> [!definition] > > Let $\mu$ be a [[Probability|probability]] measure on $\real^d$. $\mu$ is **infinitely divisible** if for every $n \in \nat$, there exists a probability measure $\nu_n = \mu_{1/n}$ such that > $ > \mu = \nu_n^{*n} = \underbrace{\nu_{n} * \cdots * \nu_n}_{n \text{ times}} > $ > Equivalently, if for every $n \in \nat$, there exists a probability measure $\nu_n$ such that $\wh \mu = \wh \nu_n^{n}$. > > $\mathcal{I}(\real^d)$ is the family of all infinitely divisible distributions on $\real^d$. > [!theorem] > > Let $\mu, \nu \in \mathcal I(\real^d)$ and $n \in \nat$, then > - $\mu * \nu \in \mathcal I(\real^d)$ and infinitely divisible distributions are closed under [[Convolution of Measures|convolution]], and $(\mu * \nu)_{1/n} = \mu_{1/n} * \nu_{1/n}$. > - If $\seq{\mu_k} \subset \mathcal I(\real^d)$ is a family of probability measures such that $\mu_k \Rightarrow \mu$ [[Weak Convergence of Measures|weakly]] and $(\mu_{k})_{1/n} \Rightarrow \nu_n$ weakly, then $\mu \in \mathcal I(\real^d)$ with $\mu_{1/n} = \nu_n$. > [!theorem] > > If $C \ge 0$, then $\gamma_{m, C} \in \mathcal I(\real^d)$. > [!theorem] > > Let $m \in \real^d$ and $C$ be a positive definite matrix, then there exists a family of finite [[Measure Space|measures]] $\seq{M_n}$ such that $\pi_{M_n} \Rightarrow \gamma_{m, C}$ weakly. > > *Proof*. Suppose that $p = 1$ and let $\sigma^2$ be the variance. Let > $ > M_n = 2n \cdot \braks{\frac{1}{2}\delta_{m/n} + \frac{\delta_{-\sigma/\sqrt{n}} + \delta_{\sigma/\sqrt{n}}}{4}} > $ > with half of the mass at the centre and the other half distributed one standard deviation away. The [[Characteristic Function|characteristic function]] > $ > \begin{align*} > \pi_{M_n}(\xi) &= \exp\braks{\int \paren{e^{ix\xi} - 1} dM_n(x)} \\ > &= \exp \braks{ne^{i\xi m/n} - 1} \cdot \exp\braks{\frac{n}{2}\paren{e^{i \xi \sigma/\sqrt{n}} - 1}}\\ > &\cdot \exp\braks{\frac{n}{2}\paren{e^{-i \xi \sigma/\sqrt{n}} - 1}} \\ > &= \exp \braks{ne^{i\xi m/n} - 1} \cdot \exp\braks{n(\cos(\xi \sigma /\sqrt{n}) - 1)} \\ > &\to \exp(i\xi m) \cdot \exp (-\xi^2 \sigma^2/2) > \end{align*} > $ > converges to $\wh \gamma_{m, \sigma^2}(\xi)$. > [!theorem] > > Let $r, T > 0$ with $0 < r < T$, then there exists $N \in \nat$ such that for any probability measure $\mu$ where > 1. $\mu$ is infinitely divisible. > 2. $\abs{1 - \wh \mu(\xi)} < 1/2$ for all $\xi \in \ol{B(0, r)}$. > > then $\inf_{\xi \in \ol{B(0, T)}}\abs{\wh \mu(\xi)} > 2^{-N}$. As a result, > 1. $\wh \mu$ and each $\wh \mu_{1/n}$ cannot vanish. > 2. $\wh \mu$ and each $\wh \mu_{1/n}$ admits a unique, continuous principal logarithm on $\complex$. > 3. $\mu_{1/n}$ is unique and is infinitely divisible. > > *Proof*. Let $N \in \nat$ and $\mu_{1/N}$ be such that $\mu = \mu_{1/N}^{*N}$ Since $\wh \mu \ne 0$ everywhere on $\ol{B(0, r)}$, $\wh \mu_{1/N} \ne 0$ everywhere on $\ol{B(0, r)}$. Let $\ell_{\wh \mu}$ and $\ell_{\wh \mu_{1/N}}$ be their [[Principal Logarithm|principal logarithms]] on $\ol{B(0, r)}$, then $\ell_{\wh \mu_{1/N}} = \frac{1}{N}\ell_{\wh \mu}$. > > Given that $\abs{1 - \wh \mu(\xi)} < 1/2$ for all $\xi \in \ol{B(0, r)}$, we can identify $\ell_{\wh \mu_{1/N}} = \log \circ \wh \mu_{1/N}$ and $\ell_{\mu} = \log \circ \wh \mu_{1/N}$, giving the bounds $\abs{\ell_{\wh \mu}} \le 2$ and $\abs{\ell_{\wh \mu_{1/N}}} \le 2/N$. Since $\wh \mu_{1/n}(\xi) \le 1$, the real part of its principal logarithm has to be non-positive, allowing the bound > $ > \abs{1 - \wh \mu_{1/N}(\xi)} \le \abs{1 - e^{\ell_{\wh \mu_{1/N}}(\xi)}} \le \abs{\ell_{\wh \mu_{1/N}}(\xi)} \le \frac{2}{N} > $ > on $\ol{B(0, r)}$. > > Let $R > 0$ and $x$ be a unit vector, then > $ > \mu(\bracs{y \in \real^d: \abs{\angles{x, y}} > R}) \le \braks{\frac{1}{r}\int_0^r \abs{1 - \hat \mu(sx)}ds} \cdot \frac{1}{m(rR)} \le \frac{2}{Nm(rR)} > $ > Let $\tilde r \le T$, then > $ > \begin{align*} > \abs{1 - \wh \mu_{1/N}(\tilde r x)} &\le \tilde r R + 2\mu(\bracs{y \in \real^d: \abs{\angles{x, y}} > R}) \\ > &\le TR + \frac{4}{N m(rR)} > \end{align*} > $ > If $TR \le 1/4$ and $\frac{4}{N m(rR)} \le 1/4$, then on $\ol{B(0, T)}$, > $ > \abs{1 - \wh \mu_{1/N}(\xi)} \le \frac{1}{2} \Rightarrow \abs{\wh \mu(\xi)} \ge 2^{-N} > $ > [!theorem] > > Let $\seq{\mu_k}$ be a family of infinitely divisible distributions such that $\mu_k \Rightarrow \mu$, then $\mu$ is also infinitely divisible. > > *Proof*. Let $r_0 > 0$ such that $\abs{1 - \wh\mu} > 1/2$ on $\ol{B(0, r_0)}$. Since $\mu_k \Rightarrow \mu$, $\wh \mu_k \Rightarrow \wh\mu$ uniformly on compact sets. Let $K \in \nat$ such that $\abs{1 - \wh \mu_k} > 1/2$ on $\ol{B(0, r_0)}$ for all $k \ge K$. By choosing a minimum, let $r > 0$ such that $\abs{1 - \wh \mu_k} > 1/2$ on $\ol{B(0, r)}$ for all $k \in \nat$. > > Let $T > r > 0$, then there exists $N \in \nat$ such that $\abs{1 - \wh \mu_k} > 1/2^N$ on $\ol{B(0, T)}$ for all $k$. Therefore $\wh \mu$ never vanishes, and it admits a principal logarithm $\ell$. Let $\ell_k$ be the principal logarithm of $\wh \mu_k$, then $\ell_k \to \ell$ uniformly on compact sets. > > Let $n \in \nat$, then > $ > (\wh \mu_k)_{1/n} = e^{\frac{1}{n}\ell_k} \to e^{\frac{1}{n}\ell} > $ > uniformly on a neighbourhood of $0$. By [[Lévy's Continuity Theorem]], there exists a probability measure corresponding to $e^{\ell/n}$, which will be $\mu_{1/n}$. > [!theorem] > > The weak closure of compound Poisson distributions is the set of all infinitely divisible laws. > > Let $\mu$ be an infinitely divisible law. Let $M_n$ such that for each $B \in \cb(\real^d)$, $M_n(B) = n\mu_{1/n}(B \setminus \bracs{0})$. Then > $ > \begin{align*} > \wh \pi_{M_n}(\xi) &= \exp\braks{\int e^{i\angles{x, \xi}} - 1 dM_n(x)} \\ > &= \exp\braks{n\int e^{i\angles{x, \xi}} - 1 d\mu_{1/n}(x)} \\ > &= \exp\braks{n(\wh \mu_{1/n}(\xi) - 1)} \\ > &= \exp\braks{n \paren{e^{\frac{1}{n}\ell_{\wh \mu}(\xi)} - 1}} \\ > &= \exp \ell_{\wh \mu}(\xi) = \wh \mu(\xi) > \end{align*} > $