Chapter 9. Integration with Respect to a Probability Measure1

南京审计大学统计学研究生第一学期课程，《高等概率论》。

欢迎大家来我的github下载源码呀，https://github.com/Berry-Wen/statistics-note-system

Background

Let (Ω,A,P)(\Omega,\mathcal{A},P)(Ω,A,P) be a probability space.
We want to define the expectation, or what is equivalent, the “integral”, of general r.v.r.v.r.v..
We have of course already done this for r.v.sr.v.sr.v.s on a countable space Ω\OmegaΩ.
The general case (for arbitrary Ω\OmegaΩ) is more delicate.

Definition 9.1

A r.v.Xr.v. \ Xr.v. X is called simple if it takes on only a finite number of values and hence can be written in the form
X=∑i=1naiIAi(1)X = \sum_{i=1}^{n} a_i I_{A_i} \tag{1} X=i=1∑naiIAi(1)
Where ai∈Ra_i \in \Rai∈R, and Ai∈A,1≤i≤nA_i \in \mathcal{A},1\le i \le nAi∈A,1≤i≤n

若 Ak∈F,k=1,2,...,nA_k \in \mathcal{F},k=1,2,...,nAk∈F,k=1,2,...,n两两不交，且
∪k=1nAk=Ω,ak∈R^(1),k=1,2,...,n\cup_{k=1}^n A_k = \Omega,a_k \in \hat{\mathcal{R}}^{(1)},k=1,2,...,n∪k=1nAk=Ω,ak∈R^(1),k=1,2,...,n
则称函数
KaTeX parse error: Undefined control sequence: \notag at position 56: …d w \in \Omega \̲n̲o̲t̲a̲g̲ ̲
为 (Ω,F)(\Omega,\mathcal{F})(Ω,F)上的简单函数

Such an XXX is clearly measurable; (Why?) 思考这里为什么XXX是可测的

∀B∈BkX−1(B)={w:X(w)∈B}=∪ak∈B{w:X(w)=ak}=∪ak∈BAk∈F逆象的定义简单函数的定义简单函数的定义简单函数中Ak∈F⇒∪Ak∈BAk∈F\forall B \in \mathcal{B}^k \\ \begin{array}{ll} \begin{aligned} X^{-1}(B) &= \{w: X(w) \in B\} \\ &= \cup_{a_k \in B} \{w:X(w) = a_k\} \\ &= \cup_{a_k \in B} A_k \\ &\in \mathcal{F} \end{aligned} & \begin{array}{ll} \text{逆象的定义} \\ \text{简单函数的定义} \\ \text{简单函数的定义} \\ \text{简单函数中} A_k \in \mathcal{F} \Rightarrow \cup_{A_k \in B} A_k \in \mathcal{F} \end{array} \end{array} ∀B∈BkX−1(B)={w:X(w)∈B}=∪ak∈B{w:X(w)=ak}=∪ak∈BAk∈F逆象的定义简单函数的定义简单函数的定义简单函数中Ak∈F⇒∪Ak∈BAk∈F

由定理8.1可知XXX可测。

Conversely, if XXX is measurable and takes on the values a1,...,ana_1,...,a_na1,...,an it must have the representation (1) with Ai={X=ai}A_i = \{X=a_i\}Ai={X=ai} ;
A simple r.v.r.v.r.v. has of course many different representation of the form (1).

If XXX is simple, its expectation (or “integral” with respect to PPP) is the number
E{X}=∑i=1naiP(Ai)(2)E\{X\} = \sum_{i=1}^{n} a_i P(A_i) \tag{2} E{X}=i=1∑naiP(Ai)(2)
- This is also written ∫X(w)P(dw)\int X(w) P(dw)∫X(w)P(dw) and even more simply ∫XdP\int X dP∫XdP;
- A little algebra shows that E{X}E\{X\}E{X} does not depend on the particular representation (1) chosen for XXX. 练习：

Exercise

Let (Ω,A,P)(\Omega,\mathcal{A},P)(Ω,A,P) be a probability space
Let X:Ω→RX:\Omega \to \RX:Ω→R be such that it admits two representations
X=∑i=1naiIAiandX=∑j=1mbjIBjX = \sum_{i=1}^{n} a_i I_{A_i} \quad \text{and} \quad X = \sum_{j=1}^{m} b_j I_{B_j} X=i=1∑naiIAiandX=j=1∑mbjIBj
where ai,bj∈Ra_i,b_j \in \Rai,bj∈R , and Ai,Bj∈AA_i,B_j \in \mathcal{A}Ai,Bj∈A for all i,j.i,j.i,j. Show that
∑i=1naiP(Ai)=∑j=1mbjP(Bj)\sum_{i=1}^{n} a_i P(A_i) = \sum_{j=1}^{m} b_j P(B_j) i=1∑naiP(Ai)=j=1∑mbjP(Bj)

First, prove that ∪i=1nAi=∪j=1mBj\cup_{i=1}^{ n}A_i = \cup_{j=1}^{m}B_j∪i=1nAi=∪j=1mBj.

Assume
ai≠0Ai∩Aj=∅bj≠0Bi∩Bj=∅(i≠j)\begin{array}{lll} a_i \neq 0 & A_i \cap A_j = \empty \\ b_j \neq 0 & B_i \cap B_j = \empty \end{array} (i\neq j) ai=0bj=0Ai∩Aj=∅Bi∩Bj=∅(i=j)

∀w∈∪i=1nAi∃i0∈{1,2,...,n}s.t.X(w)=ai0≠0sow∈∪j=1mBjelseX(w)=0∴∪i=1nAi⊂∪j=1mBj∀w∈∪j=1mBj∃j0∈{1,2,...,n}s.t.X(w)=bj0≠0sow∈∪i=1nAielseX(w)=0∴∪j=1mBj⊂∪i=1nAi⇓∪j=1mBj=∪i=1nAi\begin{array}{lll} \hline \begin{array}{lll} \forall w \in \cup_{i=1}^{n} A_i \\ \exists i_0 \in \left\{ 1,2,...,n \right\} \\ s.t. \ X(w) = a_{i_0} \neq 0 \\ so \ w \in \cup_{j=1}^{m}B_j \\ \quad else \ X(w) = 0 \\ \therefore \ \cup_{i=1}^{n}A_i \subset \cup_{j=1}^{m}B_j \\ \end{array} & \begin{array}{lll} \forall w \in \cup_{j=1}^{m}B_j \\ \exists j_0 \in \left\{ 1,2,...,n \right\} \\ s.t. \ X(w) = b_{j_0} \neq 0 \\ so \ w \in \cup_{i=1}^{n}A_i \\ \quad else \ X(w) = 0 \\ \therefore \ \cup_{j=1}^{m}B_j \subset \cup_{i=1}^{n}A_i \end{array} \Downarrow \\ \cup_{j=1}^{m}B_j = \cup_{i=1}^{n}A_i\\ \hline \end{array} ∀w∈∪i=1nAi∃i0∈{1,2,...,n}s.t. X(w)=ai0=0so w∈∪j=1mBjelse X(w)=0∴ ∪i=1nAi⊂∪j=1mBj∪j=1mBj=∪i=1nAi∀w∈∪j=1mBj∃j0∈{1,2,...,n}s.t. X(w)=bj0=0so w∈∪i=1nAielse X(w)=0∴ ∪j=1mBj⊂∪i=1nAi⇓

Second, if AiBj≠∅A_i B_j \neq \emptysetAiBj=∅, for w∈AiBjX(w)=ai=bjw \in A_i B_j \quad X(w) = a_i = b_jw∈AiBjX(w)=ai=bj

X=∑i=1naiIAi=∑i=1naiIAi∩(∪i=1nAi)=∑i=1naiIAi∩(∪j=1nBj)=∑i=1naiI∪j=1mAiBj=∑i=1n∑j=1maiIAiBjX=∑j=1mbjIBj=∑j=1mbjIBj∩(∪j=1mBj)=∑j=1mbjIBj∩(∪i=1nAi)=∑j=1mbjI∪i=1nAiBj=∑i=1n∑j=1mbjIAiBj⇓for w∈AiBj≠∅,X(w)=ai=bj\begin{array}{lll} \hline \begin{aligned} X &= \sum_{i=1}^{n} a_i I_{A_i} \\ &= \sum_{i=1}^{n} a_i I_{A_i \cap ( \cup_{i=1}^n A_i)} \\ &= \sum_{i=1}^{n} a_i I_{A_i \cap ( \cup_{j=1}^n B_j)} \\ &= \sum_{i=1}^{n} a_i I_{ \cup_{j=1}^m A_i B_j} \\ &= \sum_{i=1}^{n} \sum_{j=1}^{m} a_i I_{A_i B_j} \\ \end{aligned} & \begin{aligned} X &= \sum_{j=1}^{m} b_j I_{B_j} \\ &= \sum_{j=1}^{m} b_j I_{B_j \cap ( \cup_{j=1}^m B_j)} \\ &= \sum_{j=1}^{m} b_j I_{B_j \cap ( \cup_{i=1}^n A_i)} \\ &= \sum_{j=1}^{m} b_j I_{ \cup_{i=1}^n A_i B_j} \\ &= \sum_{i=1}^{n} \sum_{j=1}^{m} b_j I_{A_i B_j} \\ \end{aligned}\\ \Downarrow \\ \text{for } w \in A_iB_j\neq \emptyset, X(w) = a_i = b_j \\ \hline \end{array} X=i=1∑naiIAi=i=1∑naiIAi∩(∪i=1nAi)=i=1∑naiIAi∩(∪j=1nBj)=i=1∑naiI∪j=1mAiBj=i=1∑nj=1∑maiIAiBj⇓for w∈AiBj=∅,X(w)=ai=bjX=j=1∑mbjIBj=j=1∑mbjIBj∩(∪j=1mBj)=j=1∑mbjIBj∩(∪i=1nAi)=j=1∑mbjI∪i=1nAiBj=i=1∑nj=1∑mbjIAiBj

如果 AiBj=∅A_iB_j = \emptyAiBj=∅ , aiP(AiBj)=bjP(AiBj)=0a_iP(A_iB_j)=b_jP(A_iB_j)=0aiP(AiBj)=bjP(AiBj)=0，不影响计算

Last,Prove EX=EYEX=EYEX=EY

EX=∑i=1naiP(Ai)=∑i=1naiP(Ai∩(∪i=1nAi))=∑i=1naiP(Ai∩(∪j=1mBj))=∑i=1naiP(∪j=1mAiBj)=∑i=1n∑j=1maiP(AiBj)EY=∑j=1mbjP(Bj)=∑j=1mbjP(Bj∩(∪j=1mBj))=∑j=1mbjP(Bj∩(∪i=1nAi))=∑j=1mbjP(∪i=1nAiBj)pairwise disjoint=∑i=1n∑j=1mbjP(AiBj)⇓∵ai=bjEX=EY\begin{array}{lll} \hline \begin{aligned} EX &= \sum_{i=1}^{n} a_i P(A_i) \\ &= \sum_{i=1}^{n} a_i P(A_i \cap (\cup_{i=1}^{n}A_i)) \\ &= \sum_{i=1}^{n} a_i P(A_i \cap ( \cup_{j=1}^{m} B_j)) \\ &= \sum_{i=1}^{n} a_i P( \cup_{j=1}^{m}A_i B_j) \\ &= \sum_{i=1}^{n} \sum_{j=1}^{m} a_i P(A_i B_j) \\ \end{aligned} & \begin{aligned} EY &= \sum_{j=1}^{m} b_j P(B_j) \\ &= \sum_{j=1}^{m} b_j P(B_j \cap (\cup_{j=1}^m B_j)) \\ &= \sum_{j=1}^{m} b_j P(B_j \cap ( \cup_{i=1}^{n}A_i)) \\ &= \sum_{j=1}^{m} b_j P( \cup_{i=1}^{n}A_i B_j) \ \text{ pairwise disjoint}\\ &= \sum_{i=1}^{n} \sum_{j=1}^{m} b_j P(A_i B_j) \\ \end{aligned} \\ \Downarrow \because a_i = b_j \\ EX=EY \\ \hline \end{array} EX=i=1∑naiP(Ai)=i=1∑naiP(Ai∩(∪i=1nAi))=i=1∑naiP(Ai∩(∪j=1mBj))=i=1∑naiP(∪j=1mAiBj)=i=1∑nj=1∑maiP(AiBj)⇓∵ai=bjEX=EYEY=j=1∑mbjP(Bj)=j=1∑mbjP(Bj∩(∪j=1mBj))=j=1∑mbjP(Bj∩(∪i=1nAi))=j=1∑mbjP(∪i=1nAiBj) pairwise disjoint=i=1∑nj=1∑mbjP(AiBj)

Remark 测度与概率—2.3节期望与积分 - 知乎 (zhihu.com)

Let X,YX,YX,Y be two simple r.v.sr.v.sr.v.s and β\betaβ a real number. We clearly have

X=∑i=1naiIAiY=∑j=1mbjIBjX = \sum_{i=1}^n a_i I_{A_i} \quad Y = \sum_{j=1}^m b_j I_{B_j}X=∑i=1naiIAiY=∑j=1mbjIBj
EX=∑i=1naiP(Ai)EY=∑j=1mbjP(Bj)EX = \sum_{i=1}^n a_i P(A_i) \quad EY = \sum_{j=1}^m b_j P(B_j)EX=∑i=1naiP(Ai)EY=∑j=1mbjP(Bj)
- E{βX}=βE{X}E\{\beta X\} = \beta E\{X\}E{βX}=βE{X};
  
  E{βX}=∑i=1nβaiP(Ai)=β∑i=1naiP(Ai)=βE{X}E\{\beta X\}=\sum_{i=1}^n \beta a_i P(A_i)=\beta \sum_{i=1}^n a_i P(A_i)=\beta E\{X\}E{βX}=∑i=1nβaiP(Ai)=β∑i=1naiP(Ai)=βE{X}
- E{X+Y}=E{X}+E{Y}E\{X+Y\}=E\{X\}+E\{Y\}E{X+Y}=E{X}+E{Y};
  
  E{X+Y}=∑i=1n∑j=1m(ai+bj)P(AiBj)=∑i=1n∑j=1maiP(AiBj)+∑i=1n∑j=1mbjP(AiBj)=∑i=1naiP(Ai)+∑j=1mbjP(Bj)=E{X}+E{Y}\begin{aligned} E\{X+Y\} &= \sum_{i=1}^n \sum_{j=1}^m (a_i + b_j) P(A_i B_j) \\ &= \sum_{i=1}^n \sum_{j=1}^m a_i P(A_i B_j) + \sum_{i=1}^n \sum_{j=1}^m b_j P(A_i B_j) \\ &= \sum_{i=1}^n a_i P(A_i) + \sum_{j=1}^m b_j P(B_j) \\ &= E\{X\} +E\{Y\} \end{aligned}E{X+Y}=i=1∑nj=1∑m(ai+bj)P(AiBj)=i=1∑nj=1∑maiP(AiBj)+i=1∑nj=1∑mbjP(AiBj)=i=1∑naiP(Ai)+j=1∑mbjP(Bj)=E{X}+E{Y}
- If X≤YX\le YX≤Y,then E{X}≤E{Y}E\{X\}\le E\{Y\}E{X}≤E{Y}.
  
  X≤Y⇒ai≤bjX\le Y \Rightarrow a_i \le b_jX≤Y⇒ai≤bj
  E{X}=∑i=1naiP(Ai)=∑i=1n∑j=1maiP(AiBj)≤∑i=1n∑j=1mbjP(AiBj)=∑j=1mbjP(Bj)=E{Y}\begin{aligned} E\{X\} = \sum_{i=1}^n a_i P(A_i) &= \sum_{i=1}^n \sum_{j=1}^m a_i P(A_i B_j) \\ &\le \sum_{i=1}^n \sum_{j=1}^m b_j P(A_i B_j) \\ &= \sum_{j=1}^m b_j P( B_j) \\ &= E\{Y\} \end{aligned}E{X}=i=1∑naiP(Ai)=i=1∑nj=1∑maiP(AiBj)≤i=1∑nj=1∑mbjP(AiBj)=j=1∑mbjP(Bj)=E{Y}
Thus, expectation is linear on the vector space of all simple r.v.sr.v.sr.v.s.

因此，对于向量空间中的简单随机变量，期望是线性的。
Next, we define expectation for positive r.v.sr.v.sr.v.s.
定义正的随机变量

For XXX positive,
- By this, we assume that XXX may take all values in [0,∞][0,\infty][0,∞], including +∞+\infty+∞;
  在这种假设下，随机变量XXX的取值为[0,∞][0,\infty][0,∞]
- This innocuous extension is necessary for the coherence of some of our further results.
  这种无害的扩展对于我们某些进一步结果的一致性是必需的。
  
  Let
  E{X}=sup⁡{E{Y}:Ya simple r.v. with 0≤Y≤X}(3)E\{X\} = \sup \{E\{Y\}: Y \text{ a simple r.v. with } 0\le Y \le X\} \tag{3} E{X}=sup{E{Y}:Y a simple r.v. with 0≤Y≤X}(3)
- This supremum always exists in [0,∞][0,\infty][0,∞].
  这个上确界在 [0,∞][0,\infty][0,∞] 上总是存在的
  
  Since expectation is a positive operator on the set of simple r.v.′sr.v.'sr.v.′s,
  既然期望是在简单随机变量集合上的正的运算
  
  it is clear that the definition above for E{X}E\{X\}E{X} coincides with Definition 9.1.
  则上面的关于期望的定义和定义9.1是一致的
  
  定义9.1里面的关于期望的定义为 E{X}=∑i=1naiP(Ai)E\{X\} = \sum_{i=1}^{n} a_i P(A_i)E{X}=∑i=1naiP(Ai)
  思考这里的一致性是为什么？

Remark

Note that E{X}≥0E\{X\}\ge 0E{X}≥0,but we can have E{X}=∞E\{X\}=\inftyE{X}=∞,even when XXX is never equal +∞+\infty+∞.
注意到 E{X}≥0E\{X\}\ge 0E{X}≥0 ,但是我们可以有 E{X}=∞E\{X\}=\inftyE{X}=∞，即使随机变量XXX不等于无穷。
Finally, let XXX be an arbitrary r.v.r.v.r.v..
最后，令 XXX 为任意的随机变量

Let X+=max(X,0)X−=−min(X,0)X^+ = max(X,0) \quad X^- = - min(X,0)X+=max(X,0)X−=−min(X,0).

Then
X=X+−X−∣X∣=X++X−X = X^+ - X^- \quad |X| = X^+ + X^- X=X+−X−∣X∣=X++X−
and X+,X−X^+,X^-X+,X− are positive r.v.sr.v.sr.v.s.

Definition 9.2

A r.v.Xr.v. \ Xr.v. X has a finite expectation (is “integrable”) if both E{X+}E\{X^+\}E{X+} and E{X−}E\{X^-\}E{X−} are finite.
若 E{X+}E\{X^+\}E{X+} 和 E{X−}E\{X^-\}E{X−} 都是有限的，则随机变量 XXX 期望有限（也叫做可积）

In this case, its expectation is the number 期望是两数之和
E{X}=E{X+}+E{X−}(4)E\{X\} = E\{X^+\} + E\{X^-\} \tag{4} E{X}=E{X+}+E{X−}(4)

also written ∫X(w)dP(w)\int X(w) dP(w)∫X(w)dP(w) or ∫XdP\int XdP∫XdP

If X>0X>0X>0 then X−=0X^- =0X−=0 and X+=XX^+=XX+=X and, since obviously E{0}=0E\{0\}=0E{0}=0,this definition coincides with (3)
如果 X>0X>0X>0 ，则 X−=0X^- =0X−=0 、X+=XX^+=XX+=X 、E{0}=0E\{0\}=0E{0}=0，在这种情况下，和式(3)一致

We write L1\mathcal{L}^1L1 to denote the set of all integrable r.v.sr.v.sr.v.s. (Sometimes we write L1(Ω,A,P)\mathcal{L}^1 (\Omega,\mathcal{A},P)L1(Ω,A,P) to remove any possible ambiguity.)
用 L1\mathcal{L}^1L1 代表所有可积的随机变量的集合

A r.v.Xr.v. \ Xr.v. X admits an expectation if E{X+}E\{X^+\}E{X+} and E{X−}E\{X^-\}E{X−} are not both equal to +∞+\infty+∞.
随机变量XXX有期望，则 E{X+}E\{X^+\}E{X+} 和 E{X−}E\{X^-\}E{X−} 不都等于正无穷。
- Then the expectation of XXX is still given by (4), with the conventions +∞+a=+∞+\infty+a=+\infty+∞+a=+∞ and −∞+a=−∞-\infty+a=-\infty−∞+a=−∞ when a∈Ra\in \Ra∈R.
  则XXX的期望还是和(4)式一样，因为 +∞+a=+∞+\infty+a=+\infty+∞+a=+∞ 、 −∞+a=−∞-\infty+a=-\infty−∞+a=−∞ a∈Ra\in \Ra∈R.
- If X≥0X\ge 0X≥0 this definition again coincides with (3)
  如果 X≥0X\ge 0X≥0 ，则(4)和(3)是一致的
- Note that if XXX admits an expectation, then E{X}∈[−∞,+∞]E\{X\} \in [-\infty,+\infty]E{X}∈[−∞,+∞], and XXX is integrable if and only if its expectation is finite.
  如果XXX有期望，则 E{X}∈[−∞,+∞]E\{X\} \in [-\infty,+\infty]E{X}∈[−∞,+∞],
  XXX 可积 ⇔\Leftrightarrow⇔ 期望有限

Remark 9.1

When Ω\OmegaΩ is finite or countable we have thus two different definitions for the expectation of a r.v.Xr.v. \ Xr.v. X, the one above and the one given in Chapter 5.
当 Ω\OmegaΩ 是有限或可数，则有两个不同的关于随机变量 XXX 的期望定义，一个是上面给出的，另一个是第五章给出的

In fact these two definitions coincides: it is enough to verify this for a simple r.v.Xr.v. \ Xr.v. X, and in this case the formulas (5.1) and (9.2) are identical.
事实上，这两种定义是一致的

E{X}=∑j∈T′jP(X=j)(5.1)E\{X\} = \sum_{j \in T'} j P(X=j) \tag{5.1} \\ E{X}=j∈T′∑jP(X=j)(5.1)

E{X}=∑i=1naiP(Ai)(9.2)E\{X\} = \sum_{i=1}^n a_i P(A_i) \tag{9.2} E{X}=i=1∑naiP(Ai)(9.2)

这个留作练习，下次课讲

Theorem 9.1

(a) L1\mathcal{L}^1L1 is a vector space, and expectation is a linear map on L1\mathcal{L}^1L1, and it is also positive (i.e.X≥0⇒E{X}≥0i.e. X \ge 0 \Rightarrow E\{X\}\ge 0i.e.X≥0⇒E{X}≥0).
L1\mathcal{L}^1L1 是一个向量空间，期望是L1\mathcal{L}^1L1上的线性映射，它也是正的。

If further 0≤X≤Y0\le X \le Y0≤X≤Y are two r.v.sr.v.sr.v.s and Y∈L1Y \in \mathcal{L}^1Y∈L1 and E{X}≤E{Y}E\{X\}\le E\{Y\}E{X}≤E{Y}.

(b) X∈L1X\in \mathcal{L}^1X∈L1 iff ∣X∣∈L1|X| \in \mathcal{L}^1∣X∣∈L1 and in this case ∣E{X}∣≤E{∣X∣}|E \left\{ X \right\}| \le E \left\{ |X| \right\}∣E{X}∣≤E{∣X∣}.

In particular any bounded r.v.r.v.r.v. is integrable.
© If X=YX = YX=Y almost surely (s.a.s.a.s.a.) , then E{X}=E{Y}E \left\{ X \right\}= E \left\{ Y \right\}E{X}=E{Y}.

X=Ya.s. if P(X=Y)=P({w:X(w)=Y(w)})=1X = Y \text{ a.s. if } P(X=Y)=P(\{w:X(w)=Y(w)\})=1 X=Y a.s. if P(X=Y)=P({w:X(w)=Y(w)})=1
(d) (Monotone convergence theorem): 单调收敛定理

If the r.v.sr.v.sr.v.s XnX_nXn are positive and increasing a.s.a.s.a.s. to XXX, then lim⁡n→∞E{Xn}=E{X}\lim_{n\to \infty}E \left\{ X_n \right\} = E \left\{ X \right\}limn→∞E{Xn}=E{X} (even if E{X}=∞E \left\{ X \right\}= \inftyE{X}=∞).
(e) (Fatou’s lemma):

If the r.v.sr.v.sr.v.s XnX_nXn satisfy Xn>YX_n > YXn>Y a.s.a.s.a.s. (Y∈L1Y \in \mathcal{L}^1Y∈L1), all nnn, we have
E{lim⁡n→∞inf⁡E{Xn}}≤lim⁡n→∞inf⁡E{Xn}E \left\{ \lim_{n \to \infty} \inf E \left\{ X_n \right\} \right\} \le \lim_{n \to \infty} \inf E \left\{ X_n \right\} E{n→∞liminfE{Xn}}≤n→∞liminfE{Xn}
In particular, if Xn≥0X_n \ge 0Xn≥0 a.s.a.s.a.s. then
E{lim⁡n→∞inf⁡Xn}≤lim⁡n→∞inf⁡E{Xn}E \left\{ \lim_{n \to \infty} \inf X_n \right\} \le \lim_{n \to \infty} \inf E \left\{ X_n \right\} E{n→∞liminfXn}≤n→∞liminfE{Xn}
(f) (Lebesgue’s dominated convergence theorem): 勒贝格控制收敛定理

If the r.v.sr.v.sr.v.s XnX_nXn converge a.s.a.s.a.s. to XXX and if ∣Xn∣≤Ya.s.∈L1|X_n|\le Y \ a.s. \in \mathcal{L}^1∣Xn∣≤Y a.s.∈L1, all nnn,

then Xn∈L1,X∈L1X_n \in \mathcal{L}^1, X \in \mathcal{L}^1Xn∈L1,X∈L1 and E{Xn}→E{X}E \left\{ X_n \right\} \to E \left\{ X \right\}E{Xn}→E{X}.

Statement

The a.s.a.s.a.s. equality between r.v.sr.v.sr.v.s is clearly an equivalence relation, and two equivalent (i.e. almost surely equal) r.v.s have the same expectation:
在随机变量间的几乎必然相等时一个等价关系，两个几乎必然相等的随机变量具有相同的表示。

Thus :

one can define a space L1L^1L1 by considering " L1\mathcal{L}^1L1 modulo this equivalence relation"
In other words, an element of L1L^1L1 is an equivalence class, that is a collection of all r.v.r.v.r.v. in L1\mathcal{L}^1L1 which are pairwise a.s.a.s.a.s. equal.
L1L^1L1 是一个相等的类，是所有在 L1\mathcal{L}^1L1 上的两两相等的类
In view of © above, one may speak of the “expectation” of the equivalence class (which is the expectation of any one element belonging to this class).
鉴于上面的（c），可以说等价类的“期望”（对属于该类的任何一个元素的期望）。
Since further the addition of r.v.sr.v.sr.v.s or the product of a r.v.r.v.r.v. by a constant preserve a.s.a.s.a.s. equality, the set L1L^1L1 is also a vector space.
随机变量的加法或者乘法by a constant preserve a.s.a.s.a.s. equality，则 L1L^1L1 集合也是一个向量空间。

Therefore, we commit the (innocuous) abuse of identifying a r.v.r.v.r.v. with its equivalence class, and commonly write X∈L1X \in L^1X∈L1 instead of X∈L1X \in \mathcal{L}^1X∈L1.
If 1≤p<∞1\le p < \infty1≤p<∞, we define Lp\mathcal{L}^pLp to be the space of r.v.sr.v.sr.v.s such that ∣X∣p∈L1|X|^p \in \mathcal{L}^1∣X∣p∈L1 ;

LpL^pLp is defined analogously to L1L^1L1.
LpL^pLp和L1L^1L1的定义类似

That is, LpL^pLp is Lp\mathcal{L}^pLp modulo the equivalence relation “almost surely”.
Put more simply, two elements of Lp\mathcal{L^p}Lp that are a.s.a.s.a.s. equal are considered to be representatives of one element of LpL^pLp.

Two auxiliary results.

Result 1

For every positive r.v. XXX there exists a sequence {Xn}n≥1\{X_n\}_{n\ge 1}{Xn}n≥1 of positive simple r.v.s which increases toward XXX as nnn increases to infinity.

An example of such a sequence is given by 为什么要定义为 k2n\frac{k}{2^n}2nk
Xn(w)={k2nifk2n≤X(w)<k+12nand0≤k≤n2n−1nifX(w)≥nX_n(w) = \left\{ \begin{array}{lll} \frac{k}{2^n} & if \ \frac{k}{2^n} \le X(w) < \frac{k+1}{2^n} \ and \ 0 \le k \le n2^{n}-1 \\ n & if \ X(w) \ge n \end{array} \right. Xn(w)={2nknif 2nk≤X(w)<2nk+1 and 0≤k≤n2n−1if X(w)≥n

Result 2

If XXX is a positive r.v., and if {Xn}n≥1\{X_n\}_{n\ge 1}{Xn}n≥1 is any sequence of positive simple r.v.s increasing to XXX, then E{Xn}E\{X_n\}E{Xn} increases to E{X}E\{X\}E{X}.

高等概率论 Chapter 9. Integration with Respect to a Probability Measure1相关推荐

高等概率论 Chapter 6 Construction of a Probability Measure
Chapter 6 Construction of a Probability Measure 南京审计大学统计学研究生第一学期课程,<高等概率论>. 欢迎大家来我的github下载源码呀 ...
高等概率论 Chapter 5. Random Variables on a Countable Space
Chapter 5 Random Variables on a Countable Space 南京审计大学统计学研究生第一学期课程,<高等概率论>. 欢迎大家来我的github下载源码呀 ...
高等概率论 Chapter 2. Axioms of Probability
Chapter 2. Axioms of Probability 欢迎大家来我的github下载源码呀,https://github.com/Berry-Wen/statistics-note-sys ...
高等概率论和随机过程比较本质的介绍
关于研究生阶段的概率论主干课程,说得简单一些,就是在测度论的基础之上,将本科学过的内容重新梳理,逐步加深.先前也提到过,在没学测度论的时候,很多基本的概率概念和定理的数学论证是没有办法严格给出的,相应 ...
实变函数与高等概率论--如何理解生成的σ代数
如何理解生成的σ代数? 现代概率论是基于σ代数讨论的, 因为它本身具有概率测度也就是P(x)所必须的性质, 但在面对一个集合时, 其本身并不一定是一个σ代数, 所以需要用一种比较方便的概念来确保σ代数 ...
高等概率论的一些学习心得兼推荐一些相关书籍 zz
学习概率已经有快2年了,几乎查阅了所有跟概率相关的书籍,到目前为止没有找到我认为特别好的.有人认为Feller的概率论及其应用是经典,我买了两本中译本,对我来说帮助不大.看了程士宏的测度论与概率论基础 ...
人工智能中的概率论与统计学修炼秘籍之著名教材
概率论与统计学的学习者众多,为了迎合不同学习者的需求,各种教材种类繁多.眼花缭乱.为此,非常有必要推荐一些常用的教材给人工智能学习人员,提升学习的效率,提高学习的效果.根据学习逐渐深入的顺序,本文将按 ...
纪念我逝去的概率论基础
Photo: from book The Unravelers 在数学系的研究生阶段有一门课,名字非常谦逊,叫做<概率论基础>.没错,不是神马高等概率论,也不是神马现代概率论,而是基础,仅 ...
张志华-统计机器学习-概率论导论
统计机器学习-概率论导论文章目录统计机器学习-概率论导论一. 复习二. 参数方法和非参数方法三. 测度空间的建立本节内容延续第一节的内容,进行简短回顾,并对概率论中概率测度相关知识进行介绍 ...

高等概率论 Chapter 9. Integration with Respect to a Probability Measure1

Chapter 9. Integration with Respect to a Probability Measure1

高等概率论 Chapter 9. Integration with Respect to a Probability Measure1相关推荐

最新文章

热门文章