# On inverting characteristic functions

Let $$X$$ be a random variable in $$\mathbb{R}^n$$ with distribution $$\mu$$ and characteristic function $$\varphi$$ (i.e. $$\varphi(t)=\mathbb{E} e^{i\langle t,X\rangle}$$). The standard inversion formula asserts that $$\mu \big(\{a where the set $$\{ a is the rectangle $$(a_1,b_1)\times\cdots \times (a_n,b_n).$$

Instead of working with rectangles I am interested in working with open Euclidean balls. Is the corresponding integral relationship similar to the one above known in this case? I would be also very grateful for related references.

• Are you assuming that $\mu$ is absolutely continuous? Mar 29, 2021 at 13:44
• No, I am particularly interested in the case where $\mu$ is discrete and supported on a lattice.
– TOM
Mar 29, 2021 at 14:47
• Not sure it is related but Fefferman showed things can go wrong when looking at convergence of Fourier series in 2d on squares versus disks. See math.stackexchange.com/questions/1849763/… Mar 29, 2021 at 18:02
• A thank you to @MichaelHardy for the clarity mods. Mar 30, 2021 at 3:25

A general way to view these formulae is as a Parseval identity. If your measure were of the form $$d\mu(x)=f(x)\,dx$$, where $$f\in L^2$$, then you would have, for any set $$E$$ of finite positive measure, $$\mu(E)=\int_{\mathbb{R}^n}\mathbf{1}_Ef=\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n} \hat{\mathbf{1}}_E\hat{\mathbf{f}}=\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n} \hat{\mathbf{1}}_E\varphi_X.$$ In the case $$E$$ is the unit ball, the Fourier transform is $$\hat{\mathbf{1}}_E(\xi)=(2\pi)^{\frac{n}{2}}\lVert \xi\rVert^{-n/2}J_{n/2}(\lVert\xi\rVert)$$, where $$J$$ is the Bessel function.

Of course, if $$\mu$$ is less regular, then the RHS does not need to converge absolutely. There are several ways to make sense of the formula nevertheless, namely:

• replace $$\mathbf{1}_E$$ with a test function, then pass to the limit;
• approximate $$\mu$$ with a more regular measure, then pass to a limit (as in the answer by Iosif Pinelis);
• replace the integral by its principal value, i. e., integrate over a box or ball of size $$T$$ centered about the origin and send $$T$$ to infinity.

The reason the last approach often works is because cutting out the high frequencies has a smoothening effect on both $$\mathbf{1}_E$$ and $$f$$, so it is essentially a combination of the first two. In particular, it always works when $$\mu$$ is supported away from the boundary of $$E$$; indeed, by writing $$\mu=\mu\star\delta$$ and exchangind integrals, it suffices to check the formula for delta-measures, in which case it is just the Fourier inversion formula at a point of smoothness of the original function.

If, on the other hand, $$\mu$$ is allowed to be supported on the boundary, then the v. p. integration may fail, but so does your formula for rectangles, as far as I can see. For example, in dimension 1, RHS is additive under the union $$(a,b)\cup(b,c)$$, while the LHS is not if $$\mu(\{b\})\neq 0$$.

$$\newcommand\ep\varepsilon\newcommand\de\delta\newcommand\De\Delta\newcommand\vp\varphi\newcommand\R{\mathbb R}$$For real $$\ep>0$$, let $$\mu_\ep$$ be the distribution of $$X+\ep Z$$, where $$Z$$ is a standard Gaussian random vector in $$\R^d$$ independent of $$X$$. Then the pdf $$p_\ep$$ of $$\mu_\ep$$ is given by $$p_\ep(x)=\frac1{(2\pi)^d}\int_{\R^d}dt\,e^{-it\cdot x-\ep^2|t|^2/2}\vp_X(t)$$ for $$x\in\R^d$$.

Note that $$X+\ep Z$$ converges to $$X$$ almost surely (as $$\ep\downarrow0$$), and hence $$\mu_\ep$$ converges to $$\mu$$ weakly.

So, for any ball $$B$$ in $$\R^d$$ with $$\mu(\partial B)=0$$, \begin{aligned} \mu(B)&=\lim_{\ep\downarrow0}\mu_\ep(B) \\ &=\lim_{\ep\downarrow0}\int_B dx\, p_\ep(x) \\ &=\nu(B):=\frac1{(2\pi)^d}\lim_{\ep\downarrow0}\int_B dx\, \int_{\R^d}dt\,e^{-it\cdot x-\ep^2|t|^2/2}\vp_X(t). \end{aligned}

Take now any open ball $$B$$ of some radius $$r\in(0,\infty)$$ and let $$B_{-\de}$$ denote the open ball concentric with $$B$$ of radius $$r-\de$$, where $$\de\in(0,r)$$. Let $$\De:=\{\de\in(0,r)\colon\mu(\partial(B_{-\de}))=0\}$$. Then $$\mu(B)=\lim_{\de\downarrow0}\mu(B_{-\de})=\lim_{\de\downarrow0,\,\de\in\De}\mu(B_{-\de}),$$ in view of the monotonicity of $$\mu(B_{-\de})$$ in $$\de$$ and because the set $$(0,r)\setminus\De$$ is at most countable. Therefore and in view of the monotonicity of $$\nu(B_{-\de})$$ in $$\de$$, \begin{aligned} \mu(B)&=\lim_{\de\downarrow0,\,\de\in\De}\nu(B_{-\de}) \\ &=\lim_{\de\downarrow0}\nu(B_{-\de}) \\ &=\frac1{(2\pi)^d}\lim_{\de\downarrow0}\lim_{\ep\downarrow0}\int_{B_{-\de}} dx\, \int_{\R^d}dt\,e^{-it\cdot x-\ep^2|t|^2/2}\vp_X(t). \end{aligned}

• Thank you, this is most useful!
– TOM
Mar 29, 2021 at 16:18
• There seems to be a problem with the formula (or maybe I do not see something trivial). Consider a random variable $X$ such that $\mathbb{P}(X=\pm 1)=1/2$. Then the probability that $X$ hits the interval $(-1,1)$ is 0, but the smoothened version of $X$ assigns probability roughly $1/2$ to this interval (with limit $1/2$ as $\varepsilon \rightarrow 0$). Am I missing something here?
– TOM
Mar 30, 2021 at 13:56
• @TOM : Thank you for your comment. This is now fixed. Mar 30, 2021 at 14:53
• In the paragraph startin So, for any ball $B$ in $\mathbb R^d$ with you probably mean $\mu(\partial B)=0$. Mar 31, 2021 at 5:06
• @JochenWengenroth : Yes, of course. :-) Thank you for your comment. Mar 31, 2021 at 6:27