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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.

$\phi_1(t)=(1-|t|)_+$

$\phi_2(t)=\sum_{n=-\infty}^{\infty}\phi_1(t+2n\pi)$

$\phi_3(t)=(1-\frac{t^2}{2})e^{-t^2/2}$

Work/Thoughts

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

Assumptions:

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

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

$F(b)-F(a)=\lim_{n\to\infty}\frac1{2\pi}\int_{-\infty}^{\infty}\frac{e^{-ita}-e^{-itb}}{it}\phi(t)e^{-t^2/n}dt$

And I believe this simplifies to:

$\lim_{n\to\infty}\frac1{2\pi}\int_{-n}^{n}\frac{e^{-ita}-e^{-itb}}{it}\phi(t)$

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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Why this new question, identical to your previous one? –  Did Oct 24 '12 at 7:04
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This makes no sense. You could have corrected the typo on, and added your thoughts to, the previous question (and you did, before closing it). Note that the comments are lost as well. –  Did Oct 24 '12 at 15:06
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the other question's comments were just bickering back and forth... Surely you realize that two-thirds of these comments were mine? It seems beyond chutzpah to ask for people's help after calling their comments bickering. –  Did Oct 24 '12 at 22:15
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and here we go again. For the last time, I've told you before that I appreciate your help, but I don't see any value in this kind of exchange. You have an insatiable need to be right, so if that's what you need to hear, "you are right. Right about this comment section, right about anything I've ever disagreed on, and will be right in perpetuity." –  Justin Oct 25 '12 at 3:25
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Perfect. Then you might want to do what is right, and avoid duplicating/removing questions for no reason, which is contrary to the policy's site. –  Did Oct 25 '12 at 5:15
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1 Answer 1

up vote 1 down vote accepted
+100

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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