Suppose f(x) is a probability distribution of a random variable X, where x is real.
We may calculate the Fourier transform of f(x) and this way obtain the so-called characteristic function of X.
Let's further suppose that X is bounded, i.e. f(x) = 0 for |x| > a, so that X may obtain values only from [-a;a].
In this case, instead of Fourier transform we may calculate the coefficients of Fourier series of f(x).
Is such "characteristic series" somehow established in theory of probability? Could you, please, provide any links/references to related definitions, theorems, etc.?
Dear Adam,
I agree with George: this is quite classical stuff. It is related to Bochner's theorem. In addition to the literature suggested by George, you could also have a look at Laurent Schwartz's "Mathematics for the Physical Sciences" or his "Theorie des distributions".
If you think of a Fourier series describing f given on a finite domain [-a,a], then this Fourier series is a 2a-periodic function on the whole of \R which agrees with f only on [-a,a].
My main problem would be that it is no longer given as the expectation of \exp\{itX\}. It is different from the characteristic function and it should not have the connection to weak convergence, which is so important.
Dear Dirk,
You could consider X as a variable with values in the unit circle. The unit circle being a compact Abelian group, you can apply the Fourier machinery belonging to it. When identifying the dual of this group with the set of integers Z, the Fourier transform of X appears a function k \mapsto E{exp(ikX)}. This is how characteristic functions appear for this particular group. Generally, for a random variable X assuming values in an arbitrary locally compact Abelian group G, the characteristic function is defined on the dual group \hat{G} as the function \chi \mapsto E{\chi(X)}. The probability distribution of X is completely pinned down by this function. The latter is a consequence of Bochner's theorem.
Dear Wiebe,
Let me just clarify if I understood it correctly.
So you say that the discrete Fourier transform of the density f(x)
(in case |X| \leq a)
is just the characteristic function of X given by E{exp(i )}
where k \ in \Z is the dual group of the interval (which is identified as S^1)?
Dear Dirk,
Yes, that's what I wanted to say. You need a group structure in order to do Fourier analysis. On the interval [-a,+a] this group structure can be realized by identifying it with S^1={z \in C | |z|=1}. The dual group then consists of characters of the form z \mapsto z^k. Hence, the dual group can be identified with the additive group of all integers Z. It is on this dual group where the characteristic functions are defined. In the case of the additive group of all real numbers R, the dual group is isomorphic to the group itself. Hence the characteristic function can then be thought to be defined on R. It is this scenario where most probabilists are used to. Because the group R is self-dual, the Pontryagin duality is more or less hidden in this case. For this reason mathematicians may overlook that, generally, a Fourier transform is defined on the dual group and NOT on the group itself.
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