# Deconvolution with fourier transform or characteristic function?

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.

• 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? – Giuseppe Nov 23 '12 at 14:19
• @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 $j$ – Dilip Sarwate Jan 5 '17 at 16:24