# Density from characteristic function: Durrett example 3.3.8 and 3.3.9

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

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

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.

$$\frac{2(1-\cos t)}{t^2}$$ 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 $$(1-\left|y\right|)^+$$.

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

I'd like to clear my misunderstanding.

• 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.
– whuber
Jan 9 at 23:56

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 $$(1-\cos x) /\pi x^2.$$
$$f(y;1) =(2\pi) ^{-1}\int \varphi_{\mathrm{Tri}(1)}(s)\exp({-isy})~\mathrm ds;\tag 1\label 1$$ 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 $$-t.$$