Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 1 down vote accepted

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.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.