本章导读 本章放松布朗运动定义中的连续性约束，研究允许跳跃的更一般过程。§11.1 预备：11.1.1 泊松/指数分布（Def 11.1 指数分布、Prop 11.1 期望方差 CDF；Def 11.2 泊松分布、Prop 11.2 \(\mathbb E[N]=\mathrm{Var}(N)=\lambda\)）；11.1.2 特征函数（Def 11.3 \(\varphi_X(s)=\mathbb E[e^{isX}]\)、Prop 11.3 \(\varphi_X^{(k)}(0)=i^k\mathbb E[X^k]\)）；11.1.3 特征指数（Def 11.4 \(\Psi(s)=\ln\varphi_X\)、Prop 11.4–11.7：可加性、缩放、正态、泊松）；11.1.4 无穷可分分布（Def 11.5）。§11.2 Lévy 过程（Def 11.6 去掉 BM 的连续性与原点约束；Fact 11.1 Lévy⟹无穷可分；Rmk 11.2 逆命题）。§11.3 泊松过程（Def 11.7；Prop 11.8 跳跃概率 \(p(\Delta t)=\lambda\Delta t+o(\Delta t)\)；Prop 11.9 等待时间 \(T\sim\exp(\lambda)\)）。§11.4 复合泊松过程（Def 11.8 \(X_t=\sum_{j\le N_t}Y_j\)；Càdlàg；特征指数 (11.9)、Lévy 测度 \(\mu=\lambda\mu^{\#}\) (11.11)；均值 \(tm\)、方差 \(t\sigma^2\)；随机积分）。§11.5 补偿广义泊松过程（许多小跳、绝对值之和不有限但正负抵消；Def 11.10 \(Y_t=\lim_{\varepsilon\downarrow0}(X_{\varepsilon,t}-m_\varepsilon t)\) (11.21)；Prop 11.10 特征指数 (11.22)、Prop 11.11 鞅性；随机积分）。§11.6 Lévy-Khintchine 刻画（Thm 11.1 分解 \(X_t=mt+\sigma B_t+C_t+Y_t\) (11.23)；Prop 11.12 特征指数）。无图。

定义 11.1（指数分布）与命题 11.1 / Definition 11.1 (Exponential distribution) and Proposition 11.1 定义 11.1（指数分布）：连续随机变量 \(T\) 服从参数 \(\lambda>0\) 的指数分布，记 \(T\sim\exp(\lambda)\)，其概率密度为 \(f(t;\lambda)=\lambda e^{-\lambda t}\)（\(\forall t\ge0\)）。命题 11.1：\(T\sim\exp(\lambda)\) 的 CDF、期望、方差满足 \(\mathbb E[T]=\dfrac1\lambda\)，\(\mathrm{Var}(T)=\dfrac1{\lambda^2}\)，\(F(t;\lambda)=1-e^{-\lambda t}\)（\(\forall t\ge0\)）。Definition 11.1 (Exponential distribution): a continuous random variable \(T\) follows an exponential distribution with parameter \(\lambda>0\), denoted \(T\sim\exp(\lambda)\), with density \(f(t;\lambda)=\lambda e^{-\lambda t}\) (for all \(t\ge0\)). Proposition 11.1: the CDF, mean and variance of \(T\sim\exp(\lambda)\) satisfy \(\mathbb E[T]=\dfrac1\lambda\), \(\mathrm{Var}(T)=\dfrac1{\lambda^2}\), \(F(t;\lambda)=1-e^{-\lambda t}\) (for all \(t\ge0\)).

命题 11.1 证明 / Proof of Proposition 11.1 由 L'Hôpital 法则 \(\lim_{t\to\infty}t^2e^{-\lambda t}=0\) (11.1)（同理 \(\lim_{t\to\infty}te^{-\lambda t}=0\)）。期望 \(\mathbb E[T]=\int_0^\infty t\lambda e^{-\lambda t}\,dt=\dfrac1\lambda\) (11.2)。分部积分得 \(\int_0^\infty t^2e^{-\lambda t}\,dt=\dfrac2{\lambda^3}\) (11.3)。方差 \(\mathrm{Var}(T)=\mathbb E[T^2]-\mathbb E[T]^2=\lambda\cdot\dfrac2{\lambda^3}-\dfrac1{\lambda^2}=\dfrac1{\lambda^2}\)。CDF \(F(t;\lambda)=\int\lambda e^{-\lambda t}\,dt=-e^{-\lambda t}+C\)，由 \(\lim_{t\to\infty}F=1\) 得 \(C=1\)，故 \(F(t;\lambda)=1-e^{-\lambda t}\)。\(\blacksquare\)By L'Hôpital's rule \(\lim_{t\to\infty}t^2e^{-\lambda t}=0\) (11.1) (similarly \(\lim_{t\to\infty}te^{-\lambda t}=0\)). The mean \(\mathbb E[T]=\int_0^\infty t\lambda e^{-\lambda t}\,dt=\dfrac1\lambda\) (11.2). Integration by parts gives \(\int_0^\infty t^2e^{-\lambda t}\,dt=\dfrac2{\lambda^3}\) (11.3). The variance \(\mathrm{Var}(T)=\mathbb E[T^2]-\mathbb E[T]^2=\lambda\cdot\dfrac2{\lambda^3}-\dfrac1{\lambda^2}=\dfrac1{\lambda^2}\). The CDF \(F(t;\lambda)=\int\lambda e^{-\lambda t}\,dt=-e^{-\lambda t}+C\); from \(\lim_{t\to\infty}F=1\), \(C=1\), so \(F(t;\lambda)=1-e^{-\lambda t}\). \(\blacksquare\)

定义 11.2（泊松分布）与命题 11.2 / Definition 11.2 (Poisson distribution) and Proposition 11.2 定义 11.2（泊松分布）：离散随机变量 \(N\) 服从参数 \(\lambda>0\) 的泊松分布，记 \(N\sim\mathrm{Poisson}(\lambda)\)，其概率质量函数（脚注 11.1：PMF 是连续情形 PDF 的离散类比）为 \(f(k;\lambda)=\mathbb P\{N=k\}=\dfrac{e^{-\lambda}\lambda^k}{k!}\)。命题 11.2：\(N\sim\mathrm{Poisson}(\lambda)\) 的期望与方差满足 \(\mathbb E[N]=\mathrm{Var}(N)=\lambda\)。Definition 11.2 (Poisson distribution): a discrete random variable \(N\) follows a Poisson distribution with parameter \(\lambda>0\), denoted \(N\sim\mathrm{Poisson}(\lambda)\), with probability mass function (footnote 11.1: a PMF is the discrete analog of a PDF in the continuous case) \(f(k;\lambda)=\mathbb P\{N=k\}=\dfrac{e^{-\lambda}\lambda^k}{k!}\). Proposition 11.2: the mean and variance of \(N\sim\mathrm{Poisson}(\lambda)\) satisfy \(\mathbb E[N]=\mathrm{Var}(N)=\lambda\).

命题 11.2 证明 / Proof of Proposition 11.2 期望：\(\mathbb E[N]=\sum_{k=0}^\infty k\cdot\dfrac{e^{-\lambda}\lambda^k}{k!}=e^{-\lambda}\lambda\sum_{k=1}^\infty\dfrac{\lambda^{k-1}}{(k-1)!}=e^{-\lambda}\lambda\sum_{k=0}^\infty\dfrac{\lambda^k}{k!}\) (11.4)；由 Taylor 展开 \(e^\lambda=\sum_{k=0}^\infty\lambda^k/k!\)，得 \(\mathbb E[N]=e^{-\lambda}\lambda e^\lambda=\lambda\)。方差：\(\mathrm{Var}(N)=\mathbb E[N(N-1)]+\mathbb E[N]-\mathbb E[N]^2=\left[\sum_{k=2}^\infty k(k-1)\dfrac{e^{-\lambda}\lambda^k}{k!}\right]+\lambda-\lambda^2=\lambda^2e^{-\lambda}\underbrace{\sum_{k=0}^\infty\dfrac{\lambda^k}{k!}}_{=\,e^\lambda}+\lambda-\lambda^2=\lambda\)。\(\blacksquare\)Mean: \(\mathbb E[N]=\sum_{k=0}^\infty k\cdot\dfrac{e^{-\lambda}\lambda^k}{k!}=e^{-\lambda}\lambda\sum_{k=1}^\infty\dfrac{\lambda^{k-1}}{(k-1)!}=e^{-\lambda}\lambda\sum_{k=0}^\infty\dfrac{\lambda^k}{k!}\) (11.4); by the Taylor expansion \(e^\lambda=\sum_{k=0}^\infty\lambda^k/k!\), \(\mathbb E[N]=e^{-\lambda}\lambda e^\lambda=\lambda\). Variance: \(\mathrm{Var}(N)=\mathbb E[N(N-1)]+\mathbb E[N]-\mathbb E[N]^2=\left[\sum_{k=2}^\infty k(k-1)\dfrac{e^{-\lambda}\lambda^k}{k!}\right]+\lambda-\lambda^2=\lambda^2e^{-\lambda}\underbrace{\sum_{k=0}^\infty\dfrac{\lambda^k}{k!}}_{=\,e^\lambda}+\lambda-\lambda^2=\lambda\). \(\blacksquare\)

定义 11.3（特征函数）与命题 11.3 / Definition 11.3 (Characteristic function) and Proposition 11.3 定义 11.3（特征函数）：一维随机变量 \(X\) 的特征函数 \(\varphi_X:\mathbb R\to\mathbb C\) 定义为 \(\varphi_X(s)\equiv\mathbb E[e^{isX}]\)（\(\forall s\in\mathbb R\)）。命题 11.3：\(\forall k\in\mathbb N_+\)，\(\varphi_X^{(k)}(0)=i^k\mathbb E[X^k]\)，其中 \(\varphi_X^{(k)}(0)\) 是关于 \(s\) 的 \(k\) 阶导数在 \(s=0\) 处取值。证明：\(\varphi_X^{(k)}(0)=\left[\dfrac{\partial^k\mathbb E[e^{isX}]}{(\partial s)^k}\right]_{s=0}=\mathbb E[(iX)^k e^{i0X}]=i^k\mathbb E[X^k]\)。Definition 11.3 (Characteristic function): the characteristic function of a one-dimensional random variable \(X\) is \(\varphi_X:\mathbb R\to\mathbb C\), \(\varphi_X(s)\equiv\mathbb E[e^{isX}]\) (for all \(s\in\mathbb R\)). Proposition 11.3: for all \(k\in\mathbb N_+\), \(\varphi_X^{(k)}(0)=i^k\mathbb E[X^k]\), where \(\varphi_X^{(k)}(0)\) is the \(k\)th derivative w.r.t. \(s\) evaluated at \(s=0\). Proof: \(\varphi_X^{(k)}(0)=\left[\dfrac{\partial^k\mathbb E[e^{isX}]}{(\partial s)^k}\right]_{s=0}=\mathbb E[(iX)^k e^{i0X}]=i^k\mathbb E[X^k]\).

定义 11.4（特征指数）与命题 11.4、11.5 / Definition 11.4 and Propositions 11.4, 11.5 定义 11.4（特征指数）：一维随机变量 \(X\) 的特征指数 \(\Psi(s)\) 定义为 \(e^{\Psi(s)}\equiv\mathbb E[e^{isX}]\)（\(\forall s\in\mathbb R\)），即 \(\Psi(s)=\ln\varphi_X(s)\)。命题 11.4：若 \(X,Y\) 独立，则 \(\varphi_{X+Y}(s)=\varphi_X(s)\varphi_Y(s)\)，\(\Psi_{X+Y}=\Psi_X+\Psi_Y\)。命题 11.5：\(X\) 是随机变量、\(c\) 是常数，则 \(\varphi_{cX}(s)=\varphi_X(cs)\)，\(\Psi_{cX}(s)=\Psi_X(cs)\)。Definition 11.4 (Characteristic exponent): the characteristic exponent \(\Psi(s)\) of a one-dimensional random variable \(X\) is defined by \(e^{\Psi(s)}\equiv\mathbb E[e^{isX}]\) (for all \(s\in\mathbb R\)), i.e. \(\Psi(s)=\ln\varphi_X(s)\). Proposition 11.4: if \(X,Y\) are independent, then \(\varphi_{X+Y}(s)=\varphi_X(s)\varphi_Y(s)\) and \(\Psi_{X+Y}=\Psi_X+\Psi_Y\). Proposition 11.5: \(X\) a random variable, \(c\) a constant; then \(\varphi_{cX}(s)=\varphi_X(cs)\), \(\Psi_{cX}(s)=\Psi_X(cs)\).

命题 11.4、11.5 证明 / Proof of Propositions 11.4, 11.5 命题 11.4：\(e^{\Psi_{X+Y}(s)}=\mathbb E[e^{is(X+Y)}]=\mathbb E[e^{isX}e^{isY}]\overset{\text{indep}}{=}\mathbb E[e^{isX}]\mathbb E[e^{isY}]=e^{\Psi_X(s)}e^{\Psi_Y(s)}=e^{\Psi_X(s)+\Psi_Y(s)}\)，故 \(\varphi_{X+Y}=\varphi_X\varphi_Y\)、\(\Psi_{X+Y}=\Psi_X+\Psi_Y\)。命题 11.5：\(\varphi_{cX}(s)=\mathbb E[e^{is(cX)}]=\mathbb E[e^{i(cs)X}]=\varphi_X(cs)\)，取对数得 \(\Psi_{cX}(s)=\Psi_X(cs)\)。\(\blacksquare\)Proposition 11.4: \(e^{\Psi_{X+Y}(s)}=\mathbb E[e^{is(X+Y)}]=\mathbb E[e^{isX}e^{isY}]\overset{\text{indep}}{=}\mathbb E[e^{isX}]\mathbb E[e^{isY}]=e^{\Psi_X(s)}e^{\Psi_Y(s)}=e^{\Psi_X(s)+\Psi_Y(s)}\), so \(\varphi_{X+Y}=\varphi_X\varphi_Y\), \(\Psi_{X+Y}=\Psi_X+\Psi_Y\). Proposition 11.5: \(\varphi_{cX}(s)=\mathbb E[e^{is(cX)}]=\mathbb E[e^{i(cs)X}]=\varphi_X(cs)\), taking logs gives \(\Psi_{cX}(s)=\Psi_X(cs)\). \(\blacksquare\)

命题 11.6（正态）与命题 11.7（泊松）/ Propositions 11.6 (normal) and 11.7 (Poisson) 命题 11.6：若 \(X\sim\mathcal N(\mu,\sigma^2)\)，则 \(\Psi_X(s)=i\mu s-\dfrac{\sigma^2}2s^2\) (11.6)。证明：\(e^{\Psi_X(s)}=\mathbb E[e^{isX}]=e^{is\mu+\frac{\sigma^2}2s^2i^2}=e^{is\mu-\frac{\sigma^2}2s^2}\)。命题 11.7：若 \(X\sim\mathrm{Poisson}(\lambda)\)，则 \(\Psi_X(s)=\lambda(e^{is}-1)\)。证明：\(e^{\Psi_X(s)}=\mathbb E[e^{isX}]=\sum_{k=0}^\infty e^{isk}\dfrac{e^{-\lambda}\lambda^k}{k!}=e^{-\lambda}\sum_{k=0}^\infty\dfrac{(e^{is}\lambda)^k}{k!}\overset{\text{Taylor}}{=}e^{-\lambda}e^{e^{is}\lambda}\)，故 \(\Psi_X(s)=e^{is}\lambda-\lambda=\lambda(e^{is}-1)\)。Proposition 11.6: if \(X\sim\mathcal N(\mu,\sigma^2)\), then \(\Psi_X(s)=i\mu s-\dfrac{\sigma^2}2s^2\) (11.6). Proof: \(e^{\Psi_X(s)}=\mathbb E[e^{isX}]=e^{is\mu+\frac{\sigma^2}2s^2i^2}=e^{is\mu-\frac{\sigma^2}2s^2}\). Proposition 11.7: if \(X\sim\mathrm{Poisson}(\lambda)\), then \(\Psi_X(s)=\lambda(e^{is}-1)\). Proof: \(e^{\Psi_X(s)}=\mathbb E[e^{isX}]=\sum_{k=0}^\infty e^{isk}\dfrac{e^{-\lambda}\lambda^k}{k!}=e^{-\lambda}\sum_{k=0}^\infty\dfrac{(e^{is}\lambda)^k}{k!}\overset{\text{Taylor}}{=}e^{-\lambda}e^{e^{is}\lambda}\), so \(\Psi_X(s)=e^{is}\lambda-\lambda=\lambda(e^{is}-1)\).

定义 11.5（无穷可分分布）/ Definition 11.5 (Infinitely divisible distribution) 随机变量 \(X\) 具有无穷可分分布，若 \(\forall n\in\mathbb N_+\)，都能找到独立同分布的 \(X_1^{(n)},X_2^{(n)},\dots,X_n^{(n)}\) 使 \(X\overset{d}{=}X_1^{(n)}+X_2^{(n)}+\dots+X_n^{(n)}\)。A random variable \(X\) has an infinitely divisible distribution if for all \(n\in\mathbb N_+\) one can find independent and identically distributed \(X_1^{(n)},X_2^{(n)},\dots,X_n^{(n)}\) such that \(X\overset{d}{=}X_1^{(n)}+X_2^{(n)}+\dots+X_n^{(n)}\).

定义 11.6（Lévy 过程）/ Definition 11.6 (Lévy process) 回顾定义 1.2，布朗运动由条件 (1)–(4) 定义。若去掉条件 (4)（连续路径）与条件 (1)（从原点出发）以求一般性，便得到更一般的随机过程——Lévy 过程。一维 Lévy 过程是满足以下条件的随机过程 \(\{X_t\}\)：(1) 增量独立：\(rRecall from Definition 1.2 that Brownian motion is defined by conditions (1)–(4). If we remove condition (4) (continuous path) and condition (1) (starting from the origin) for generality, we obtain a more general stochastic process — the Lévy process. A one-dimensional Lévy process is a stochastic process \(\{X_t\}\) satisfying: (1) independent increments: for \(r

注 11.1、事实 11.1、注 11.2 / Remark 11.1, Fact 11.1, Remark 11.2 注 11.1：布朗运动是连续路径的 Lévy 过程特例；泊松过程是另一特例（下文讨论）。事实 11.1：若 \(\{X_t\}\) 是 Lévy 过程，则 \(X_t\) 对 \(\forall t\) 无穷可分，因为总可写 \(X_t=\underbrace{(X_{\frac tn}-X_0)}_{\equiv X_{1,t}^{(n)}}+\underbrace{(X_{\frac{2t}n}-X_{\frac tn})}_{\equiv X_{2,t}^{(n)}}+\dots+\underbrace{(X_{\frac{nt}n}-X_{\frac{(n-1)t}n})}_{\equiv X_{n,t}^{(n)}}\)。注 11.2：事实 11.1 的逆命题也成立——若 \(X_t\) 对 \(\forall t\) 无穷可分，则 \(\{X_t\}\) 是 Lévy 过程。Remark 11.1: Brownian motion is a special case of a Lévy process with continuous path; the Poisson process is another (discussed below). Fact 11.1: if \(\{X_t\}\) is a Lévy process, then \(X_t\) is infinitely divisible for all \(t\), because we can always write \(X_t=\underbrace{(X_{\frac tn}-X_0)}_{\equiv X_{1,t}^{(n)}}+\underbrace{(X_{\frac{2t}n}-X_{\frac tn})}_{\equiv X_{2,t}^{(n)}}+\dots+\underbrace{(X_{\frac{nt}n}-X_{\frac{(n-1)t}n})}_{\equiv X_{n,t}^{(n)}}\). Remark 11.2: the converse of Fact 11.1 is also true — if \(X_t\) is infinitely divisible for all \(t\), then \(\{X_t\}\) is a Lévy process.

定义 11.7（泊松过程）/ Definition 11.7 (Poisson process) 过程 \(\{N_t\}\) 是速率 \(\lambda\) 的泊松过程，若满足：(1) 从原点出发：\(N_0=0\)；(2) 增量独立：\(rA process \(\{N_t\}\) is a Poisson process with rate \(\lambda\) if it satisfies: (1) starts from the origin: \(N_0=0\); (2) independent increments: for \(r

命题 11.8（跳跃概率）与命题 11.9（等待时间）/ Propositions 11.8, 11.9 命题 11.8：记 \([t,t+\Delta t]\) 内有一次跳跃的概率为 \(p(\Delta t)\)，则 \(p(\Delta t)=\lambda\Delta t+o(\Delta t)\)。命题 11.9：两次跳跃间的等待时间 \(T\) 服从参数 \(\lambda\) 的指数分布，即 \(T\sim\exp(\lambda)\)。Proposition 11.8: denote the probability of having a jump during \([t,t+\Delta t]\) by \(p(\Delta t)\); then \(p(\Delta t)=\lambda\Delta t+o(\Delta t)\). Proposition 11.9: the waiting time \(T\) between two jumps follows an exponential distribution with parameter \(\lambda\), i.e. \(T\sim\exp(\lambda)\).

命题 11.8、11.9 证明 / Proof of Propositions 11.8, 11.9 命题 11.8：\(p(\Delta t)\) 的 \(t\)-无关性来自条件 (3)；由 \(N_t-N_s\sim\mathrm{Poisson}(\lambda(t-s))\)，\(p(\Delta t)=\dfrac{e^{-\lambda\Delta t}[\lambda\Delta t]^1}{1}=(1-\lambda\Delta t+o(\Delta t))\lambda\Delta t=\lambda\Delta t+o(\Delta t)\)。命题 11.9：由等待时间定义、条件 (2) 与命题 11.8，\(\mathbb P\{T>t\}=\prod_{j=1}^n\mathbb P\{\text{no jump in }[\tfrac{j-1}n t,\tfrac jn t]\}=\prod_{j=1}^n\left[1-p\!\left(\tfrac tn\right)\right]=\left[1-\lambda\tfrac tn+o\!\left(\tfrac tn\right)\right]^n=\left(e^{-\lambda\frac tn}\right)^n=e^{-\lambda t}\)，故 \(\mathbb P\{T\le t\}=1-e^{-\lambda t}\)，即 \(T\sim\exp(\lambda)\)。\(\blacksquare\)Proposition 11.8: the \(t\)-independence of \(p(\Delta t)\) comes from condition (3); by \(N_t-N_s\sim\mathrm{Poisson}(\lambda(t-s))\), \(p(\Delta t)=\dfrac{e^{-\lambda\Delta t}[\lambda\Delta t]^1}{1}=(1-\lambda\Delta t+o(\Delta t))\lambda\Delta t=\lambda\Delta t+o(\Delta t)\). Proposition 11.9: by the definition of waiting time, condition (2), and Proposition 11.8, \(\mathbb P\{T>t\}=\prod_{j=1}^n\mathbb P\{\text{no jump in }[\tfrac{j-1}n t,\tfrac jn t]\}=\prod_{j=1}^n\left[1-p\!\left(\tfrac tn\right)\right]=\left[1-\lambda\tfrac tn+o\!\left(\tfrac tn\right)\right]^n=\left(e^{-\lambda\frac tn}\right)^n=e^{-\lambda t}\), so \(\mathbb P\{T\le t\}=1-e^{-\lambda t}\), i.e. \(T\sim\exp(\lambda)\). \(\blacksquare\)

定义 11.8（复合泊松过程）与注 11.3 / Definition 11.8 (Compound Poisson process) and Remark 11.3 定义 11.8（复合泊松过程）：设 \(\{N_t\}\) 是速率 \(\lambda\) 的泊松过程，\(Y_1,Y_2,\dots\overset{d}{=}Y\) 是 i.i.d. 随机变量（脚注 11.2：\(Y_1,Y_2,\dots\) 也独立于 \(N_t\)，\(\forall t\)），分布测度为 \(\mu^{\#}\)（脚注 11.3：\(\mu^{\#}\) 定义为 \(\forall V\subset\mathbb R\)，\(\mu^{\#}(V)=\mathbb P\{Y_j\in V\}\)）。则定义 \(\{X_t\}\) 为复合泊松过程，\(X_t=\sum_{j\le N_t}Y_j\) (11.7)。注 11.3：把 (11.7) 中 \(Y_j=1\)（\(\forall j\le N_t\)）则退化为泊松过程。Definition 11.8 (Compound Poisson process): let \(\{N_t\}\) be a Poisson process with rate \(\lambda\), and \(Y_1,Y_2,\dots\overset{d}{=}Y\) be i.i.d. random variables (footnote 11.2: \(Y_1,Y_2,\dots\) are also independent of \(N_t\) for all \(t\)) with distribution measure \(\mu^{\#}\) (footnote 11.3: \(\mu^{\#}\) is defined by, for all \(V\subset\mathbb R\), \(\mu^{\#}(V)=\mathbb P\{Y_j\in V\}\)). Then define \(\{X_t\}\) as a compound Poisson process with \(X_t=\sum_{j\le N_t}Y_j\) (11.7). Remark 11.3: setting \(Y_j=1\) (for all \(j\le N_t\)) in (11.7) degenerates to the Poisson process.

定义 11.9（Càdlàg）与 (11.8) / Definition 11.9 (Càdlàg) and (11.8) 定义 11.9（Càdlàg 函数）：函数 \(f\) 是 Càdlàg 的（脚注 11.4：法语 "continue à droite, limite à gauche"，意为"右连续、有左极限"），若 \(f\) 右连续且有左极限。复合泊松过程 \(\{X_t\}\) 是 Càdlàg 的。记 \(X_{t-}\equiv\lim_{\varepsilon\downarrow0}X_{t-\varepsilon}\)、\(X_{t+}\equiv\lim_{\varepsilon\downarrow0}X_{t+\varepsilon}\)，由 Càdlàg 性 \(X_{t-}\) 良定义且 \(X_t=X_{t+}\)。设 \(X_0=0\) 且跳跃只发生在 \([0,t]\) 内的 \(s_1,\dots,s_{N_t}\)（\(\{N_t\}\) 是泊松过程），则 \(X_t=\sum_{j=1}^{N_t}\underbrace{(X_{s_j}-X_{s_j-})}_{\text{jump at }s_j}\) (11.8)。Definition 11.9 (Càdlàg function): a function \(f\) is Càdlàg (footnote 11.4: French "continue à droite, limite à gauche", meaning "right continuous with left limits") if \(f\) is right continuous and has left limits. The compound Poisson process \(\{X_t\}\) is Càdlàg. Denote \(X_{t-}\equiv\lim_{\varepsilon\downarrow0}X_{t-\varepsilon}\), \(X_{t+}\equiv\lim_{\varepsilon\downarrow0}X_{t+\varepsilon}\); by the Càdlàg property \(X_{t-}\) is well-defined and \(X_t=X_{t+}\). Suppose \(X_0=0\) and jumps happen only at \(s_1,\dots,s_{N_t}\) in \([0,t]\) (with \(\{N_t\}\) a Poisson process); then \(X_t=\sum_{j=1}^{N_t}\underbrace{(X_{s_j}-X_{s_j-})}_{\text{jump at }s_j}\) (11.8).

特征指数与 Lévy 测度：(11.9)–(11.11) / Characteristic exponent and Lévy measure: (11.9)–(11.11) 由命题 11.4（不失一般性设 \(t\in\mathbb N_+\)），\(\varphi_{X_t}(s)=(\varphi_{X_1}(s))^t\)。特征指数 \(\Psi_{X_1}(s)\) 满足By Proposition 11.4 (WLOG \(t\in\mathbb N_+\)), \(\varphi_{X_t}(s)=(\varphi_{X_1}(s))^t\). The characteristic exponent \(\Psi_{X_1}(s)\) satisfies

$$e^{\Psi_{X_1}(s)}=\sum_{k=0}^\infty\underbrace{\left(\mathbb E[e^{isY}]\right)^k}_{=\,\varphi_Y(s)^k}\frac{e^{-\lambda}\lambda^k}{k!}=e^{-\lambda}e^{\varphi_Y(s)\lambda}\;\Rightarrow\;\Psi_{X_1}(s)=\lambda(\varphi_Y(s)-1),\tag{11.9}$$

即 \(\varphi_{X_1}(s)=e^{\lambda(\varphi_Y(s)-1)}\) (11.10)。由分布测度定义 \(\varphi_Y(s)=\int_{-\infty}^\infty e^{isy}\,d\mu^{\#}(y)\)。定义 Lévy 测度（脚注 11.5：一般地，随机变量 \(X\) 的 Lévy 测度 \(\mu\) 满足 (1) \(\mu(\{0\})=0\)；(2) \(\int(x^2\wedge1)\,d\mu(x)<\infty\)）\(\mu\equiv\lambda\mu^{\#}\)，则 (11.9) 改写为i.e. \(\varphi_{X_1}(s)=e^{\lambda(\varphi_Y(s)-1)}\) (11.10). By the distribution measure, \(\varphi_Y(s)=\int_{-\infty}^\infty e^{isy}\,d\mu^{\#}(y)\). Define the Lévy measure (footnote 11.5: in general the Lévy measure \(\mu\) for a random variable \(X\) satisfies (1) \(\mu(\{0\})=0\); (2) \(\int(x^2\wedge1)\,d\mu(x)<\infty\)) by \(\mu\equiv\lambda\mu^{\#}\); then (11.9) is rewritten as

$$\Psi_{X_1}(s)=\lambda\int(e^{isy}-1)\,d\mu^{\#}(y)=\int(e^{isy}-1)\underbrace{(\lambda\,d\mu^{\#}(y))}_{\equiv\,d\mu(y)}=\int_{-\infty}^\infty(e^{isy}-1)\,d\mu(y).\tag{11.11}$$

复合泊松过程的均值与方差 / Mean and variance of the compound Poisson process 由命题 11.3，\(\varphi_{X_1}^{(k)}(0)=i^k\mathbb E[X_1^k]\) (11.12)。\(k=1\) 时，由 (11.10) 求导（并用 \(\int d\mu^{\#}=1\) 使指数项 \(e^{\lambda(\int d\mu^{\#}-1)}=e^0=1\)）：By Proposition 11.3, \(\varphi_{X_1}^{(k)}(0)=i^k\mathbb E[X_1^k]\) (11.12). For \(k=1\), differentiating (11.10) (using \(\int d\mu^{\#}=1\) so the exponential factor \(e^{\lambda(\int d\mu^{\#}-1)}=e^0=1\)):

$$\varphi_{X_1}^{(1)}(0)=\lambda\left(\int iy\,d\mu^{\#}(y)\right)=i\left(\int y\,\lambda d\mu^{\#}(y)\right)=i\left(\int y\,d\mu(y)\right).\tag{11.14}$$

\(k=2\) 时，由 (11.10) 同理得 \(\varphi_{X_1}^{(2)}(0)=i^2\left(\int y\,d\mu(y)\right)^2+i^2\left(\int y^2\,d\mu(y)\right)\) (11.15)。均值：把 (11.14) 代入 (11.12)（\(k=1\)）：\(i\int y\,d\mu(y)=i\mathbb E[X_1]\Rightarrow\int y\,d\mu(y)=\mathbb E[X_1]\)；又 \(\mathbb E[X_t]=t\mathbb E[X_1]\)，故 \(\mathbb E[X_t]=t\underbrace{\int y\,d\mu(y)}_{\equiv\,m}=tm\)。方差：把 (11.15) 代入 (11.12)（\(k=2\)）：\(\mathbb E[X_1^2]=\left(\int y\,d\mu(y)\right)^2+\int y^2\,d\mu(y)\Rightarrow\mathrm{Var}(X_1)=\int y^2\,d\mu(y)\)；又 \(\mathrm{Var}(X_t)=t\mathrm{Var}(X_1)\)，故 \(\mathrm{Var}(X_t)=t\underbrace{\int y^2\,d\mu(y)}_{\equiv\,\sigma^2}=t\sigma^2\)。\(\blacksquare\)For \(k=2\), differentiating (11.10) likewise gives \(\varphi_{X_1}^{(2)}(0)=i^2\left(\int y\,d\mu(y)\right)^2+i^2\left(\int y^2\,d\mu(y)\right)\) (11.15). Mean: plug (11.14) into (11.12) (\(k=1\)): \(i\int y\,d\mu(y)=i\mathbb E[X_1]\Rightarrow\int y\,d\mu(y)=\mathbb E[X_1]\); and \(\mathbb E[X_t]=t\mathbb E[X_1]\), so \(\mathbb E[X_t]=t\underbrace{\int y\,d\mu(y)}_{\equiv\,m}=tm\). Variance: plug (11.15) into (11.12) (\(k=2\)): \(\mathbb E[X_1^2]=\left(\int y\,d\mu(y)\right)^2+\int y^2\,d\mu(y)\Rightarrow\mathrm{Var}(X_1)=\int y^2\,d\mu(y)\); and \(\mathrm{Var}(X_t)=t\mathrm{Var}(X_1)\), so \(\mathrm{Var}(X_t)=t\underbrace{\int y^2\,d\mu(y)}_{\equiv\,\sigma^2}=t\sigma^2\). \(\blacksquare\)

11.4.3 关于复合泊松过程的随机积分 / Stochastic integration w.r.t. the compound Poisson process 把 \(\{X_t\}\) 按 (11.8) 表示 \(X_t=\sum_{j=1}^{N_t}(X_{s_j}-X_{s_j-})\)，则定义关于 \(\{X_t\}\) 的随机积分 \(\int_0^t A_s\,dX_s\equiv\sum_{j=1}^{N_t}A_{s_j}(X_{s_j}-X_{s_j-})\)。Representing \(\{X_t\}\) as in (11.8), \(X_t=\sum_{j=1}^{N_t}(X_{s_j}-X_{s_j-})\), the stochastic integral w.r.t. \(\{X_t\}\) is defined by \(\int_0^t A_s\,dX_s\equiv\sum_{j=1}^{N_t}A_{s_j}(X_{s_j}-X_{s_j-})\).

动机 / Motivation 我们想构造一个有许多小跳的跳跃过程，其跳幅绝对值之期望和并不有限，但正负跳跃彼此良好抵消。We want to construct a jump process with many small jumps, whose expected sum of absolute values is not finite but positive and negative jumps cancel well.

受限 Lévy 测度 \(\mu_\varepsilon\) 的假设：(11.16)–(11.19) / Assumptions on the restricted Lévy measure \(\mu_\varepsilon\) 设 \(\mu\) 是复合泊松过程中 (11.11) 的 Lévy 测度，\(\mu_\varepsilon\) 是满足如下假设的 Lévy 测度：(i) 对任意 \(\varepsilon>0\)，\(\mu_\varepsilon\) 是把 \(\mu\) 限制到 \(\{|Y|>\varepsilon\}\) 的跳跃上，即 \(\mu_\varepsilon\{y:|y|>\varepsilon\}=\mu\{y:|y|>\varepsilon\}\)、\(\mu_\varepsilon\{y:|y|\le\varepsilon\}=0\)，则 \(\mu_\varepsilon(\mathbb R)=\mu\{y:|y|>\varepsilon\}<\infty\) (11.16)；(ii) 所有跳幅以 1 为界：\(\mu_\varepsilon\{y:|y|>1\}=\mu\{y:|y|>1\}=0\) (11.17)；(iii) 二阶矩有限：\(\sigma^2\equiv\int_{-\infty}^\infty y^2\,d\mu_\varepsilon(y)<\infty\) (11.18)。定义 \(m_\varepsilon\equiv\int_{-\infty}^\infty y\,d\mu_\varepsilon(y)=\int_{\varepsilon<|y|\le1}y\,d\mu(y)\) (11.19)。Let \(\mu\) be the Lévy measure in (11.11) for the compound Poisson process, and \(\mu_\varepsilon\) a Lévy measure satisfying: (i) for any \(\varepsilon>0\), \(\mu_\varepsilon\) is \(\mu\) restricted to jumps \(\{|Y|>\varepsilon\}\), i.e. \(\mu_\varepsilon\{y:|y|>\varepsilon\}=\mu\{y:|y|>\varepsilon\}\), \(\mu_\varepsilon\{y:|y|\le\varepsilon\}=0\), so \(\mu_\varepsilon(\mathbb R)=\mu\{y:|y|>\varepsilon\}<\infty\) (11.16); (ii) all jumps are bounded by 1: \(\mu_\varepsilon\{y:|y|>1\}=\mu\{y:|y|>1\}=0\) (11.17); (iii) the second moment is finite: \(\sigma^2\equiv\int_{-\infty}^\infty y^2\,d\mu_\varepsilon(y)<\infty\) (11.18). Define \(m_\varepsilon\equiv\int_{-\infty}^\infty y\,d\mu_\varepsilon(y)=\int_{\varepsilon<|y|\le1}y\,d\mu(y)\) (11.19).

定义 11.10（补偿广义泊松过程）与注 11.4 / Definition 11.10 and Remark 11.4 对每个 \(\varepsilon>0\)，存在关联到 Lévy 测度 \(\mu_\varepsilon\) 的复合泊松过程 \(\{X_{\varepsilon,t}\}\)（由 (11.7) \(X_t=\sum_{j\le N_t}Y_j\) (11.20)，取 \(Y^\varepsilon\) 使 \(\mu^{\#}(Y^\varepsilon\le\varepsilon)=0\)、\(\mu^{\#}(Y^\varepsilon>1)=0\)，则该 \(Y^\varepsilon\) 定义的 \(X_t\) 即 \(X_{\varepsilon,t}\)）。定义 11.10（补偿广义泊松过程）：设 \(\mu_\varepsilon\) 是对每个 \(\varepsilon>0\) 满足 (11.16)–(11.18) 的 Lévy 测度。则定义关于 Lévy 测度 \(\mu\) 的补偿广义泊松过程 \(\{Y_t\}\) 为 \(Y_t=\lim_{\varepsilon\downarrow0}M_{\varepsilon,t}=\lim_{\varepsilon\downarrow0}(X_{\varepsilon,t}-m_\varepsilon t)\) (11.21)，\(m_\varepsilon\) 由 (11.19) 定义。注 11.4："补偿 (compensated)" 在此意为去均值 (demeaned)。For each \(\varepsilon>0\), there is a compound Poisson process \(\{X_{\varepsilon,t}\}\) associated to the Lévy measure \(\mu_\varepsilon\) (from (11.7) \(X_t=\sum_{j\le N_t}Y_j\) (11.20), take \(Y^\varepsilon\) with \(\mu^{\#}(Y^\varepsilon\le\varepsilon)=0\), \(\mu^{\#}(Y^\varepsilon>1)=0\); the \(X_t\) defined by such \(Y^\varepsilon\) is the associated \(X_{\varepsilon,t}\)). Definition 11.10 (Compensated generalized Poisson process): let \(\mu_\varepsilon\) be a Lévy measure satisfying (11.16)–(11.18) for every \(\varepsilon>0\). Then define the compensated generalized Poisson process \(\{Y_t\}\) w.r.t. the Lévy measure \(\mu\) by \(Y_t=\lim_{\varepsilon\downarrow0}M_{\varepsilon,t}=\lim_{\varepsilon\downarrow0}(X_{\varepsilon,t}-m_\varepsilon t)\) (11.21), where \(m_\varepsilon\) is defined in (11.19). Remark 11.4: the word "compensated" here means demeaned.

命题 11.10（特征指数）/ Proposition 11.10 (Characteristic exponent) 设 \(\{Y_t\}\) 是由 (11.21) 定义的补偿广义泊松过程，则 \(\Psi_{Y_t}(s)=t\int_{-1}^1(e^{isy}-1-isy)\,d\mu(y)\)。Suppose \(\{Y_t\}\) is the compensated generalized Poisson process defined by (11.21); then \(\Psi_{Y_t}(s)=t\int_{-1}^1(e^{isy}-1-isy)\,d\mu(y)\).

命题 11.10 证明 / Proof of Proposition 11.10 由 (11.5)，\(\Psi_{m_\varepsilon}(s)=\ln\mathbb E[e^{ism_\varepsilon}]=\ln e^{ism_\varepsilon}=ism_\varepsilon\)。\(\mu_\varepsilon,\mu\) 可互换（\(Y^\varepsilon\) 使 \(\mu_\varepsilon(Y^\varepsilon\in V)=\mu(Y^\varepsilon=V)\)，\(\forall V\subset\mathbb R\)）。\(X_{\varepsilon,t}\) 仍是复合泊松，故 (11.11) 适用：\(\Psi_{X_{\varepsilon,1}}=\int_{\varepsilon<|y|\le1}(e^{isy}-1)\,d\mu(y)\)。\(X_{\varepsilon,t}\) 与 \(m_\varepsilon\) 独立，由命题 11.4：By (11.5), \(\Psi_{m_\varepsilon}(s)=\ln\mathbb E[e^{ism_\varepsilon}]=\ln e^{ism_\varepsilon}=ism_\varepsilon\). We can use \(\mu_\varepsilon,\mu\) interchangeably (\(Y^\varepsilon\) defined so \(\mu_\varepsilon(Y^\varepsilon\in V)=\mu(Y^\varepsilon=V)\), for all \(V\subset\mathbb R\)). \(X_{\varepsilon,t}\) is still compound Poisson, so (11.11) applies: \(\Psi_{X_{\varepsilon,1}}=\int_{\varepsilon<|y|\le1}(e^{isy}-1)\,d\mu(y)\). Since \(X_{\varepsilon,t}\) and \(m_\varepsilon\) are independent, by Proposition 11.4:

$$\begin{aligned}\Psi_{M_{\varepsilon,1}}(s)&=\Psi_{X_{\varepsilon,1}}-\Psi_{m_\varepsilon}=\int_{\varepsilon<|y|\le1}(e^{isy}-1)\,d\mu(y)-\int_{\varepsilon<|y|\le1}isy\,d\mu_\varepsilon(y)\\&=\int_{\varepsilon<|y|\le1}(e^{isy}-1-isy)\,d\mu(y),\end{aligned}$$

由独立性 \(\Psi_{M_{\varepsilon,t}}(s)=t\int_{\varepsilon<|y|\le1}(e^{isy}-1-isy)\,d\mu(y)\)；由 \(Y_t=\lim_{\varepsilon\downarrow0}M_{\varepsilon,t}\)，得 \(\Psi_{Y_t}(s)=t\int_{-1}^1(e^{isy}-1-isy)\,d\mu(y)\) (11.22)。\(\blacksquare\)By independence \(\Psi_{M_{\varepsilon,t}}(s)=t\int_{\varepsilon<|y|\le1}(e^{isy}-1-isy)\,d\mu(y)\); and since \(Y_t=\lim_{\varepsilon\downarrow0}M_{\varepsilon,t}\), \(\Psi_{Y_t}(s)=t\int_{-1}^1(e^{isy}-1-isy)\,d\mu(y)\) (11.22). \(\blacksquare\)

命题 11.11（鞅性）/ Proposition 11.11 (Martingale) 设补偿广义泊松过程 \(\{Y_t\}\) 即 (11.21)，\(Y_t=X_t-mt\)，其中 \(X_t\equiv\lim_{\varepsilon\downarrow0}X_{\varepsilon,t}\)、\(m\equiv\lim_{\varepsilon\downarrow0}m_\varepsilon\)。则 \(\{Y_t\}\) 是关于 \(\mathcal F_t\equiv\sigma\{X_s:0\le s\le t\}\) 的鞅。Let the compensated generalized Poisson process \(\{Y_t\}\) be (11.21), \(Y_t=X_t-mt\), where \(X_t\equiv\lim_{\varepsilon\downarrow0}X_{\varepsilon,t}\), \(m\equiv\lim_{\varepsilon\downarrow0}m_\varepsilon\). Then \(\{Y_t\}\) is a martingale w.r.t. \(\mathcal F_t\equiv\sigma\{X_s:0\le s\le t\}\).

命题 11.11 证明 / Proof of Proposition 11.11 只需验证 \(s\le t\) 时 \(\mathbb E[Y_t\mid\mathcal F_s]=Y_s\)：It suffices to check that for \(s\le t\), \(\mathbb E[Y_t\mid\mathcal F_s]=Y_s\):

$$\begin{aligned}\mathbb E[Y_t\mid\mathcal F_s]&=\mathbb E[X_t-mt\mid\mathcal F_s]=X_s-ms+\mathbb E[(X_t-X_s)-m(t-s)\mid\mathcal F_s]\\&\overset{\text{incr indep}}{=}X_s-ms+\mathbb E[X_{t-s}]-m(t-s)\overset{\text{incr identical}}{=}X_s-ms+m(t-s)-m(t-s)=Y_s,\end{aligned}$$

其中用到增量独立与增量同分布（\(\mathbb E[X_{t-s}]=m(t-s)\)）。\(\blacksquare\)using increment independence and identical increments (\(\mathbb E[X_{t-s}]=m(t-s)\)). \(\blacksquare\)

11.5.3 关于补偿广义泊松过程的随机积分 / Stochastic integration w.r.t. the compensated generalized Poisson process 把 \(\{Y_t\}\) 类似 (11.8) 表示为 \(Y_t=\sum_{j=1}^{N_t}(X_{s_j}-X_{s_j-})-mt\)，则定义随机积分 \(\int_0^t A_s\,dY_s\equiv\sum_{j=1}^{N_t}A_{s_j}(X_{s_j}-X_{s_j-})-\int_0^t mA_s\,dt\)。Representing \(\{Y_t\}\) similarly to (11.8) as \(Y_t=\sum_{j=1}^{N_t}(X_{s_j}-X_{s_j-})-mt\), the stochastic integral is defined by \(\int_0^t A_s\,dY_s\equiv\sum_{j=1}^{N_t}A_{s_j}(X_{s_j}-X_{s_j-})-\int_0^t mA_s\,dt\).

定理 11.1（Lévy-Khintchine 表示）与注 11.5 / Theorem 11.1 (Lévy-Khintchine Representation) and Remark 11.5 定理 11.1：设 \(\{X_t\}\) 是 \(X_0=0\) 的 Lévy 过程。则对 \(\forall t\)，\(X_t\) 可分解为 \(X_t=mt+\sigma B_t+C_t+Y_t\) (11.23)，其中 \(m\in\mathbb R\)、\(\sigma\ge0\)，\(B_t,C_t,Y_t\) 是独立随机变量，满足：(1) \(\{B_t\}\) 是标准布朗运动；(2) \(\{C_t\}\) 是 Lévy 测度 \(\mu_C\) 的复合泊松过程，\(\mu_C(\{|x|\le1\})=0\) 且 \(\mu_C(\mathbb R)=\mu_C(\{|x|>1\})<\infty\)；(3) \(\{Y_t\}\) 是 Lévy 测度 \(\mu_Y\) 的补偿广义泊松过程，\(\mu_Y(\{0\})=0\) 且 \(\int_{-\infty}^\infty y^2\,d\mu_Y(y)<\infty\)。注 11.5：Lévy-Khintchine 表示把 Lévy 过程分解为连续部分 \((mt+\sigma B_t)\) 与纯跳跃部分 \((C_t+Y_t)\)；纯跳跃部分中 \(C_t\) 管辖较大跳幅（不小于 1），\(Y_t\) 管辖较小跳幅（小于 1）。Theorem 11.1: suppose \(\{X_t\}\) is a Lévy process with \(X_0=0\). Then for all \(t\), \(X_t\) can be decomposed into \(X_t=mt+\sigma B_t+C_t+Y_t\) (11.23), where \(m\in\mathbb R\), \(\sigma\ge0\), and \(B_t,C_t,Y_t\) are independent random variables such that: (1) \(\{B_t\}\) is a standard Brownian motion; (2) \(\{C_t\}\) is a compound Poisson process with Lévy measure \(\mu_C\) such that \(\mu_C(\{|x|\le1\})=0\) and \(\mu_C(\mathbb R)=\mu_C(\{|x|>1\})<\infty\); (3) \(\{Y_t\}\) is a compensated generalized Poisson process with Lévy measure \(\mu_Y\) such that \(\mu_Y(\{0\})=0\) and \(\int_{-\infty}^\infty y^2\,d\mu_Y(y)<\infty\). Remark 11.5: the Lévy-Khintchine representation decomposes a Lévy process into a continuous part \((mt+\sigma B_t)\) and a pure jump part \((C_t+Y_t)\); in the pure jump part, \(C_t\) governs jumps of higher value (no less than 1) while \(Y_t\) governs jumps of lower value (less than 1).

命题 11.12（\(X_t\) 的特征指数）/ Proposition 11.12 (Characteristic exponent of \(X_t\)) 设 \(\{X_t\}\) 是满足 (11.23) 的 Lévy 过程，则其特征指数为Let \(\{X_t\}\) be a Lévy process satisfying (11.23); then its characteristic exponent is

$$\Psi_{X_t}(s)=t\left[ism-\frac{\sigma^2}2s^2+\left(\int_{-\infty}^\infty(e^{isy}-1)\,d\mu_C(y)\right)+\left(\int_{-\infty}^\infty(e^{isy}-1-isy)\,d\mu_Y(y)\right)\right].$$

证明：由 (11.6) \(\Psi_{mt+\sigma B_t}(s)=t(ism-\frac{\sigma^2}2s^2)\)；由 (11.11) \(\Psi_{C_t}(s)=t\int(e^{isy}-1)\,d\mu_C(y)\)；由 (11.22) \(\Psi_{Y_t}(s)=t\int_{-1}^1(e^{isy}-1-isy)\,d\mu_Y(y)\)。因 \(B_t,C_t,Y_t\) 独立，由命题 11.4 相加即得。\(\blacksquare\)Proof: by (11.6) \(\Psi_{mt+\sigma B_t}(s)=t(ism-\frac{\sigma^2}2s^2)\); by (11.11) \(\Psi_{C_t}(s)=t\int(e^{isy}-1)\,d\mu_C(y)\); by (11.22) \(\Psi_{Y_t}(s)=t\int_{-1}^1(e^{isy}-1-isy)\,d\mu_Y(y)\). Since \(B_t,C_t,Y_t\) are independent, summing by Proposition 11.4 gives the result. \(\blacksquare\)