Characteristic function and Fourier transform

Question

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: $$\phi_X(t) = E(\exp(itX)) = \int_{\mathbf{R}} e^{\mathrm{i}tx}f_X(x)\, dx.$$

I read that

[...] 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)?

Answer 1

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.

By the same taken, you can define the characteristic functions via the Fourier transform or the inverse Fourier transform depending on your choice.