Lecture 17 - 2025 / 4 / 17

Martingale

Definition (filter): =F0F1F2Fn\emptyset = \cal F_0 \sube F_1 \sube F_2 \sube \cdots \sube F_n 是一个概率空间上的递增 σ\sigma-代数。

例如 Fn=Z1,,Zn\mathcal F_n = Z_1, \cdots, Z_n,其中 ZiZ_i 是随机变量。

Definition (martingale): (Xi)(X_i) 是关于 (Fi)(\cal F_i) 的鞅,如果满足
E[XiFi1]=Xi1\mathbb E[X_i \mid \mathcal F_{i-1}] = X_{i-1}

Azuma Inequality

Lemma: 对于 r.v. XX,若 X1,E[X]=0|X| \le 1, \mathbb E[X] = 0,则 E[etX]et2/2\mathbb E[e^{tX}] \le e^{t^2/2}

根据凸性和 Taylor 展开,E[etX]12(et+et)et2/2\mathbb E[e^{tX}] \le \frac{1}{2}\left( e^t + e^{-t}\right) \le e^{t^2/2}

Theorem:(Xi)(X_i) 是关于 (Fi)(\mathcal F_i) 的鞅,Yi=XiXi1Y_i = X_i - X_{i-1} 是“差异”序列,如果 ci>0c_i > 0 使得 Yici|Y_i| \le c_i,则
Pr[XnX0+λ]Pr[XnX0λ]  exp(λ22i=1nci2)\begin{aligned} \Pr[X_n \ge X_0 + \lambda] \\ \Pr[X_n \le X_0 - \lambda] \end{aligned} \quad \le \ \ \exp\left( -\frac{\lambda^2}{2 \sum_{i=1}^{n} c_i^2} \right)

n=1n = 1 时,X1X0c1|X_1 - X_0| \le c_1,则
Pr[X1X0+λ]=mintPr[et(X1X0)etλ]mintE[et(X1X0)]etλmintexp(c12t22tλ)=exp(λ22c12)\begin{aligned} \Pr[X_1 \ge X_0 + \lambda] & = \min_t \Pr[e^{t(X_1 - X_0)} \ge e^{t \lambda}]\\ & \le \min_t \frac{\mathbb E[e^{t(X_1 - X_0)}]}{e^{t\lambda}}\\ & \le \min_t \exp \left(\frac{c_1^2t^2}{2} - t \lambda \right)= \exp\left(-\frac{\lambda^2}{2c_1^2} \right) \qquad \\ \end{aligned}

接下来归纳,
Pr[XnX0+λ]mintE[et(XnXn1)et(Xn1X0)]etλ=mintEFn1[E[et(XnXn1)Fn1]et(Xn1X0)]etλmintecn2t2/2E[et(Xn1X0)]etλmintecn2t2/2exp(λ2/2i=1n1ci2)etλexp(λ22i=1nci2)\begin{aligned} \Pr[X_n \ge X_0 + \lambda ] & \le \min_t \frac{\mathbb E[e^{t(X_n - X_{n-1})} \cdot e^{t(X_{n - 1} - X_0)}]}{e^{t \lambda}} \\ & = \min_t \frac{\mathbb E_{\mathcal F_{n-1}}[\mathbb E[e^{t(X_n - X_{n-1})} \mid \mathcal F_{n-1}] \cdot e^{t(X_{n - 1} - X_0)}]}{e^{t \lambda}}\\ & \le \min_t \frac{e^{c_{n}^2 t^2 /2} \cdot\mathbb E[ e^{t(X_{n - 1} - X_0)}]}{e^{t \lambda}}\\ & \le \min_t \frac{e^{c_{n}^2 t^2 /2} \cdot \exp(-\lambda^2 /2\sum_{i=1}^{n-1} c_i^2) }{e^{t \lambda}}\\ & \le \exp\left( -\frac{\lambda^2}{2 \sum_{i=1}^{n} c_i^2} \right) \end{aligned}

Doob Martingale

Claim:A,(Zi)A, (Z_i) 是 r.v.,则 Xi=E[AZ1,,Zi]X_i = \mathbb E[A \mid Z_1, \cdots, Z_i] 是鞅,称之为 AA 的 Doob 鞅

验证定义即可:
E[XiZ1,,Zi1]=EZi[E[XiZ1,,Zi]Z1,,Zi1]=EZi[E[AZ1,,Zi]Z1,,Zi1]=E[AZ1,,Zi1]=Xi1\begin{aligned} \mathbb E[X_i \mid Z_1, \cdots, Z_{i-1}] & = \mathbb E_{Z_i}[ \mathbb E[X_i \mid Z_1, \cdots, Z_{i}] \mid Z_1, \cdots, Z_{i-1} ]\\ & = \mathbb E_{Z_i}[\mathbb E[A \mid Z_1, \cdots, Z_{i}] \mid Z_1, \cdots, Z_{i-1} ]\\ & = \mathbb E[A \mid Z_1, \cdots, Z_{i-1}] = X_{i-1} \end{aligned}

Definition: f(Z1,,Zn)f(Z_1, \cdots, Z_n)cc-Lipschitz 函数,当且仅当改变 ff 的任何一个坐标值,ff 的变化绝对值不超过 ±c\pm c

Lemma: 如果 ffcc-Lipschitz 函数,给定 Z1,,Zi1Z_1, \cdots, Z_{i-1} 的条件下,ZiZ_iZi+1,,ZnZ_{i+1}, \cdots, Z_n 相互独立,则 ff 关于 ZiZ_i 的 Doob 鞅 (Xi)(X_i) 满足 XiXi1c|X_i - X_{i-1}| \le c

我们根据定义对 XiXi1|X_i - X_{i-1}| 进行展开
=EZi+1,,Zn[fZ1,,Zi]EZi,,Zn[fZ1,,Zi1]=EZi+1,,Zn[fZ1,,Zi]EZi+1,,Zn[EZi[fZ1,,Zi1,Zi+1,,Zn]Z1,,Zi1]=EZi+1,,Zn[f(Z1,,Zi,,Zn)Z1,,Zi1]EZi+1,,Zn[EZi[f(Z1,,Zn)Z1,,Zi1,Zi+1,,Zn]Z1,,Zi1]=EZi+1,,Zn[f(Z1,,Zi,,Zn)EZi[f(Z1,,Zi,,Zn)Z1,,Zi1,Zi+1,Zn]Z1,,Zi1]=EZi+1,,Zn[EZi[f(Z1,,Zi,,Zn)f(Z1,,Zi,,Zn)Z1,,Zi1,Zi+1,Zn]Z1,,Zi1]\begin{aligned} & = | \mathbb E_{Z_{i+1}, \cdots, Z_n}[f \mid Z_1, \cdots, {\color{blue}Z_i}] - \mathbb E_{{\color{red}Z_i}, \cdots, Z_{n}}[f \mid Z_1, \cdots, Z_{i-1}]|\\ & = | \mathbb E_{Z_{i+1}, \cdots, Z_n}[f \mid Z_1, \cdots, {\color{blue}Z_i}]- \mathbb E_{Z_{i+1}, \cdots, Z_n} [\mathbb E_{{\color{red}Z_i}}[f \mid Z_1, \cdots, Z_{i-1}, Z_{i+1}, \cdots, Z_n] \mid Z_1, \cdots, Z_{i-1}]|\\ & = | \mathbb E_{Z_{i+1}, \cdots, Z_n}[f(Z_1, \cdots, {\color{blue}Z_i}, \cdots, Z_n) \mid Z_1, \cdots, Z_{i-1}]\\ & \quad - \mathbb E_{Z_{i+1}, \cdots, Z_n} [\mathbb E_{{\color{red}Z_i}}[f(Z_1, \cdots, Z_n) \mid Z_1, \cdots, Z_{i-1}, Z_{i+1}, \cdots, Z_n] \mid Z_1, \cdots, Z_{i-1}]|\\ & = | \mathbb E_{Z_{i+1}, \cdots, Z_n}[f(Z_1, \cdots, {\color{blue}Z_i}, \cdots, Z_n)\\ & \quad - \mathbb E_{\color{red}Z_i}[f(Z_1,\cdots,{\color{red}Z_i}, \cdots, Z_n) \mid Z_1, \cdots, Z_{i-1}, Z_{i+1}\cdots, Z_n] \mid Z_1, \cdots, Z_{i-1}]|\\ & = | \mathbb E_{Z_{i+1}, \cdots, Z_n}[ \mathbb E_{\color{red} Z_i} [f(Z_1, \cdots, {\color{blue}Z_i}, \cdots, Z_n) - f(Z_1,\cdots,{\color{red}Z_i}, \cdots, Z_n) \\ &\qquad \mid Z_1, \cdots, Z_{i-1}, Z_{i+1}\cdots, Z_n] \mid Z_1, \cdots, Z_{i-1}]|\\ \end{aligned}

注意这里 Zi\color{blue} Z_i 是已知量,而 Zi\color{red} Z_i 是未知量,可以看作两者是独立同分布的变量。从而每一项均 c\le c,由此结论成立。

Applications: Balls and Bins

mm 个球 nn 个桶,ZiZ_iii 号球选择的桶,X=f(Z1,,Zm)X = f(Z_1, \cdots, Z_m) 是空桶的个数。容易看出 ff11-Lipschitz 的,从而

Pr[XE[X]λ]2exp(λ22m)\Pr[| X - \mathbb E[X] | \ge \lambda] \le 2 \exp\left ( - \frac{\lambda^2}{2m} \right )

这是 Chernoff bound 所不能得到的结论。

Applications: Chromatic Number of Gn,1/2\mathcal G_{n, 1/2}

染色数 χ(G)\chi(G) 代表最少需要的颜色数量,使得存在一组同色不相邻的方案。

对于随机图我们有两种常见的鞅。

Edge Exposure Martingale:Zi=0/1Z_i = 0/1 表示第 ii 条边是否在图中出现,则 A=f(Z1,,Z(n2))A = f\left(Z_1, \cdots, Z_{\binom n 2}\right) 的 Doob 鞅是 edge exposure maringle。

Vertex Exposure Martingle:Zi{0,1}niZ_i \in \{0, 1\}^{n-i} 代表是否 iijj(满足 j>ij>i)的边是存在的,则 A=f(Z1,,Zn)A = f(Z_1, \cdots, Z_n) 的 Doob 鞅是 vertex exposure martingle。

这里我们使用后者,用 X=f(Z1,,Zn)X = f(Z_1, \cdots, Z_n) 代表 χ(G)\chi(G),则容易看出 ff11-Lipschitz 的。从而

Pr[XE[X]λ]2exp(λ22n)\Pr[| X - \mathbb E[X] | \ge \lambda] \le 2 \exp\left ( - \frac{\lambda^2}{2n} \right )

注意我们不依赖任何关于 E[X]\mathbb E[X] 的知识,给出了一个 concentration bound。