> [!theorem] > > Let $\gamma_{m, C}$ be the [[Normal Distribution|Gaussian distribution]] with mean $m$ and $C \ge 0$, and $M \in \mathfrak{M}_2$ be a [[Lévy Measure|Lévy measure]]. Define > $ > \ell_{m, C, M}(\xi) = i(m, \xi) - \frac{\xi C \xi}{2} + \int_{\real^d}\paren{e^{i\angles{x, \xi}} - 1 - i\angles{x, \xi} \cdot \one_{B(0, 1)}}dM(x) > $ > then $\pi_{m, C, M}$ with [[Characteristic Function|characteristic function]] $\exp(\ell_{m, C, M})$ is an [[Infinitely Divisible Distribution|infinitely divisible law]]. The measure $\pi_{m, C, M}$ is known as a **Lévy system**. > > *Proof.* Let $M \in M_2$, $r > 0$. Define $M_r = M|_{\ol{B(0, r)}}$ and > $ > \ell_{m, C, M_r}(\xi) = i(m, \xi) - \frac{\xi C \xi}{2} + \int_{\real^d}\paren{e^{i\angles{x, \xi}} - 1 - i\angles{x, \xi} \cdot \one_{B(0, 1)}}dM_r(x) > $ > Note that this allows factoring > $ > \begin{align*} > \ell_{m, C, M_r}(\xi) &= i\angles{\braks{m - \int_{B(0, 1)}xdM_r(x)}, \xi} - \frac{\xi C \xi}{2} \\ > &+ \int_{\real^d}\paren{e^{i\angles{x, \xi}} - 1 - i\angles{x, \xi}}dM_r(x) > \end{align*} > $ > where > $ > m_r = m - \int_{B(0, 1)}xdM_r(x) > $ > identifying a new mean for the Gaussian yields a similar form for the logarithm. If $m' = m - \int_{B(0, 1)} xdM(x)$, define > $ > \ell_{m', C, M}(\xi) = i\angles{m', \xi} - \frac{\xi C \xi}{2} + \int_{\real^d}\paren{e^{i\angles{x, \xi}} - 1 - i\angles{x, \xi}}dM(x) > $ > Similar to the extension to $M_1$, applying a Taylor expansion on the term inside the integral yields that $\ell_{n, C, M_r} \to \ell_{n, C, M}$ as $r \to 0$. > [!theorem] > > Let $(m, C, M)$ be a Lévy system, then for any $t > 0$, $\pi_{(tm, tC, tM)}$ is also a Lévy system with, where $\ell_{tm, tC, tM} = t\ell_{m, C, M}$. > [!theorem] Levy-Khintchine Formula (Task 4) > > Let $\mu \in \mathcal I(\real^d)$ be a [[Infinitely Divisible Distribution|infinitely divisible law]], then there exists a Lévy system $(m, C, M)$ such that $\mu = \pi_{m, C, M}$. > > Hint given: $\wh \mu_{1/n} = e^{\ell_{\wh \mu}/n}$. Therefore > $ > \begin{align*} > \ell_{\wh \mu} &= \limv{n}(e^{\ell_{\wh \mu}/n} - 1) = \limv{n}\frac{\wh \mu_{1/n} - \wh \delta_0}{1/n} > \end{align*} > $ > so > $ > \begin{align*} > \ell_{\wh \mu}(\xi) &= \limv{n}\frac{\mu_{1/n}(e^{i \angles{\cdot, \xi}}) - \delta_0(e^{i\angles{\cdot, \xi}})}{1/n} > \end{align*} > $ > This induces an operator on $C_c^\infty(\real^d)$, > $ > A(\phi) = \limv{n}\frac{\mu_{1/n}(\phi) - \phi(0)}{1/n} > $ > which turns out to be > $ > \begin{align*} > A\phi &= \sum_{j = 1}^d m_j\partial_j\varphi(0) + \frac{1}{2}\sum_{i, j = 1}^dC_{j}\partial_i\partial_j\varphi \\ > &+ \int_{\real^d}(\phi(x) - \phi(0) \cdot \one_{B(0, 1)}(x)\sum_{j = 1}^dx_j\partial_j\varphi(0))dM(x) > \end{align*} > $ > [!theorem] > > Let $M \in M_2$ be a Lévy measure, then > $ > \lim_{\abs{\xi} \to \infty}\frac{1}{\abs{\xi}^2}\int_{\real^d}\paren{e^{i\angles{x, \xi}} - 1 - \one_{B(0, 1)}(x) \cdot i\angles{x, \xi}}dM(x) = 0 > $ > In particular, if $\mu$ is an infinitely divisible law with canonical representation $(m, C, M)$, then > $ > (\xi C \xi) = -2\lim_{t \to \infty}\frac{\ell_{\wh \mu}(t\xi)}{t^2} > $ > *Proof*. First suppose that the limit holds, then > $ > \begin{align*} > \ell_{\wh \mu}(t\xi) &= \frac{i\angles{m, \xi}}{t^2}- \frac{t^2}{2}(\xi C\xi)+ \int_{\real^d}\paren{e^{i\angles{x, t\xi}} - 1 - \one_{B(0, 1)}(x) \cdot i\angles{x, t\xi}}dM(x) \\ > \frac{\ell_{\wh \mu}(t\xi)}{t^2} &= \frac{i\angles{m, \xi}}{t^2}- \frac{(\xi C\xi)}{2}+ \frac{1}{t^2}\int_{\real^d}\paren{e^{i\angles{x, t\xi}} - 1 - \one_{B(0, 1)}(x) \cdot i\angles{x, t\xi}}dM(x) \\ > \end{align*} > $ > Sending $t \to \infty$ and applying the limit yields the desired result. Now for the limit, let $r \in (0, 1)$, then the integral term can be decomposed into two parts. For the part near the origin, the integral is bounded by $c_1\abs{\xi}^2\int_{B(0, r)}\abs{x}^2dM(x)$ via a Taylor expansion. For the second part, the integrand is bounded by > $ > c_2 \int_{B(0, r)^c}(2 + \abs{\xi})dM(x) \le c_2(2 + \abs{\xi}) \cdot M(B(0, r)^c) > $ > This bounds the limit to be $\int_{B(0, r)} \abs{x}^2 dM(x)$. Sending $r \to 0$ yields the desired result.