高等概率论第七讲

记录本校的高等概率论课程笔记，参考教材《测度与概率》。

本讲介绍了Fubini定理与无穷乘积转移概率测度。

2.Fubini定理

定理4.2.1

$$设(E_1, \Sigma_1, \mu_1)和(E_2, \Sigma_2, \mu_2)是两个概率空间，\\ f是(E_1\times E_2, \Sigma_1\times \Sigma_2)上非负可测函数，则\\ \begin{aligned} \underset{E_1\times E_2}{\int \int} f(x_1,x_2)d(\mu_1\times \mu_2) =\int_{E_1} \mu_1(dx_1) \int_{E_2} f(x_1,x_2) \mu_2(dx_2) =\int_{E_2} \mu_2(dx_2) \int_{E_1} f(x_1,x_2) \mu_1(dx_1) \end{aligned}$$

证明：$f=1_B ,B\in \Sigma_1\times \Sigma_2$时，

$$\begin{aligned} \underset{E_1\times E_2}{\int \int} f(x_1,x_2)d(\mu_1\times \mu_2) &=\underset{E_1\times E_2}{\int \int} 1_B(x_1,x_2)d(\mu_1\times \mu_2) \\ &=\int_{E_2} \mu_1(B(x_2))\mu_2(dx_2) (乘积测度的定义)\\ &=\int_{E_2} \mu_2(dx_2) \int_{E_1} 1_{B{(x_2)}}(x_1) \mu_1(dx_1) \\ &=\int_{E_2} \mu_2(dx_2) \int_{E_1}1_{B}(x_1,x_2)\mu_1(dx_1) \\ &=\int_{E_2} \mu_2(dx_2) \int_{E_1} f(x_1,x_2) \mu_1(dx_1) \end{aligned}$$

当$f= \sum_{i=1}^n b_i 1_{B_i}$时，

$$\begin{aligned} \underset{E_1\times E_2}{\int \int} f(x_1,x_2)d(\mu_1\times \mu_2) &= \sum_{i=1}^nb_i \underset{E_1\times E_2}{\int \int} 1_{B_i}d(\mu_1\times \mu_2)\\ &=\sum_{i=1}^nb_i\int_{E_2} \mu_2(dx_2) \int_{E_1}1_{B_i}(x_1,x_2)\mu_1(dx_1)\\ &=\int_{E_2} \mu_2(dx_2) \int_{E_1} f(x_1,x_2) \mu_1(dx_1) \end{aligned}$$

当$f$为非负可测函数时，存在非负简单函数$f_n \uparrow f$，则

$$\begin{aligned} \underset{E_1\times E_2}{\int \int} f(x_1,x_2)d(\mu_1\times \mu_2) &= \lim_{n\to\infty}\underset{E_1\times E_2}{\int \int} f_nd(\mu_1\times \mu_2) \\ &=\lim_{n\to\infty}\int_{E_2} \mu_2(dx_2) \int_{E_1} f_n(x_1,x_2) \mu_1(dx_1) \\ &=\int_{E_2} \mu_2(dx_2) \lim_{n\to\infty}\int_{E_1} f_n(x_1,x_2) \mu_1(dx_1) \\ &=\int_{E_2} \mu_2(dx_2) \int_{E_1} f(x_1,x_2) \mu_1(dx_1) \end{aligned}$$

由上一定理可以得到如下定理：

定理4.2.2

$$设(E_1, \Sigma_1, \mu_1)和(E_2, \Sigma_2, \mu_2)是两个概率空间，\\ f是(E_1\times E_2, \Sigma_1\times \Sigma_2)上实可测函数，若\underset{E_1\times E_2}{\int \int} fd(\mu_1\times \mu_2)存在，则\\ \begin{aligned} \underset{E_1\times E_2}{\int \int} f(x_1,x_2)d(\mu_1\times \mu_2) =\int_{E_1} \mu_1(dx_1) \int_{E_2} f(x_1,x_2) \mu_2(dx_2) =\int_{E_2} \mu_2(dx_2) \int_{E_1} f(x_1,x_2) \mu_1(dx_1) \end{aligned}$$

证明：由上一定理可知

$$\underset{E_1\times E_2}{\int \int} f^{\pm}d(\mu_1\times \mu_2) =\int_{E_1} \mu_1(dx_1) \int_{E_2} f^{\pm}\mu_2(dx_2)$$

所以

$$\begin{aligned} \underset{E_1\times E_2}{\int \int} fd(\mu_1\times \mu_2) &= \underset{E_1\times E_2}{\int \int} f^+d(\mu_1\times \mu_2) -\underset{E_1\times E_2}{\int \int} f^-d(\mu_1\times \mu_2)\\ &=\int_{E_1} \mu_1(dx_1) \int_{E_2} f^+\mu_2(dx_2)- \int_{E_1} \mu_1(dx_1) \int_{E_2} f^-\mu_2(dx_2) \\ &=\int_{E_1} \mu_1(dx_1) \int_{E_2} (f^+- f^-)\mu_2(dx_2) \\ &=\int_{E_1} \mu_1(dx_1) \int_{E_2} f\mu_2(dx_2) \end{aligned}$$

例1

设$X$是$(\Omega, \mathcal F, \mathbb P)$上非负$r.v$列，则

$$\begin{aligned} \mathbb E [X] &=\int_{0}^{+\infty} \mathbb P(X> x) dx\\ &=\int_{0}^{+\infty}(1-F(x)) dx\\ &=\int_{0}^{+\infty} \mathbb P(X\ge x) dx \\ \end{aligned}$$

最后一个等号是因为$\mathbb P(X> x)$单调，最多有可数多个不连续点，从而

$$\int_{0}^{+\infty} \mathbb P(X> x) dx = \int_{0}^{+\infty} \mathbb P(X\ge x) dx$$

该例子的证明见一下题。

例2

$X$是非负随机变量，那么$\forall r >0$，

$$\mathbb E[X^r] = r \int_{0}^{+\infty} x^{r-1} \mathbb P(X >x) dx$$

证明：

$$\begin{aligned} \mathbb E[X^r] &=\int_{-\infty}^{+\infty} x^{r}d(F(x)) \\ &=\int_{-\infty}^{+\infty} x^{r}\mu_X(dx) \\ &=\int_{-\infty}^{0} x^{r}\mu_X(dx) +\int_{0}^{+\infty} x^{r}\mu_X(dx)\\ &=\int_{0}^{+\infty} x^{r}\mu_X(dx)\\ &=\int_{0}^{+\infty} \mu_X(dx) \int_{0}^x ry^{r-1}dy\\ &=\int_{0}^{+\infty} ry^{r-1}dy \int_y ^{\infty} \mu_X(dx)\\ &=\int_{0}^{+\infty} ry^{r-1}\mathbb P(X\ge y)dy \end{aligned}$$

例3

$$如果S(x)=\sum_{n=1}^{\infty} a_n(x),a_n(x)\ge 0\\ 那么\int_{a}^b S(x) dx = \int_{a}^b \sum_{n=1}^{\infty} a_n(x) dx= \sum_{n=1}^{\infty}\int_{a}^b a_n(x) dx$$

证明：首先将$ a_n(x)$改写为$ a(n,x)$，那么

$$\sum_{n=1}^{\infty} a_n(x)= \sum_{n=1}^{\infty} a(n,x)$$

$a$的映射关系如下

$$a: \mathbb N \times [a,b] \to \mathbb R^+$$

考虑$\mathbb N$上的计数测度

$$\mu(B)= \sharp B= |B|$$

所以

$$\int_{\{m\}} a(n,x)\mu(dn)=a(m,x)\int_{\{m\}} \mu(dn) =a(m,x)$$

从而

$$\begin{aligned} \sum_{n=1}^{\infty} a(n,x) &= \int_{\mathbb N} a(n,x)\mu(dn) \end{aligned}$$

因此由Fubini定理可得

$$\begin{aligned} \int_{a}^b \sum_{n=1}^{\infty} a_n(x) dx &= \int_{a}^b\int_{\mathbb N} a(n,x)\mu(dn) dx\\ &=\int_{\mathbb N}\int_{a}^b a(n,x)\mu(dn) dx\\ &=\sum_{n=1}^{\infty}\int_{a}^b a_n(x) dx \end{aligned}$$

定理4.2.3

$$(E_1, \Sigma_1),(E_2, \Sigma_2)是两个可测空间，\\ \mu_1是(E_1, \Sigma_1)上概率测度，\mathbb P(X,B)是E_1\times \Sigma_2上转移概率测度，\\ \mu是\Sigma_1\times \Sigma_2上如下定义的概率测度:\\ \mu(B)= \int_{E_1} \mathbb P(x,B) \mu_1(dx),B\in \Sigma_1\times \Sigma_2\\ f是\Sigma_1\times \Sigma_2上可测函数，若 \underset{E_1\times E_2}{\int \int}fd\mu存在，则\\ \underset{E_1\times E_2}{\int \int}fd\mu =\int_{E_1} \mu_1(dx)\int_{E_2}f(x,y)\mathbb P(x,dy)$$

证明思路依旧是从示性函数到简单函数再到非负可测函数最后到一般可测函数的步骤处理。

3.无穷乘积转移概率测度

定理4.3.1

$$设(E_n,\Sigma_n,\mu_n)(n\ge 1)是一列概率空间，\\则在 (\times_{n=1}^{\infty}E_n,\times_{n=1}^{\infty}\Sigma_n) 上存在唯一的概率测度\mu，满足\\ \mu(B_1\times ...\times B_n \times E_{n+1} \times E_{n+2}...) =\mu_1(B_1)\times...\times \mu_n(B_n)\\ \forall B_i \in \Sigma_i, 1\le i \le n ,\forall n \ge 1\\ 或等价地\\ \mu(B^{(n)}\times E_{n+1} \times E_{n+2}...)= (\times_{k=1}^n \mu_k)(B^{(n)})\\ \forall B^{(n)} \in \times_{i=1}^{n}E_i\\ 称\mu为\{\mu_n\}的乘积测度，记为\\ \mu= \mu_1\times \mu_2...=\times_{n=1}^{\infty}\mu_n$$

例1

$$设(E_n,\Sigma_n,\mu_n)(n\ge 1)是一列概率空间,\\ 定义(\Omega,\mathcal F, \mathbb P)= (\times_{n=1}^{\infty}E_n,\times_{n=1}^{\infty}\Sigma_n，\times_{n=1}^{\infty}\mu_n)\\ 第n个分量X_{n}(w)= w_n\\ 则\{X_n\}在\mathbb P下相互独立，X_n取值于E_n，分布是\mu_n$$

证明：取值于$E_n$以及独立性显然，计算概率

$$p=\mathbb P((w_1,...,w_n,w_{n+1},...) \in (E_1,...,E_{n-1},B_n,E_{n+1},...))$$

该概率等于

$$\begin{aligned} p &= \mu_1(E_1)...\mu_{n-1}(E_{n-1}) \mu_n(B_n)\mu_{n+1}(E_{n+1})...\\ &=\mu_n(B_n) \end{aligned}$$

习题

习题1

（课本P153/6.2/1）

证明：

$$\begin{aligned} \mathbb E[\xi^n] &= \int_{-\infty}^{+\infty} x^n d(F(x))\\ &= \int_{-\infty}^{+\infty} x^n \mu_X(dx)\\ &= \int_{0}^{+\infty} x^n \mu_X(dx) +\int_{-\infty}^{0} x^n \mu_X(dx)\\ &=\int_{0}^{+\infty}\mu_X(dx)\int_{0}^x n t^{n-1}dt -\int_{-\infty}^{0}\mu_X(dx)\int_{x}^0 nt^{n-1} dt\\ &=\int_{0}^{+\infty}\mu_X(dx)\int_{0}^{+\infty} nt^{n-1}1_{t\le x}dt- \int_{-\infty}^{0}\mu_X(dx)\int_{-\infty}^0 nt^{n-1}1_{x\le t} dt\\ &=\int_{0}^{+\infty}dt\int_{0}^{+\infty} nt^{n-1}1_{t\le x}\mu_X(dx)- \int_{-\infty}^{0}dt\int_{-\infty}^0 nt^{n-1}1_{x\le t} \mu_X(dx) \\ &=\int_{0}^{+\infty}dt\int_{t}^{+\infty} nt^{n-1}\mu_X(dx)- \int_{-\infty}^{0}dt\int_{-\infty}^t nt^{n-1} \mu_X(dx)\\ &=n\int_{0}^{+\infty}t^{n-1}(1-F(t))dt -n\int_{-\infty}^{0}t^{n-1}F(t)dt \end{aligned}$$

习题2

（课本P153/6.2/2）

证明：

$$\begin{aligned} \mathbb E[|X|] &= c\mathbb E[|\frac Xc|] \\ &=c\int_{0}^{+\infty}| \frac x c| \mu_X(dx) +c\int_{-\infty}^{0} |\frac x c|\mu_X(dx)\\ &=c\int_{0}^{+\infty} \frac x c \mu_X(dx) -c\int_{-\infty}^{0} \frac x c\mu_X(dx)\\ &=c\int_{0}^{+\infty}\mu_X(dx)\int_{0}^{\frac x c}dt +c\int_{-\infty}^{0}\mu_X(dx)\int_{\frac x c}^{0}dt\\ &=c\int_{0}^{+\infty}dt\int_{ct}^{+\infty} \mu_X(dx)- c\int_{-\infty}^{0}dt\int_{-\infty}^{ct} \mu_X(dx)\\ &=c\int_{0}^{+\infty}\mathbb P(X\ge ct)dt- c\int_{-\infty}^{0}\mathbb P(X\le ct)dt\\ &\overset{对第二个式子令t=-t}= c\int_{0}^{+\infty}\mathbb P(X\ge ct)dt- c\int_{0}^{+\infty}\mathbb P(X\le -ct)dt\\ &=c\int_{0}^{+\infty}\mathbb P(|X|\ge ct)dt\\ &=c\sum_{n=0}^{\infty} \int_{n}^{n+1}\mathbb P(|X|\ge ct)dt \end{aligned}$$

注意到

$$\mathbb P(|X|\ge c(n+1)) \le \int_{n}^{n+1} \mathbb P(|X|\ge ct) dt \le \mathbb P(|X|\ge cn)$$

所以

$$\sum_{n=1}^{\infty} \mathbb P(|X|\ge cn)=\sum_{n=0}^{\infty} \mathbb P(|X|\ge c(n+1)) \le \mathbb E[|X|] \le \sum_{n=1}^{\infty} \mathbb P(|X|\ge cn) +1$$

从而结论成立。

习题3

$$\mu(\prod_{i=1}^{\infty} B_i) = \prod_{i=1}^{\infty}\mu(B_i)$$

证明：记$A_n =\prod_{i=1}^{n} B_i$，则$A_n \downarrow \prod_{i=1}^{\infty} B_i$，由定义可知

$$\mu(A_n) = \prod_{i=1}^{n}\mu(B_i)$$

因为$\mu$是概率测度，所以

$$\begin{aligned} \lim_{n\to \infty} \mu(A_n) &=\lim_{n\to \infty} \prod_{i=1}^{n}\mu(B_i) \\ &=\mu(\lim_{n\to \infty} A_n)\\ &=\mu(\prod_{i=1}^{\infty} B_i) \end{aligned}$$

因此

$$\mu(\prod_{i=1}^{\infty} B_i) = \prod_{i=1}^{\infty}\mu(B_i)$$