求证：该表达式为何是无限可分分布的特征函数？

Hey there, let's walk through proving that this expression is the characteristic function of an infinitely divisible distribution. We'll lean on the Lévy-Khintchine representation (the gold standard for infinite divisibility) and use the truncated measure hint to build our argument step by step.

We need to show the following is the 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 on $\mathbb{R}$
• The integrand is defined to be $-t^2/2$ at $x=0$ (to ensure continuity there)

The hint tells us to consider truncated measures $\mu_n$ with density $I_{[-n,n]}(x)\cdot(1+x^2)$ with respect to $\nu$—we'll expand on that below.

1. Construct Truncated Lévy Measures #

First, define a sequence of truncated measures for each $n \in \mathbb{N}$:
$$\mu_n(dx) = I_{[-n,n]}(x) \cdot \frac{1+x^2}{x2}\nu(dx)$$
This measure is finite for every $n$: since $\nu$ is finite, and $\frac{1+x^2}{x2}$ is bounded on $[-n,n]\setminus{0}$ (and $\nu$ assigns 0 mass to ${0}$, since the integrand is well-defined there).

2. Write Truncated Characteristic Functions #

For each $n$, we can write a characteristic function in the standard Lévy-Khintchine form using $\mu_n$:
$$\phi_n(t) = \exp\left[i\left(\gamma + \int_{[-n,n]}x\nu(dx)\right)t + \int_{[-n,n]}\left(e^{itx}-1-itx\right)\mu_n(dx)\right]$$
This is clearly the characteristic function of an infinitely divisible distribution: it corresponds to a sum of independent Poisson-type random variables (governed by $\mu_n$) plus a linear drift term. Every such $\phi_n(t)$ is infinitely divisible by definition of the Lévy-Khintchine representation.

3. Take the Limit as $n \to \infty$ #

We need to show $\phi_n(t) \to \phi(t)$ for all $t \in \mathbb{R}$, and that the limit is a valid characteristic function.

First, rewrite the integrand in $\phi(t)$ to match the truncated form:
$$\left(e^{{itx}-1-\frac{itx}{1+x}2}\right)\frac{1+x^2}{x2} = \left(e^{{itx}-1-itx\right)\frac{1+x}2}{x^2} + itx$$
Substitute this into $\phi(t)$'s exponent:
$$i\gamma + \int_{-\infty}^\infty \left[\left(e^{{itx}-1-itx\right)\frac{1+x}2}{x^2} + itx\right]\nu(dx) = i\left(\gamma + \int_{-\infty}^\infty x\nu(dx)\right)t + \int_{-\infty}^\infty \left(e^{{itx}-1-itx\right)\frac{1+x}2}{x^2}\nu(dx)$$

Now, apply the Lévy Continuity Theorem:

• $\phi_n(t)$ converges pointwise to $\phi(t)$: the truncated integrals converge to the full integrals because $\nu$ is finite, and the integrand is bounded by a $\nu$-integrable function (for $|x| > 1$, the integrand is bounded by $4 + |t|/|x|$; for $|x| < 1$, it's bounded by $t^2/2 + C|x|$ for some constant $C$).
• $\phi(0) = \exp[i\gamma + 0] = 1$, so the limit is continuous at $t=0$.

By the theorem, $\phi(t)$ is a characteristic function. Additionally, since the limit of infinitely divisible characteristic functions (that is itself a characteristic function) is infinitely divisible, $\phi(t)$ corresponds to an infinitely divisible distribution.

The core reason ties directly to the Lévy-Khintchine Representation Theorem, which states:

A function $\phi(t)$ is the characteristic function of an infinitely divisible distribution if and only if it can be written as:
$$\phi(t) = \exp\left[i\alpha t - \frac{1}{2}\sigma^2 t^2 + \int_{-\infty}^\infty \left(e^{itx} - 1 - itxI_{|x|<1}\right)\tilde{\nu}(dx)\right]$$
where $\alpha \in \mathbb{R}$, $\sigma^2 \geq 0$, and $\tilde{\nu}$ is a $\sigma$-finite measure on $\mathbb{R}\setminus{0}$ satisfying $\int_{-\infty}^\infty \min(1,x^2)\tilde{\nu}(dx) < \infty$.

Our expression fits this framework perfectly:

• $\sigma^2 = 0$ (no Gaussian component)
• The drift term $\alpha = \gamma + \int_{-\infty}^\infty x\nu(dx)$
• The Lévy measure $\tilde{\nu}(dx) = \frac{1+x^2}{x2}\nu(dx)$, which satisfies the integrability condition:
  • For $|x| \geq 1$, $\min(1,x^2)=1$, so $\int_{|x|\geq1}\tilde{\nu}(dx) \leq 2\nu(\mathbb{R}) < \infty$
  • For $|x| < 1$, $\min(1,x^2)=x2$, so $\int_{|x|<1}\min(1,x^2)\tilde{\nu}(dx) = \int_{|x|<1}(1+x^2)\nu(dx) \leq 2\nu(\mathbb{R}) < \infty$

Since all conditions of the theorem are satisfied, $\phi(t)$ must be the characteristic function of an infinitely divisible distribution.

内容的提问来源于stack exchange，提问作者Squird37