distribution domain, the Weibull
distribution and many others are with 2-3 parameters are quite
flexible and good for data fitting. Gaussian is a good one, but I am
looking for positive valued probability distributions.
I am looking for a characteristic function model, with 2-3
parameters,
and then it can be fit to data flexibly...
Any more comments? Thanks again!
You could start from the (2 parameter) Gamma distribution, which has a
simple cf. If you were looking for a generalisation of this you could
try having a model where a power-transformation of the observations
(with an unknown power) has a Gamma distribution. I don't think you
would find an expression for the characteristic function of a
power-transformed Gamma, but you should be able to slightly modify the
approach so that you match the empirical cf of transformed
observations to the cf of a Gamma rv, instead of the basic approach
where you would match the empirical cf of untransformed observations
to the cf of a transformed Gamma rv.
The idea of transforming the observations can of course be extended.
For example you could take the logarithms of the data and then choose
a family of distributions to match the distribution of the transformed
observations, which would have the whole real line for their range ...
this gives you access to this entire class of distributions.
You seem to be starting rather with the mathematical considerations,
rather than data-fitting as such. Applied statistics has a number of
data-analytic approaches to help in choosing appropriate families of
distributions. Alternatively it may be that the data being considered
are of a general type where the use of particular families of
distributions has already be investigated.
David Jones
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