Let us consider the following model:

$$Y_j = X_j + \epsilon_j \hspace{15pt} j=1, ..., n$$

Where $Y_j$ is a noisy signal, $\epsilon_j$ is the noise which is independend from the signal $X_j$. We have only i.i.d. samples of $Y_j$ and $\epsilon_j$ and are interested in the distribution of $X_j$. The density $f_{\epsilon}$ is assumed to be unknown. Comte and Lacour suggest a method based on fourier transform to solve this problem (see section 2.2).

Let's call $\varphi_X(t)$ the characteristic function for $X$ and $f^{*}_X$ the fourier transform of the density $f_X$.

Here is my question: The main idea in deconvolution is to use the independence assumption for $X$ and $\epsilon$ and then use fourier transform to solve the equation $f^{*}_X = f^{*}_Y/f^{*}_{\epsilon}$. Applying the inverse fourier transform leads to an estimate for $f_X$.

Can I use the characteristic function instead of the fourier transform? Does this give me any advantages or disadvantages? I assume that both fourier transform or characteristic function could be used but would like to know what other people think.


1 Answer 1


From Page 342, of Probability and Measure. P. Billingsley 3rd editon.

The characteristic function in nonprobabilistic contexts is called the Fourier transform.

So yes both can be used as they are the same thing.

  • $\begingroup$ Okay this sounds like what I was looking for. My issue is that I can't see from their definition that both characteristic function and fourier transform are the same thing: fourier transform is defined as $\int \exp(-itx) \cdot f(x) \; dx$ link and the characteristic function $\int \exp(itx) \cdot f(x) \; dx$ link As we can see the definition differ with a minus. Why can we than say that both are the same thing? Is there any proof? $\endgroup$
    – Giuseppe
    Nov 23, 2012 at 14:19
  • $\begingroup$ They are conceptually the same thing, this differences of signs is a plague but insignificant conceptually, they contain the same information. $\endgroup$ Jan 5, 2017 at 16:13
  • $\begingroup$ @Giuseppe There are two square roots of $-1$. If you label one of them as $i$, then the other equals $-i$. But, someone else (e.g. an electrical engineer such as myself) might well label the other square root as $j$ in which case, what you call as $i$ is my $-j$. So, my Fourier transforms are your characteristic functions and vice versa, and yet, my math works in exactly the same way as your math and I can use all your theorems and tables of Fourier transforms as long as I simply replace $i$, wherever I find it, with $j$ $\endgroup$ Jan 5, 2017 at 16:24

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