Characteristic function question

**melies** · May 7th 2009, 09:08 AM

Hi,

I am interested in clarifying the meaning and uses of the characteristic functions. I want to create a likelihood distribution in the Fourier Domain bearing in mind that it is a conditioned Poisson (or Gaussian) distribution in the inverse-Fourier domain.

1º) The characteristic function is defined as E[exp(itx)] and can be seen as the conjugate of the Fourier Transform of the density function f(x) if I am not wrong. If such a density function is the P(X=x) or P(a<X<b). Can I consider the characteristic function as the equivalent probabilities P(W=w) or P(A<W<B)?

I mean, is the characteristic function a kind of density function in the Fourier Domain for the distribution X?

2º) Suppose I define a likelihood function as a Poisson distribution.

L=exp(-lambda) [lambda^x]/x!

Where the probability P(X=p|h) = exp(-h) [h^p]/p!

And I have the p dataset both in the Fourier (P) and non-Fourier domain (p).

What would be the equivalent expression for the charateristic function? or how can I get it?

At the beginning I thought I could say that, if the characteristic function for a Poisson distribution is exp[lambda(e^iw-1)], the equivalent expression could be something similar to exp[H(P-1)], bearing in mind that P are complex numbers, H as well. Or exp[|H|(e^i|P|-1) in order to keep lambda and w reals.
But now I believe I am misunderstanding things, especially related with point number 1. What is the role of the characteristic function?

If anybody can help would be great.
Thanks

**cl85** · May 8th 2009, 07:35 PM

I won't claim that I have studied characteristic functions in depth but my take on your question is as follows.

Basically, the characteristic function provides an equivalent representation of a distribution. We can specify a distribution either by its cumulative distribution function, its probability density function, all its moments, its characteristic function, etc. All these are equivalent representations of a distribution.

Just like if we have piece of signal, taking its Fourier transform gives us its equivalent representation in the frequency domain. It's a 1-to-1 mapping. But of course I don't think it's enlightening at all to think about frequency of a distribution.

So, my view on the characteristic function is that it is the sum of all the moments of the distribution. This can be observed by taking the power series expansion:
$\int_{-\infty}^{\infty}f_X(x)e^{itx}dx = \int_{-\infty}^{\infty}f_X(x)\sum_{k=0}^{\infty}\frac{(it )^k}{k!}x^k dx =\sum_{k=0}^{\infty}\frac{(it)^k}{k!}E[X^k]$
Each of the moments can be extracted from the characteristic function through differentiation.

**Moo** · May 10th 2009, 09:20 AM

We can specify a distribution either by its [...], its probability density function

IF it exists, which is not always true.
So it's not an equivalent representation of a random variable

all its moments

Not sure about this one :/

Anyway, what is a Fourier domain ?

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