# Characteristic function and Fourier transform

I understand the definition of characteristic functions used in probability theory: For a random Variable $X$ with probability density function $f_X$ the characteristic function is defined as: $$\phi_X(t) = E(\exp(itX)) = \int_{\mathbf{R}} e^{\mathrm{i}tx}f_X(x)\, dx.$$

[...] the characteristic function in nonprobabilistic contexts is called the fourier transform (Page 342, of Probability and Measure. P.Billingsley 3rd editon).

But I still can't see this from their definitions, because the fourier transform $f^*$ of a function $f$ is defined as follows:

$$f^*(t) = \int_{\mathbf{R}^n} f(x)\,e^{-\mathrm{i} t \cdot x} \,\mathrm{d} x$$

If the function $f$ is a density, for e.g. $f=f_X$, then the fourier transform can be written as: $$f^*_X = \int_{\mathbf{R}^n} f_X(x)\,e^{-\mathrm{i} t \cdot x} \,\mathrm{d} x \neq \int_{\mathbf{R}} e^{\mathrm{i}tx}f_X(x)\, dx = \phi_X(t)$$ The definitions differ with a minus, why can they than be used both for the same thing (e.g. for deconvolution)?

• Yes, as you note $f_X^* = \phi_X(-t)$. They can each be used for the things you use the other for, as long as you keep the change of sign straight (and in some definitions, different constants out the front) when going the other way (inverting the transformation). – Glen_b Dec 2 '12 at 23:32
• There are two square roots of $-1$. Mathematicians use one of them and call it $i$; engineers use the other and call it $j$. Since $j = -i$, the characteristic function $\int e^{itx} f(x) dx$ used by mathematicians is exactly the same as the Fourier transform $\int e^{-jtx} f(x) dx$ used by engineers – Dilip Sarwate Dec 3 '12 at 2:17
• – leonbloy Jan 2 '13 at 2:20

Actually, the Fourier transform can be defined in both ways by using $e^{-\mathrm{i} t \cdot x}$ or $e^{\mathrm{i} t \cdot x}$. They are essentially the same, just like you can call either $i$ or $-i$ the imaginary unit.