Why this expression is a characteristic function? (Infinitely Distributions)

Question

I'm trying to prove that the next formula is a characteristic function of an infinitely divisible distribution:

$$\phi(t)=\exp\left[i\gamma+\int_{-\infty}^{\infty}\left(e^{itx}-1-\frac{itx}{1+x^{2}}\right)\frac{1+x^2}{x^2}\nu(dx)\right],$$ where $\nu$ is a finite measure, and the integrand is equal to $-t^2/2$ at the origin.

As a suggestion to prove this is the next:

If $\mu_{n}$ has density $I_{[-n,n]}(1+x^2)$ with respect to $\nu$ then $$\exp\left(i\gamma t + it\int_{-\infty}^{\infty}\frac{x}{1+x^2}\mu_{n}(dx)+\int_{-\infty}^{\infty}(e^{itx}-1-itx)\frac{1}{x^2}\mu_{n}(dx)\right)$$ is a characteristic function.

Why that expression is a characteristic function? I can't see it.

Any kind of help is thanked in advanced.

Answer 1

Setting $\sigma^2 := \nu(\{0\})$ we can write

$$\phi(t) = \exp \left[ i \gamma t - \sigma^2 \frac{t^2}{2} + \int_{x \neq 0} \left( e^{itx}-1- \frac{itx}{1+x^2} \right) \frac{1+x^2}{x^2} \, \nu(dx) \right]. \tag{1}$$

Once we know that $\phi$ is a characteristic function for any $\gamma \in \mathbb{R}$ and any finite measure $\nu$ it follows easily that the associated distribution is infinitely divisible; just note that

$$\psi(t) := \exp \left[ i \frac{\gamma}{n} t - \frac{\sigma^2}{n} \frac{t^2}{2} + \int_{x \neq 0} \left( e^{itx}-1- \frac{itx}{1+x^2} \right) \frac{1+x^2}{x^2} \, \frac{\nu(dx)}{n} \right]. $$

is a characteristic function satisfying $\phi(t) = \psi(t)^n$ for all $t \in \mathbb{R}$.

To prove that $(1)$ is indeed a characteristic function we can use an approximation argument. For fixed $n \in \mathbb{N}$ we set

$$\mu_n := 1_{\{1/n \leq |x| \leq n\}} \frac{1+x^2}{x^2} \, \nu(dx)$$

and

$$\lambda_n := \int_{1/n \leq |x| \leq n} \frac{1+x^2}{x^2} \, \nu(dx).$$

The next step is to show that

$$\phi_n(t) := \exp \bigg[ it \underbrace{\left( \gamma - \int_{x \neq 0} \frac{x}{1+x^2} \,\mu_n(dx) \right)}_{=:\gamma_n}- \sigma^2 \frac{t^2}{2} + \int_{x \neq 0} (e^{itx}-1) \, \mu_n(dx) \bigg] \tag{2}$$

is a characteristic function. Since $\mu_n/\lambda_n$ is a probability measure, there exists a sequence of independent random variables $(X_i)_{i \in \mathbb{N}}$ such that each $X_i$ has distribution $\mu_n/\lambda_n$ (recall that $n$ is fixed). Let $N$ be a Poisson distributed random variable with parameter $\lambda_n$ and let $Y \sim N(\gamma_n,\sigma^2)$ be such that $N$, $Y$ and $(X_i)_{i \in \mathbb{N}}$ are independent. A straightforward computation shows that the characteristic function of the random variable

$$Z := Y + \sum_{i=1}^N X_i$$

equals $\phi_n$ defined in $(2)$.

Now note that we can rewrite $\phi_n$ in the following way:

$$\begin{align*} \phi_n(t) &= \exp \bigg[it \gamma - \sigma^2 \frac{t^2}{2} + \int_{x \neq 0} \left( e^{itx}-1-\frac{itx}{1+x^2} \right) \mu_n(dx) \bigg] \\ &= \exp \bigg[it \gamma - \sigma^2 \frac{t^2}{2} + \int_{1/n \leq |x| \leq n} \left( e^{itx}-1-\frac{itx}{1+x^2} \right) \frac{1+x^2}{x^2} \, \nu(dx) \bigg]. \end{align*}$$

This implies $\phi_n(t) \to \phi(t)$ as $n \to \infty$ for all $t \in \mathbb{R}$. Since $\phi$ is continuous at $t=0$, the pointwise convergence $\phi_n \to \phi$ entails that $\phi$ is a characteristic function.

Answer 2

Define finite Borel measures $\mu_n, n \geq 1$ by

$$\mu_n(A) = \int_A I_{[-n,n]}(x) (1+x^2) \nu (dx), \quad A \in \mathcal{R}$$

Then, define functions $\phi_n: \mathbb{R} \to \mathbb{C}, n \geq 1$ by

$$\phi_n(t) = \exp\left(i \gamma t + it \int_\mathbb{R} \frac{x}{1+x^2} \mu_n(dx) + \int_\mathbb{R} \frac{e^{itx}-1-itx}{x^2}\mu_n(dx)\right)$$ $$= \exp\left(i \gamma t + \int_{[-n,n]}\left(e^{itx}-1-\frac{itx}{1+x^2}\right)\frac{1+x^2}{x^2} \nu (dx)\right)$$

Clearly we have pointwise $\phi_n \to \phi$ and $\phi_n$ is the characteristic function of a random variable $\gamma_n + X_n$ where $$\gamma_n:= \gamma + \int_\mathbb{R} \frac{x}{1+x^2} \mu_n(dx)$$ and $X_n$ has characteristic function $$t \mapsto \exp\left(\int_\mathbb{R} \frac{e^{itx}-1-itx}{x^2}\mu_n(dx)\right)$$ See Billingsley, probability and measure, third edition, theorem 28.1, p372 for a proof that this is a characteristic function of some probability distribution.

Since $\phi$ is continuous at $0$ (invoke the dominated convergence theorem to prove this), we can conclude that $\phi$ is a characteristic function and as in the answer of @saz, it follows that it is the characteristic function of an infinitely divisible distribution.