> [!definition] > > Let $\mathfrak{M}_\alpha(\real^d)$ be the space of $\sigma$-finite [[Measure Space|measures]] on $\real^d$ such that $M(\bracs{0}) = 0$, and > $ > \int_{\real^d}\min\bracs{1, \abs{x}^\alpha}dM(x) < \infty \quad M(\real^d \setminus B(0, r)) < \infty \quad \forall r > 0 > $ > which is equivalent to the following two conditions: > 1. $\int \abs{x}^\alpha dM(x) < \infty$. > 2. $M(\real^d \setminus B(0, r)) < \infty$ for all all $r > 0$. > > If $\alpha = \infty$, we simply look at $\sigma$-finite measures with $M(\bracs{0}) = 0$ and $M(\real^d \setminus B(0, r)) < \infty$ for all $r > 0$. > > If $M$ has [[Dense|density]] $\frac{1}{\abs{x}^p}$ with respect to the [[Lebesgue Measure|Lebesgue measure]], then $M \in \mathfrak{M}_\alpha$ only when $p \in (d, d + \alpha)$. > > The set $\mathfrak{M}_2(\real^d)$ is known as the space of **Lévy measures**. > [!theorem] > > Let $\gamma_{m, C}$ be the [[Normal Distribution|Gaussian distribution]] with mean $m$ and $C \ge 0$, and $M \in \mathfrak{M}_0$, define > $ > \pi^{(1)}_{m, C, M} = \gamma_{m, C} * \pi_{M} \in \mathcal I(\real^d) > $ > to be the [[Convolution of Functions|convolution]] of a [[Poisson Distribution|Poisson distribution]] and a Gaussian distribution, which is [[Infinitely Divisible Distribution|infinitely divisible]] and has characteristic function > $ > \wh \pi^{(1)}_{m, C, M}(\xi) = \exp\underbrace{\braks{i\angles{m, \xi} - \frac{\xi C \xi}{2} + \int\paren{e^{i\angles{x, \xi}} - 1}dM(x)}}_{\ell^{(1)}_{m, C, M}} > $ > If $M \in \mathfrak{M}_1$, then $\pi_{m, C, M}^{(1)}$ is also an infinitely divisible law. > > *Proof*. Let $M \in \mathfrak{M}_1$, then $M_r = M|_{\ol{B(0, r)}^c} \in \mathfrak{M}_0$ for all $r > 0$. In which case, > $ > \ell_{m, C, M_r} = i\angles{m, \xi} - \frac{\xi C \xi}{2} + \int\paren{e^{i\angles{x, \xi}} - 1}dM_r(x) > $ > To send $r \to 0$, expand the part inside via a Taylor expansion > $ > (e^{i\angles{x, \xi}} - 1) \cdot \one_{\ol{B(0, r)}}dM(x) \le \one_{\ol{B(0, r)}} \cdot 2 + \one_{\ol{B(0, r)}}\abs{x}\abs{\xi} \in L^1(M) > $ > By the dominated convergence theorem, the difference $\wh M_r - 1 \to \wh M - 1$. Therefore $\ell_{m, C, M_r} \to \ell_{M, C, M}$. By the dominated convergence theorem again, this function of $\xi$ is continuous at $0$. By [[Lévy's Continuity Theorem]], there exists a probability measure $\pi_{m, C, M}$ such that $\pi_{m, C, M_r} \Rightarrow \pi_{m, C, M}$ weakly. Hence the measure is infinitely divisible.