Show that $E|X|^p$ is finite

Question

Show $\varphi(t) = \frac{1}{1 + |t|^{\beta}}$ is a characteristic function for $\beta \in (0,1]$. Let $X$ be r.v. with characteristic function $\varphi$. Prove that $E|X|^p < \infty$ for all $0 < p < \beta$.

I did the first part - Polya's criterion works since $\varphi(o) = 1$, $\varphi$ is even, its limit in infinity is 0 and for $t \in [0, \infty)$ we have

$$ \varphi^{(2)}(t) = \left(\frac{1}{1 + t^\beta}\right)'' = \cdots = \frac{\beta t^{\beta-2}(1+t^\beta)[2\beta t^\beta -(\beta - 1)(1+t^\beta) ]}{(1+t\beta)^4} \geq 0 \text { for all $\beta$ in $(0, 1]$} $$

How can I proceed with the second part? Any hints appreciated.

Answer 1

There are different ways to prove this; it depends a bit on which results you know. I will use the so-called truncation inequality:

Let $X$ be any real-valued random variable and denote by $\phi(\xi)=\mathbb{E}\exp(i \xi X)$ its characteristic function. Then $$\mathbb{P}(|X| \geq R) \leq 7R \int_0^{1/R} (1-\text{Re} \, \phi(\xi))\, d\xi.\tag{1}$$ for all $R>0$.

In your problem, we have $\phi(\xi) = \frac{1}{1+|\xi|^{\beta}}$ and so

$$1- \text{Re} \phi(\xi) = 1-\phi(\xi) = \frac{|\xi|^{\beta}}{1+|\xi|^{\beta}} \leq |\xi|^{\beta}.$$

By $(1)$, this implies

$$\mathbb{P}(|X| \geq R) \leq 7R \int_0^{1/R} |\xi|^{\beta} \, d\xi = \frac{7}{\beta} R^{-\beta}.\tag{2}$$

Since

$$\mathbb{E}(Y) = \int_0^{\infty} \mathbb{P}(Y \geq r) \, dr$$

for any random variable $Y \geq 0$, we have

\begin{align*} \mathbb{E}(|X|^p) &= \int_0^{\infty} \mathbb{P}(|X|^p \geq r) \, dr \\ &= p \int_0^{\infty} u^{p-1} \mathbb{P}(|X| \geq u) \, du. \end{align*}

Hence, by $(2)$,

$$\mathbb{E}(|X|^p) \leq p \int_0^{1} u^{p-1} \, du + \frac{7p}{\beta} \int_1^{\infty} u^{p-1} u^{-\beta} \, du.$$

The first integral on the right-hand side is finite for any $p>0$, the second integral is finite for $p<\beta$. Hence, $\mathbb{E}(|X|^p)<\infty$ for $p \in (0,\beta)$.