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I am struggling with this concept (self-study). Could someone show me how to explicitly apply the inversion formula for these examples? I am working through about 15 examples, but these 3 seemed sufficiently different to help me do the rest.





The only reference to the inversion formula that I have found is the following theorem:


1-$\phi$ is a characteristic function of a given probability distribution $F$

2- $F$ has continuity points $a, b$ with $a<b$.


And I believe this simplifies to:


From here I am not sure what to do. Thanks for any help.

more thoughts

I have read about two kinds of invertible CFs- those that are integrable, and those that are periodic. $\phi_2(t)$ is obviously of the periodic nature.

I also understand the following properties about characteristic functions:

If $F$ and $G$ are probability distributions and $G$ is absolutely continuous, then $F*G$ has density $\int_{-\infty}^{\infty}g(u-x)F(dx)$

This this helpful at all, perhaps for number 3?

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1 Answer 1

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In Durrett's book's, we have the following inversion formula: $$\lim_{T\to +\infty}(2\pi)^{-1}\int_{-T}^T\frac{e^{-ita}-e^{-itb}}{it}\varphi(t)dt=\mu(a,b)+\frac 12\mu(\{a,b\}).$$ Let $\mu_i$ the measure associated with $\varphi_i$.

  1. As $\varphi_1$ is integrable, we have that $\mu_1$ has density $$f(y)=\frac 1{2\pi}\int_{\Bbb R}e^{-ity}(1-|t|)^+dt=\frac 1{2\pi}\int_{-1}^1e^{-ity}(1-|t|)dt.$$ We will find Polya's distribution.

  2. As $\varphi_2$ is not integrable, we have to use the classical inversion formula. We have \begin{align} \mu_2(a,b)+\frac 12\mu_2(\{a,b\})&=\lim_{T\to +\infty}(2\pi)^{-1}\int_{-T}^T\frac{e^{-ita}-e^{-itb}}{it}\varphi_2(t)dt\\ &=\lim_{n\to +\infty}(2\pi)^{-1}\int_{-(2n+1)\pi}^{(2n+1)\pi}\frac{e^{-ita}-e^{-itb}}{it}\varphi_2(t)dt\\ &=\lim_{n\to +\infty}(2\pi)^{-1}\sum_{j=-n}^{n}\int_{(2j-1)\pi}^{(2j+1)\pi}\frac{e^{-ita}-e^{-itb}}{it}\varphi_2(t)dt\\ &=\lim_{n\to +\infty}(2\pi)^{-1}\sum_{j=-n}^{n}\int_{-\pi}^\pi\frac{e^{-i(t+2j\pi)a}-e^{-i(t+2j\pi)b}}{it}(1-|t|)^+dt\\ &=\lim_{n\to +\infty}(2\pi)^{-1}\sum_{j=-n}^{n}\int_0^1\frac{e^{-i2j\pi b}\sin(tb)-e^{-i2j\pi a}\sin(ta)}{it}(1-t)dt. \end{align}

  3. As $\varphi_3$ is integrable, we have $$ f_3(y)=\frac 1{2\pi}\int_{\Bbb R}e^{-ity}\left(1-\frac{t^2}2\right)\exp(-t^2/2)dt,$$ where $f_3$ is a density of the measure with Fourier transform $\varphi_3$. The integral $$\int_{\Bbb R}e^{-ity}\exp(-t^2/2)dt$$ is classical; the other term can be found by integration by parts.
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