本章导读 本章引入 Part I 的最后一类过程——对称稳定过程。§12.1 定义（Def 12.1：特征指数 \(\Psi_{X_t}(s)=-|Cs|^\alpha\) (12.1)，\(0<\alpha\le2\)；Rmk 12.1：布朗运动是 \(\alpha=2\) 的特例，取 \(\mu=0\)、\(C=\sigma\sqrt{t/2}\)）。§12.2 性质（Prop 12.1：若 \(X_1,\dots,X_n\) 独立且都具参数 \(\alpha\) 的对称稳定特征指数，则 \(Y=(X_1+\dots+X_n)/n^{1/\alpha}\) 也是参数 \(\alpha\) 的对称稳定，体现"稳定性"——归一化求和后分布形态不变）。无图。

定义 12.1（对称稳定过程）/ Definition 12.1 (Symmetric stable process) 设 \(\{X_t\}\) 是定义 11.6 中的 Lévy 过程。称 \(\{X_t\}\) 为对称稳定过程，若其特征指数 \(\Psi_{X_t}(s)\) 满足Let \(\{X_t\}\) be a Lévy process as in Definition 11.6. \(\{X_t\}\) is called a symmetric stable process if its characteristic exponent \(\Psi_{X_t}(s)\) satisfies

$$\Psi_{X_t}(s)=-|Cs|^\alpha\tag{12.1}$$

其中 \(0<\alpha\le2\)，\(C\) 为某常数。for \(0<\alpha\le2\) and some constant \(C\).

注 12.1（布朗运动是 α=2 的特例）/ Remark 12.1 由命题 11.6，布朗运动的特征指数为：对满足 \(dW_t=\mu dt+\sigma dB_t\) 的布朗运动 \(\{W_t\}\)（\(\{B_t\}\) 标准 BM），\(\Psi_{W_t}(s)=i\mu ts-\dfrac{\sigma^2 t}2 s^2\)。故布朗运动是 \(\alpha=2\) 的对称稳定过程特例——只需令 \(\mu=0\) 且 \(C=\sigma\sqrt{\tfrac t2}\)。By Proposition 11.6, the characteristic exponent of Brownian motion is: for a Brownian motion \(\{W_t\}\) with \(dW_t=\mu dt+\sigma dB_t\) (\(\{B_t\}\) standard BM), \(\Psi_{W_t}(s)=i\mu ts-\dfrac{\sigma^2 t}2 s^2\). So Brownian motion is a special case of the symmetric stable process with \(\alpha=2\) — simply set \(\mu=0\) and \(C=\sigma\sqrt{\tfrac t2}\).

命题 12.1（稳定性）/ Proposition 12.1 (Stability) 若 \(X_1,\dots,X_n\) 是独立随机变量，且都具参数 \(0<\alpha\le2\) 的对称稳定特征指数，即 \(\Psi_{X_j}(s)=-|Cs|^\alpha\)（\(j=1,\dots,n\)），则 \(Y\equiv\dfrac{X_1+\dots+X_n}{n^{1/\alpha}}\) 也是参数 \(0<\alpha\le2\) 的对称稳定，即 \(\Psi_Y(s)=-|Cs|^\alpha\)。If \(X_1,\dots,X_n\) are independent random variables that all have the symmetric stable characteristic exponent with parameter \(0<\alpha\le2\), i.e. \(\Psi_{X_j}(s)=-|Cs|^\alpha\) (\(j=1,\dots,n\)), then \(Y\equiv\dfrac{X_1+\dots+X_n}{n^{1/\alpha}}\) is also symmetric stable with parameter \(0<\alpha\le2\), i.e. \(\Psi_Y(s)=-|Cs|^\alpha\).

命题 12.1 证明 / Proof of Proposition 12.1 直接计算：Straightforward:

$$\begin{aligned}e^{\Psi_Y(s)}=\mathbb E\!\left[e^{isY}\right]&=\mathbb E\!\left[e^{is\frac{X_1+\dots+X_n}{n^{1/\alpha}}}\right]\overset{\text{indep}}{=}\left(\mathbb E\!\left[e^{is\frac{X_j}{n^{1/\alpha}}}\right]\right)^n=\left(e^{\Psi_{X_j}\left(\frac s{n^{1/\alpha}}\right)}\right)^n\\&=\left(e^{-\left|C\frac s{n^{1/\alpha}}\right|^\alpha}\right)^n=\left(e^{-\frac{|Cs|^\alpha}n}\right)^n=e^{-|Cs|^\alpha}=e^{\Psi_{X_j}(s)},\end{aligned}$$

故 \(\Psi_Y(s)=\Psi_{X_j}(s)=-|Cs|^\alpha\)。\(\blacksquare\)so \(\Psi_Y(s)=\Psi_{X_j}(s)=-|Cs|^\alpha\). \(\blacksquare\)