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Consider the function $\phi$ defined by $$\phi(t) = \left\{\begin{array}{ccc}t^2+1-2|t|& \rm for &|t|\le 1- \frac{1}{\sqrt{2}}\\ \frac{1}{2}&\rm for& |t|> 1- \frac{1}{\sqrt{2}}\end{array}\right. $$ Is $\phi$ a characteristic function?

I assume it is not, as it do not have a derivative at $t=1- \frac{1}{\sqrt{2}}$, but how to prove that?

Did
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llama
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  • @LordSharktheUnknown that's not right, look at characteristic function of random variable with Dirac distribution – llama May 06 '17 at 07:00

1 Answers1

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If we define

$$\varphi_1(t) := \begin{cases} 2t^2-4 |t| +1, & |t| \leq 1- \frac{1}{\sqrt{2}} \\ 0, & |t| > 1- \frac{1}{\sqrt{2}}\end{cases},$$

then $\varphi_1$ is an even continuous real-valued function which satisfies $$\varphi_1(0)=1 \quad \text{and} \quad \lim_{|t| \to \infty} \varphi_1(t)=0.$$ Moreover, $\varphi_1|_{(0,\infty)}$ is convex. It follows from Polya's theorem that $\varphi_1$ is a characteristic function. On the other hand,

$$\varphi_2(t) := 1$$

is clearly a characteristic function. This implies that

$$\varphi(t) = \frac{1}{2} \varphi_1(t) + \frac{1}{2} \varphi_2(t)$$

is a characteristic function.

saz
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  • @saz can you please explain with more details, why $\varphi_1$ is characteristic? – llama May 06 '17 at 08:16
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    @Ilama well, as I explained in my answer it follows for instance from Polya's theorem. if you don't know it, then it would be a great idea to look it up, right? – saz May 06 '17 at 08:17