概率论基础第七次作业

1.称一列随机变量$X_1,X_2,...$是独立的，如果对任意$n\in\mathbb{N}, X_1,X_2,...,X_n$是独立的。若$\mathscr{F} = \sigma(X_1), \mathscr{G} = \sigma(X_2,...,X_n,...)$，请证明$\mathscr{F}$与$\mathscr{G}$相互独立。

证明：Let \begin{equation} \mathscr{A} = \{\bigcap_{i\geq 2}A_i;\quad A_i\in \sigma(X_i)\}\notag \end{equation} First we check that $\mathscr{A}$ is a $\pi$-system:

$\forall B_1, B_2\in\mathscr{A}$, \begin{equation} B_1 = \bigcap_{i\geq 2}A_i^{(1)},\quad A_i^{(1)}\in \sigma(X_i)\\ B_2 = \bigcap_{i\geq 2}A_i^{(2)},\quad A_i^{(2)}\in \sigma(X_i)\\ B_1\cap B_2 = \bigcap_{i\geq 2}(A_i^{(1)}\cap A_i^{(2)}) \in\mathscr{A} \notag  \end{equation} since $\sigma(X_i)$ is closed under intersection.

It's obvious to see $\Omega\in\mathscr{A}$, now check that $\bigcup_{i\geq 2}\sigma(X_i)\subset\mathscr{A}$,

$\forall B\in \bigcup_{i\geq 2}\sigma(X_i)$, $\exists i_0\geq 2$ s.t. $B\in \sigma(X_{i_0})$, so $B\in\mathscr{A}$ which implies $\bigcup_{i\geq 2}\sigma(X_i)\subset\mathscr{A}$(this implies $\mathscr{G}\subset\sigma(\mathscr{A})$). So using (i) in Theorem 2.1.1 and Theorem 2.1.7, $\mathscr{F}$ and $\sigma(\mathscr{A})$ are independent. As $\mathscr{G}\subset\sigma(\mathscr{A})$, $\mathscr{F}$ and $\mathscr{G}$ are independent.

2.非空集合类$\mathscr{A}$与$\mathscr{B}$相互独立，$\sigma(\mathscr{A})$与$\sigma(\mathscr{B})$是否也一定相互独立呢？说明理由。

解：不一定。

Let $X_1,X_2,X_3$ be independent random variables with \begin{equation} P(X_i = 0) = P(X_i = 1) = 1/2\notag \end{equation} Let $A_1 = \{X_2 = X_3\},A_2 = \{X_3 = X_1\}$,and $A_3 = \{X_1 = X_2\}$.

Let $\mathscr{A} = \{A_1,A_2\}$,$\mathscr{B} = \{A_3\}$. $\mathscr{A}$ and $\mathscr{B}$ are independent since \begin{equation} P(A_1 \cap A_3) = P(A_2 \cap A_3) = 1/4 = P(A_1)P(A_3) = P(A_2)P(A_3)\notag \end{equation} Notice that $A_1\cap A_2 \in \sigma(\mathscr{A})$, $A_3\in \sigma(\mathscr{B})$ \begin{equation} P(A_1\cap A_2\cap A_3) = 1/4 \neq 1/8 = P(A_1\cap A_2)P(A_3)\notag \end{equation} So $\sigma(\mathscr{A})$ and $\sigma(\mathscr{B})$ are not independent.

3.设A和B是两个事件，令$X = I_A$，$Y = I_B$，证明X和Y不相关等价于它们独立。

证明： \begin{equation} \begin{aligned} E(XY) = E(I_A I_B) = E(I_{A\cap B}) = P(A\cap B)\\ E(X)E(Y) = E(I_A)E(I_B) = P(A)P(B)\notag \end{aligned} \end{equation} By above, X and Y are uncorrelated $\Leftrightarrow$ A and B are independent. A and B are independent $\Leftrightarrow$ $I_A$ and $I_B$ are independent by (ii) in Theorem 2.1.2 .

设$(\Omega,\mathscr{F},P)$是一个概率空间，X,Y是其上的随机变量，X $\sim$ Poisson(2), Y$\sim$Exp(2), 定义乘积空间$\Omega\times \Omega$上的函数 \begin{equation} \tilde{X}(\omega_1,\omega_2) = X(\omega_1),\qquad \tilde{Y}(\omega_1,\omega_2) = Y(\omega_2).\notag \end{equation}

4.证明$\tilde{X}$与$\tilde{Y}$都是乘积概率空间$(\Omega\times\Omega,\mathscr{F}\times\mathscr{F},P\times P)$上的随机变量。

证明：$\forall A,B\subset \mathbb{R}$, \begin{equation} \begin{aligned} {\tilde{X}}^{-1}(A) = \{(\omega_1,\omega_2) ; \quad\tilde{X}(\omega_1,\omega_2)\in A \} &= \{(\omega_1,\omega_2) ; \quad X(\omega_1)\in A \}\\ &= X^{-1}(A)\times\Omega \in \mathscr{F}\times\mathscr{F}\\ {\tilde{Y}}^{-1}(B) = \{(\omega_1,\omega_2) ; \quad\tilde{Y}(\omega_1,\omega_2)\in B \} &= \{(\omega_1,\omega_2) ; \quad Y(\omega_2)\in B \}\\ &= \Omega\times Y^{-1}(B) \in \mathscr{F}\times\mathscr{F}\notag \end{aligned} \end{equation}

5.计算${\tilde{X}}^2\tilde{Y}$的期望。

解：By Fubini's Theorem, \begin{equation} \begin{aligned} E{\tilde{X}}^2\tilde{Y} &= \int_{\Omega\times\Omega}{\tilde{X}}^2(\omega_1,\omega_2)\tilde{Y}(\omega_1,\omega_2)d(P\times P)\\ &= \int_{\Omega\times\Omega}X^2(\omega_1)Y(\omega_2)d(P\times P)\\ &= \int_{\Omega}X^2dP\int_{\Omega}YdP\\ &=EX^2EY= (2+2^2)\times\frac{1}{2} = 3\notag \end{aligned} \end{equation}

6.假设随机变量X和Y独立，且X $\sim$ Poisson(5), Y$\sim$N(0,1)，请计算X+Y的密度函数。

解：The density of Y is denoted $f$, the distribution function of X is denoted $G$, by Theorem 2.1.16, X+Y has density \begin{equation} \begin{aligned} h(z) &= \int f(z-y)dG(y) = \sum_{k=0}^{+\infty}f(z-k)G(k)\\ &= \sum_{k=0}^{+\infty}\frac{1}{\sqrt{2\pi}}\exp(-\frac{(z-k)^2}{2})\frac{5^k}{k!}\exp(-5) \notag \end{aligned} \end{equation} 

令F和G表示两个Stieltjes函数，$\mu$和$\nu$分别是它们对应的$(\mathbb{R},\mathscr{R})$上的测度。

7.证明 \begin{equation} \begin{aligned} \int_{(a,b]}F(y)dG(y)+\int_{(a,b]}G(y)dF(y) &= F(b)G(b) - F(a)G(a)\\ &+\sum_{x\in (a,b]}\mu(\{x\})\nu(\{x\}).\notag \end{aligned} \end{equation} 且将上等式与微积分中的“分部积分公式”进行比较分析。

证明：Let $f = I_{\{a<x\leq y\leq b\}}$, by Fubini's Theorem, \begin{equation} \begin{aligned} (\mu\times\nu)(\{(x,y);\quad a<x\leq y\leq b\}) = \int fd(\mu\times\nu) &= \iint f d\mu d\nu\\ &= \int_{(a,b]}(F(y)-F(a))dG(y)\notag \end{aligned} \end{equation} Similarly, let $g = I_{\{a<y\leq x\leq b\}}$, there will be \begin{equation} \begin{aligned} (\mu\times\nu)(\{(x,y);\quad a<y\leq x\leq b\}) = \int gd(\mu\times\nu) &= \iint g d\nu d\mu\\ &= \int_{(a,b]}(G(x)-G(a))dF(x)\\ &=\int_{(a,b]}(G(y)-G(a))dF(y)\notag \end{aligned} \end{equation} Add the two equations together, then we get \begin{equation} \begin{aligned} (\mu\times\nu)(\{(x,y);\quad (a,b]\times(a,b]\}) +\sum_{x\in (a,b]}\mu(\{x\})\nu(\{x\}) =& \int_{(a,b]}(F(y)-F(a))dG(y)\\ &+\int_{(a,b]}(G(y)-G(a))dF(y)\notag \end{aligned} \end{equation} Since $(\mu\times\nu)(\{(x,y);\quad (a,b]\times(a,b]\}) = (F(b)-F(a))(G(b)-G(a))$, the proposition is proved.

Integration by parts: Suppose $F$ and $G$ are differentiable functions on [a,b], $F' = f, G' = g$, $f$ and $g$ are Riemann integrable functions, then \begin{equation} \begin{aligned} \int_{a}^{b}F(x)g(x)dx + \int_{a}^{b}G(x)f(x)dx = F(b)G(b)-F(a)G(a)\notag \end{aligned} \end{equation} Compare the two formulas, notice there is one more part on the right of the target proof \begin{equation} \begin{aligned} \sum_{x\in (a,b]}\mu(\{x\})\nu(\{x\}) = \iint_{\{(x,x);\quad a<x\leq b\}}d(\mu\times\nu)\notag \end{aligned} \end{equation} When $F$ and $G$ fit the conditions of Integration by parts, this part is equal to 0. (Consider the geometric significance of integration, it's the volume of a 2D object)

8.若$\mu$是$\mathbb{R}$上的一个有限测度，请证明 \begin{equation} \begin{aligned} \int(F(x+c)-F(x))dx = c\mu(\mathbb{R})\notag \end{aligned} \end{equation} 如果将$dx$换成$\nu(dx)$，且$\nu$满足$\nu(A+x) = \nu(A), \forall A\in\mathscr{R}, x\in\mathbb{R}$，计算对应的积分值。

证明：Set $A_x = (x,x+c]$, Lebesgue measure is denoted $\lambda$ \begin{equation} \begin{aligned} \int(F(x+c)-F(x))dx &= \int(F(x+c)-F(x))\lambda(dx)\\ &=\int_{\mathbb{R}}\int_{\mathbb{R}}I_{A_x}\mu(dy)\lambda(dx)\\ &=\int_{\mathbb{R}}\int_{\mathbb{R}}I_{A_x}\lambda(dx)\mu(dy)\qquad (\mbox{Fubini's theorem})\\ &=\int_{\mathbb{R}}c\mu(dy)\\ &= c\mu(\mathbb{R})\notag \end{aligned} \end{equation} \begin{equation} \begin{aligned} \int(F(x+c)-F(x))\nu(dx) &=\int_{\mathbb{R}}\int_{\mathbb{R}}I_{A_x}\mu(dy)\nu(dx)\\ &=\int_{\mathbb{R}}\int_{\mathbb{R}}I_{A_x}\nu(dx)\mu(dy)\qquad (\mbox{Fubini's theorem})\\ &=\int_{\mathbb{R}}\nu(A_0)\mu(dy)\\ &= \nu(A_0)\mu(\mathbb{R})\\ &= \nu((0,c])\mu(\mathbb{R})\notag \end{aligned} \end{equation}