For $f \in L^1(\mathbb{R}^n)$ its Fourier transform is defined as $$\hat{f}(\xi) = \int_{\mathbb{R}^n} f(x) e^{-i\xi\cdot x} dx$$ up to a choice of normalization. The inverse Fourier transform is defined as $$\check{f}(x) = \int_{\mathbb{R}^n} \hat{f}(\xi) e^{i\xi\cdot x} d\xi.$$

However when defining characteristic functions in probability things are defined the other way around. The characteristic function of a probability measure $\mu$ is defined as $$\hat{\mu}(\xi) = \int e^{i\xi \cdot x} d\mu(x).$$

In probability books they say $\hat{\mu}$ is the Fourier transform of $\mu$. But compared to the definition for $L^1$ functions, or even for general Radon measures, the exponential has the opposite sign.

Would it be more appropriate to call this the inverse Fourier transform of $\mu$? I suspect this is largely due to convention, as answers to other similar questions suggest. If this is convention, what is the reason or history behind it?


1 Answer 1


The Fourier Transform is based on the Fourier Series, which can be computed for a real-valued (integrable) function defined on a bounded interval $[a,b]$.

In the Fourier Series, the coefficients are calculated by doing the scalar product between the given function and the functions $e^{inx}$, for $n\in\mathbb{Z}$ (normalized appropriately).

Recall that the scalar product between two complex-valued functions $f,g$ on $[a,b]$ is given by $$\langle f,g\rangle=\int_a^b f(x)\overline{g(x)}\,dx$$

In particular, if $g(x)=e^{inx}$ we get $$\langle f,g\rangle=\int_a^b f(x)\overline{e^{inx}}\,dx=\int_a^b f(x)e^{-inx}\,dx$$

If now $f$ is a function defined on the whole real line and assuming that $f$ decays rapidly (so that the integral converges), we can extend this definition naturally, giving us the 'common' formula $$\hat{f}(\xi) = \int_{\mathbb{R}} f(x) e^{-i\xi\cdot x} dx$$

which can obviously be generalized for $\mathbb{R^n}$.

This is just a convention and the sign can be changed as one desires, as this is equivalent to defining $\sqrt{-1}=i$ or $\sqrt{-1}=-i$.

  • $\begingroup$ Thank you, is there any reason probabilists use the convention $e^{i\xi\cdot x}$? $\endgroup$
    – CBBAM
    Commented May 23 at 18:31
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    $\begingroup$ I am not sure, but it is most probably a historical reason (some author possibly used the positive sign and others followed). As I said, it does not really affect as long as the choice is consistent within the text. $\endgroup$ Commented May 23 at 18:36
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    $\begingroup$ @CBBAM My guess is the same as Julio's. Moreover, according to the Wikipedia page for characteristic functions, Bochner was an example of someone who used the convention $\varphi_X(t) = \mathbb{E}[e^{-i2\pi tX}]$, so it's not like other conventions have never been considered I suppose. $\endgroup$
    – Bruno B
    Commented May 23 at 18:43

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