Letting $\varphi(t)$ be the characteristic function for the probability measure $\mu$, we know if $\int \left|\varphi(t)\right|dt < \infty$, then $\mu$ has density function

$$f(y) = \frac{1}{2\pi} \int e^{-ity} \varphi(t)dt.$$

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Example 3.3.8 states that what is denoted as "Polya's distribution", its pdf is $\frac{1-\cos x}{\pi x^2}$ and its characteristic function is $(1-\left|t\right|)^+.$

Now, there is a proof below. But I'm not certain what this is proving.

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$\frac{2(1-\cos t)}{t^2}$ is the characteristic function for the triangular distribution(whose pdf is given by $1-\left|x\right|$ for $x \in (-1,1)$), so by theorem 3.3.5, it seems that

$$\frac{1}{2\pi} \int \frac{2(1-\cos s)}{s^2}e^{-isy} ds = (1-\left|y\right|), $$ not $(1-\left|y\right|)^+$.

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I'm not even sure what the proof is proving.

I'd like to clear my misunderstanding.

  • $\begingroup$ As a quick check, recall the definition of the cf is $\phi_X(t)=E\left[e^{itX}\right],$ whence $$|\phi_X(t)| \le E\left[|e^{itX}\right|] = E[1] = 1.$$ Thus, a function like $1-|y|,$ which exceeds $1$ in magnitude whenever $|y|\gt 2,$ cannot possibly be a cf. $\endgroup$
    – whuber
    Jan 9 at 23:56

1 Answer 1


For $\mathrm{Tri}(a) $ distribution (support on $[-a, a]$), the characteristic function is $$\varphi_{\mathrm{Tri}(a)}(t) =2\left[\frac{1-\cos(at)}{a^2t^2}\right], $$ the density being

$$f(x;a)= \frac1a\left(1-\frac{|x|}{a}\right) ^+.$$

Durrett talks about $\mathrm{Tri}(1).$

Then as an application of Fourier Inversion formula, he focusses on "Polya's distribution" (owing to it being the simplest example satisfying Polya's criterion) whose density is $(1-\cos x) /\pi x^2.$

Then, by the Inversion formula

$$f(y;1) =(2\pi) ^{-1}\int \varphi_{\mathrm{Tri}(1)}(s)\exp({-isy})~\mathrm ds;\tag 1\label 1$$ however to yield characteristic function of Polya's distribution, we will use the definition of characteristic function - this is easily achieved in $\eqref 1$ by switching $s$ to $x$ and $y$ to $-t.$


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