# Why is the Fourier transform in probability defined with the opposite sign?

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?

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 $$\sqrt{-1}=-i$$.

• Thank you, is there any reason probabilists use the convention $e^{i\xi\cdot x}$? Commented May 23 at 18:31
• 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. Commented May 23 at 18:36
• @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. Commented May 23 at 18:43