the meaning of bound of characteristic function in the neighborhood of zero

Question

Let $\{X_n: n=1,2,\ldots\}$ be a sequence of integrable random variables. Let $\{\phi_n: n=1,2,\ldots\}$ be the corresponding characteristic functions. Suppose that we have $$ |1-\phi_n(t)|\leq A |t|^{1+\delta} \quad \text{for all } t \in R $$ for some real values $A$ and $\delta>0$ that do not depend on $n$. Is the family of random variables $\{X_n: n=1,2,\ldots\}$ dominated by an integrable random variable?

Answer 1

By Lévy's truncation inequality,

$$\mathbb{P}(|X_n| \geq R) \leq 7R \int_0^{\frac{1}{R}} |\text{Re}(1-\phi_n(t))| \, dt.$$

Using the given estimate for the characteristic functions, we find

$$\mathbb{P}(|X_n| \geq R) \leq c \cdot \frac{1}{R^{1+\delta}}$$

for some constant $c=c(A,\delta)$. Since

$$\mathbb{E}(|X_n|^p) = \int_{0}^{\infty} \mathbb{P}(|X_n| \geq r^{1/p}) \, dr$$

for any $p \geq 1$, we get

$$\mathbb{E}(|X_n|^p) \leq 1+ \int_1^{\infty} \frac{1}{r^{(1+\delta)/p}} \, dr<\infty$$

if we choose $p>1$ sufficiently small. This shows

$$\sup_{n \in \mathbb{N}} \mathbb{E}(|X_n|^p)< \infty;$$

in particular the family of random variables is uniformly integrable.

Note, however, that we can in general not expect that there exists $Y \in L^1$ such that $|X_n| \leq Y$ for all $n \in \mathbb{N}$. To see this, consider a sequence of independent identically distributed (e.g. Gaussian with mean $0$) random variables.