Probability Theory – Prove That ?(t) is Not a Characteristic Function

Question

$\phi(t)=\sqrt{1-t^2} \ $ if $|t|<1$ and $\phi(t)=0$ if $|t| \geq 1.$
Prove that $\phi(t)$ is not a characteristic function.

$\phi(t)$ checks obvious signs of a characteristic function ($\phi(0)=1, \ |\phi(t)| \leq 1$).

$\phi(t)$ is an even real-valued function, so if it is a characteristic function, then its corresponding random variable $X$ has a distribution that is symmetric about zero.

Assuming $E|X|^k$ exists, we should have
$\phi^k(0) = i^k E X^k.$

$\phi'(0) = \frac{-2t}{2\sqrt{1-t^2}} |_{t=0}=0$ (since $X$ is assumed to be symmetric, odd moments of $X$ would be $0$, so it's correct).

$\phi''(0) = \frac{-\phi(t) + \phi ' (t)(t)}{\phi^2(t)} |_{t=0} = \frac{-1+0}{1} = i^2 EX^2 = -EX^2 \Longrightarrow EX^2 = 1$.

$\phi'''(0)$ is also $0$, and I can't calculate further. Wolfram says $\phi''''(0)=3$; I see no use for it.

Finding pdf from the "characteristic function" (Wolfram calculated the integral):

$p(x)=\frac{1}{2\pi} \int_{t=-1}^{t=1} \cos(tx) \sqrt{1-t^2} dt = \frac{\pi J_1(x)}{x},$ where $J_1(x)$ is "Bessel function of the first kind" (we haven't studied it).

Could someone please help me see how $\phi(t)$ is not a characteristic function?

Answer 1

No, $\phi(t)=\sqrt{(1-t^2)_+}$ is not a characteristic function since the function $\phi$ has a finite derivative of second order at $0$, but $\phi\notin\mathcal{C}^2(\mathbb{R})$.

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The argument is encapsulated in two known facts about the Fourier transform of measures (for probability measures, see Wikipedia here).

For a complex measure $\mu$ on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ define $$\widehat{\mu}(t)=\int_\mathbb{R} e^{itx}\,\mu(dx)$$

Theorem 1: If $\mu$ is a complex measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ ((real finite is included in this case) and $$\int_{\mathbb{R}}|x|^m|\mu|(dx)<\infty$$ for some $m\in\mathbb{N}$ ($|\mu|$ is the variation measure of $\mu$), then $\widehat{\mu}$ is $m$-th differentiable, $\widehat{\mu}^{(m)}$ is uniformly continuous, and $$\widehat{\mu}^{(m)}(t)=i^m \int_\mathbb{R} x^m e^{ixt}\,\mu(du)$$

Theorem 2: If $\mu$ is a finite positive measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$, and $\widehat{\mu}^{(n)}(0)$ exists and is finite for some $n\in 2\mathbb{N}$, then $\widehat{\mu}\in \mathcal{C}^{n}(\mathbb{R})$, $\int_{\mathbb{R}}|x|^n\mu(dx)<\infty$, and $$\widehat{\mu}^{(n)}(t)=i^n \int_\mathbb{R} x^n e^{ixt}\,\mu(du)$$

• Theorem 1 follows by dominated convergence an induction on the order of derivatives.
• Theorem 2 follows by dominated convergence, the identity $$-\frac{e^{ihx}-2+e^{-ixh}}{h^2}=\frac{2(1-\cos(hx))}{h^2}$$ and some induction arguments on half the order of the derivative, and by Theorem 1.

These results appear either as theorems or as exercises in many textbooks (see for example, Kallenberg, O. Foundations of modern Probability , Springer-Verlag, Berlin, 2nd Edition, 2002, pp. 101, or Klenke, A., Probability Theory: A comprehensive course, Universitext, Springer-Verlag, London 2008, pp. 312-315)

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