事件空间 Fsub2^(Omega)\mathcal{F} \subset 2^{\Omega} 是我们希望分配概率的样本空间的子集集合。一个集合 A inFA \in \mathcal{F} 称为事件;
P\mathbb{P} 是一种概率测度,它为每个事件 A inFA \in \mathcal{F} 指派一个概率 P(A)in[0,1]\mathbb{P}(A) \in[0,1] 。
示例 1.1.1. 考虑抛掷一个公平硬币两次的实验。我们模型 Omega={HH,HT,TH,TT}=\Omega=\{H H, H T, T H, T T\}={omega_(i):i=1,dots,4}\left\{\omega_{i}: i=1, \ldots, 4\right\} 和
{:[F=2^(Omega)={O/","Omega","{HH}","{HT}","{TH}","{TT}","{HH","HT}","{HH","TH}","{HH","TT}","{HT","TH}","{HT","TT}","],[{TH","TT}","{HH","HT","TH}","{HH","HT","TT}","{HH","TH","TT}","{HT","TH","TT}}]:}\begin{aligned}
& \mathcal{F}=2^{\Omega}=\{\varnothing, \Omega,\{H H\},\{H T\},\{T H\},\{T T\},\{H H, H T\},\{H H, T H\},\{H H, T T\},\{H T, T H\},\{H T, T T\}, \\
&\{T H, T T\},\{H H, H T, T H\},\{H H, H T, T T\},\{H H, T H, T T\},\{H T, T H, T T\}\}
\end{aligned}
由于硬币是公平的,每个 4 个基本事件 {HH},{HT},{TH},{TT}\{H H\},\{H T\},\{T H\},\{T T\} 具有相同的概率,即 1//41 / 4 ,对于所有 omega in Omega\omega \in \Omega 来说是 P({omega})=1//4\mathbb{P}(\{\omega\})=1 / 4 。其中一个不同结果 HTH T 和 THT H 发生的概率可以通过求和直观地获得,即 P({HT,TH})=\mathbb{P}(\{H T, T H\})=P({HT})+P({TH})=1//4+1//4=1//2\mathbb{P}(\{H T\})+\mathbb{P}(\{T H\})=1 / 4+1 / 4=1 / 2 。更一般地,对于 C sub{1,dots,n}C \subset\{1, \ldots, n\} ,我们可以设定
P({omega_(i):i in C})=P(uuu_(i in C){omega_(i)})=sum_(i in C)P({omega_(i)})=|C|//4.\mathbb{P}\left(\left\{\omega_{i}: i \in C\right\}\right)=\mathbb{P}\left(\bigcup_{i \in C}\left\{\omega_{i}\right\}\right)=\sum_{i \in C} \mathbb{P}\left(\left\{\omega_{i}\right\}\right)=|C| / 4 .
注意,然而根据这个定义,事件的并集的概率(意为“至少一个事件发生”)并不总是通过相加得到的:
P({HT}uu{HT,TH})=P({HT,TH})=1//2!=3//4=P({HT})+P({HT,TH}).\mathbb{P}(\{H T\} \cup\{H T, T H\})=\mathbb{P}(\{H T, T H\})=1 / 2 \neq 3 / 4=\mathbb{P}(\{H T\})+\mathbb{P}(\{H T, T H\}) .
原因在于:这些事件不是不相交的!在上述模型中,我们可以例如将“第一次掷出正面”识别为事件 {HH,HT}\{H H, H T\} ,其概率为 1//21 / 2 ,或者将“至少出现一次正面”识别为事件 {HT,TH,HH}\{H T, T H, H H\} ,其概率为 3//43 / 4 。对此进行重新表述可以是“反面未出现两次”,其概率可以表示为
引理 1.1.3。任何 sigma\sigma -代数 F\mathcal{F} 都具有以下性质: {i} O/ inF\varnothing \in \mathcal{F}
(ii) A,B inFLongrightarrow A uu B inFA, B \in \mathcal{F} \Longrightarrow A \cup B \in \mathcal{F}
(iii) A,B inFLongrightarrow A nn B inFA, B \in \mathcal{F} \Longrightarrow A \cap B \in \mathcal{F} (四) (A_(n))_(n inN)subFLongrightarrownnn_(n inN)A_(n)inF\left(A_{n}\right)_{n \in \mathbb{N}} \subset \mathcal{F} \Longrightarrow \bigcap_{n \in \mathbb{N}} A_{n} \in \mathcal{F}
(v) A,B inFLongrightarrow A\\B inFA, B \in \mathcal{F} \Longrightarrow A \backslash B \in \mathcal{F}
练习 1.1.6. 设 Omega\Omega 是一个可数集, E={{omega}:omega in Omega}\mathcal{E}=\{\{\omega\}: \omega \in \Omega\} 。证明 sigma(E)=2^(E)\sigma(\mathcal{E})=2^{\mathcal{E}} 。
For our statistical purposes the by far most important sigma\sigma-algebra is the Borel sigma\sigma-algebra B(R)\mathcal{B}(\mathbb{R}) over the real numbers R\mathbb{R}, which is defined as 对于我们的统计目的,迄今为止最重要的 sigma\sigma -代数是实数 R\mathbb{R} 上的 Borel sigma\sigma -代数 B(R)\mathcal{B}(\mathbb{R}) ,其定义为
B(R):=sigma({O subR:O" open "})\mathcal{B}(\mathbb{R}):=\sigma(\{O \subset \mathbb{R}: O \text { open }\})
and we will always implicitly equip R\mathbb{R} with this sigma\sigma-algebra when considering it as a measurable space. The Borel sigma\sigma-algebra has the following simpler characterisation. 我们在将 R\mathbb{R} 视为可测空间时,会始终隐式地为其装备这个 sigma\sigma -代数。Borel sigma\sigma -代数有以下更简单的表述。
Lemma 1.1.7. All of the following families of sets are generators of B(R)\mathcal{B}(\mathbb{R}) : 引理 1.1.7. 以下所有集合族都是 B(R)\mathcal{B}(\mathbb{R}) 的生成元:
(i) E_(1)={(a,b):-oo < a <= b < oo}\mathcal{E}_{1}=\{(a, b):-\infty<a \leq b<\infty\}E_(1)={(a,b):-oo < a <= b < oo}\mathcal{E}_{1}=\{(a, b):-\infty<a \leq b<\infty\}
(ii) E_(2)={[a,b]:-oo < a <= b < oo}\mathcal{E}_{2}=\{[a, b]:-\infty<a \leq b<\infty\}
(iii) E_(3)={(a,b]:-oo < a <= b < oo}\mathcal{E}_{3}=\{(a, b]:-\infty<a \leq b<\infty\}
(iv) E_(4)={(-oo,a):a inR}\mathcal{E}_{4}=\{(-\infty, a): a \in \mathbb{R}\}
(v) E_(5)={(-oo,a]:a inR}\mathcal{E}_{5}=\{(-\infty, a]: a \in \mathbb{R}\}
证明。我们只证明 sigma(E_(1))=B(R)\sigma\left(\mathcal{E}_{1}\right)=\mathcal{B}(\mathbb{R}) ,其余的陈述留作练习。由于任何开区间 (a,b)(a, b) 在 R\mathbb{R} 中都是开放的,我们有 E_(1)sub{O subR:O\mathcal{E}_{1} \subset\{O \subset \mathbb{R}: O 开放 }\} ,因此
sigma(E_(1))sub sigma({O subR:O" open "})=B(R)\sigma\left(\mathcal{E}_{1}\right) \subset \sigma(\{O \subset \mathbb{R}: O \text { open }\})=\mathcal{B}(\mathbb{R})
{O subR:O" open "}sub sigma(E_(1))\{O \subset \mathbb{R}: O \text { open }\} \subset \sigma\left(\mathcal{E}_{1}\right)
因此也
B(R)=sigma({O subR:O" open "})sub sigma(E_(1))\mathcal{B}(\mathbb{R})=\sigma(\{O \subset \mathbb{R}: O \text { open }\}) \subset \sigma\left(\mathcal{E}_{1}\right)
如果 A sub BA \subset B ,则 P(B\\A)=P(B)-P(A)\mathbb{P}(B \backslash A)=\mathbb{P}(B)-\mathbb{P}(A) 成立。特别是, P(A^(c))=1-P(A)\mathbb{P}\left(A^{\mathrm{c}}\right)=1-\mathbb{P}(A) 。
(ii) P(A uu B)=P(A)+P(B)-P(A nn B)\mathbb{P}(A \cup B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A \cap B) 。特别是, P(A uu B) <= P(A)+P(B)\mathbb{P}(A \cup B) \leq \mathbb{P}(A)+\mathbb{P}(B) 。
(iii)如果 A sub BA \subset B ,则 P(A) <= P(B)\mathbb{P}(A) \leq \mathbb{P}(B) 。
备注 1.1.13。我们在处理更一般的有限测度 mu\mu 时,可以推导出类似的语句。对于无限测度,我们需要更加小心。例如,虽然 mu(A uu B) <= mu(A)+mu(B)\mu(A \cup B) \leq \mu(A)+\mu(B) 总是成立,但只有在 mu(A nn B) < oo\mu(A \cap B)<\infty 的情况下,我们才能理解 mu(A)+mu(B)-mu(A nn B)\mu(A)+\mu(B)-\mu(A \cap B) ,在此情况下, mu(A uu B)=mu(A)+mu(B)-mu(A nn B)\mu(A \cup B)=\mu(A)+\mu(B)-\mu(A \cap B) 也确实成立。
证明。(i) A sub BA \subset B 蕴含 B=A⊎(B\\A)B=A \uplus(B \backslash A) ,因此通过 sigma\sigma -可加性, P(B)=P(B\\A)+P(A)\mathbb{P}(B)=\mathbb{P}(B \backslash A)+\mathbb{P}(A) ,或者等价地, P(B\\A)=P(B)-P(A)\mathbb{P}(B \backslash A)=\mathbb{P}(B)-\mathbb{P}(A) 。令 B=OmegaB=\Omega ,我们从中得到
(ii) 设 C=A nn BC=A \cap B 。我们可以写 A uu B=(A\\C)⊎(B\\C)⊎CA \cup B=(A \backslash C) \uplus(B \backslash C) \uplus C ,这给出(使用 C sub A,C sub BC \subset A, C \subset B 和 sigma\sigma - 可加性)
P(Omega∣A)=(P(Omega nn A))/(P(A))=(P(A))/(P(A))=1\mathbb{P}(\Omega \mid A)=\frac{\mathbb{P}(\Omega \cap A)}{\mathbb{P}(A)}=\frac{\mathbb{P}(A)}{\mathbb{P}(A)}=1
which establishes (ii). For (iii), note that we have A nnuuu_(n)A_(n)=uuu_(n)(A_(n)nn A)A \cap \bigcup_{n} A_{n}=\bigcup_{n}\left(A_{n} \cap A\right) and since (A_(n))_(n)\left(A_{n}\right)_{n} are pairwise disjoint, the same remains true for (A_(n)nn A)_(n)\left(A_{n} \cap A\right)_{n}. Thus, sigma\sigma-additivity of P\mathbb{P} yields 建立(ii)。对于(iii),注意到我们有 A nnuuu_(n)A_(n)=uuu_(n)(A_(n)nn A)A \cap \bigcup_{n} A_{n}=\bigcup_{n}\left(A_{n} \cap A\right) ,并且由于 (A_(n))_(n)\left(A_{n}\right)_{n} 是成对不相交的,因此 (A_(n)nn A)_(n)\left(A_{n} \cap A\right)_{n} 的情况仍然成立。因此, P\mathbb{P} 的 sigma\sigma -可加性得出
定理 1.2.5(贝叶斯公式)。设 A,B inFA, B \in \mathcal{F} 使得 P(A),P(B) > 0\mathbb{P}(A), \mathbb{P}(B)>0 。那么,对于任意 B inFB \in \mathcal{F} ,
obrace(P(B∣A))^("posterior ")=(P(A∣B) obrace(P(B))^("prior "))/(P(A))=(P(A∣B)P(B))/(P(A∣B)P(B)+P(A∣B^(c))P(B^(c)))\overbrace{\mathbb{P}(B \mid A)}^{\text {posterior }}=\frac{\mathbb{P}(A \mid B) \overbrace{\mathbb{P}(B)}^{\text {prior }}}{\mathbb{P}(A)}=\frac{\mathbb{P}(A \mid B) \mathbb{P}(B)}{\mathbb{P}(A \mid B) \mathbb{P}(B)+\mathbb{P}\left(A \mid B^{\mathrm{c}}\right) \mathbb{P}\left(B^{c}\right)}
证明。我们有 P(A∣B)=P(A nn B)//P(B)\mathbb{P}(A \mid B)=\mathbb{P}(A \cap B) / \mathbb{P}(B) ,或者等价地, P(A nn B)=P(A∣B)P(B)\mathbb{P}(A \cap B)=\mathbb{P}(A \mid B) \mathbb{P}(B) 。将这个表达式代入 P(A∣B)=P(A nn B)//P(B)\mathbb{P}(A \mid B)=\mathbb{P}(A \cap B) / \mathbb{P}(B) 中的 P(A nn B)\mathbb{P}(A \cap B) 得到第一个所述的等式。第二个等式是从第一个推导出来的,经过分解 A=(A nn B)⊎(A nnB^(c))A=(A \cap B) \uplus\left(A \cap B^{\mathrm{c}}\right) ,得出
1_(A):E rarrR,quad x|->{[1",",x in A],[0",",x!in A]:}\mathbf{1}_{A}: E \rightarrow \mathbb{R}, \quad x \mapsto \begin{cases}1, & x \in A \\ 0, & x \notin A\end{cases}
对于任何集合 B inB(R)B \in \mathcal{B}(\mathbb{R}) ,我们有
1_(A)^(-1)(B)={x in E:1_(A)(x)in B}={[A",",1in B","0!in B],[A^(c)",",0in B","1!in B],[E",",0","1in B],[O/",",0","1!in B]:}\mathbf{1}_{A}^{-1}(B)=\left\{x \in E: \mathbf{1}_{A}(x) \in B\right\}= \begin{cases}A, & 1 \in B, 0 \notin B \\ A^{\mathrm{c}}, & 0 \in B, 1 \notin B \\ E, & 0,1 \in B \\ \varnothing, & 0,1 \notin B\end{cases}
证明。(i) P_(X)(O/)=P(X^(-1)(O/))=P(O/)=0\mathbb{P}_{X}(\varnothing)=\mathbb{P}\left(X^{-1}(\varnothing)\right)=\mathbb{P}(\varnothing)=0 。
(ii) P_(X)(R)=P(X^(-1)(R))=P(Omega)=1\mathbb{P}_{X}(\mathbb{R})=\mathbb{P}\left(X^{-1}(\mathbb{R})\right)=\mathbb{P}(\Omega)=1.
(iii) It is easily checked that for a pairwise disjoint sequence (A_(n))_(n inN)subB(R)\left(A_{n}\right)_{n \in \mathbb{N}} \subset \mathcal{B}(\mathbb{R}), the sequence (X^(-1)(A_(n)))_(n inN)\left(X^{-1}\left(A_{n}\right)\right)_{n \in \mathbb{N}} is also disjoint and by measurability of XX it is a sequence in F\mathcal{F}. Consequently, (iii) 很容易检查,对于一对成对不相交的序列 (A_(n))_(n inN)subB(R)\left(A_{n}\right)_{n \in \mathbb{N}} \subset \mathcal{B}(\mathbb{R}) ,序列 (X^(-1)(A_(n)))_(n inN)\left(X^{-1}\left(A_{n}\right)\right)_{n \in \mathbb{N}} 也是不相交的,并且由于 XX 的可测性,它是序列 F\mathcal{F} 中的一个序列。因此,
where we used sigma\sigma-additivity of P\mathbb{P} for the third line. 我们在第三行中使用了 sigma\sigma 对 P\mathbb{P} 的加法性。
Definition 2.1.11. A random variable XX is called discrete if there exists N inNuu{oo}N \in \mathbb{N} \cup\{\infty\}, weights p_(1),dots,p_(N) >=p_{1}, \ldots, p_{N} \geq 0 and points x_(1),dots,x_(N)inRx_{1}, \ldots, x_{N} \in \mathbb{R} (both interpreted as sequences if N=ooN=\infty ) such that sum_(k=1)^(N)p_(k)=1\sum_{k=1}^{N} p_{k}=1 and 定义 2.1.11. 如果存在 N inNuu{oo}N \in \mathbb{N} \cup\{\infty\} 、权重 p_(1),dots,p_(N) >=p_{1}, \ldots, p_{N} \geq 0 和点 x_(1),dots,x_(N)inRx_{1}, \ldots, x_{N} \in \mathbb{R} (在 N=ooN=\infty 被解释为序列),则随机变量 XX 被称为离散的,使得 sum_(k=1)^(N)p_(k)=1\sum_{k=1}^{N} p_{k}=1 。
f_(X)(x)=P_(X)({x})=P(X=x)=sum_(k=1)^(N)p_(k)1_({x_(k)})(x),quad x inRf_{X}(x)=\mathbb{P}_{X}(\{x\})=\mathbb{P}(X=x)=\sum_{k=1}^{N} p_{k} \mathbf{1}_{\left\{x_{k}\right\}}(x), \quad x \in \mathbb{R}
either the discrete density function or probability mass function (pmf). 离散密度函数或概率质量函数 (pmf)。
Example 2.1.12. Consider a simple coin toss with probability of heads equal to p in[0,1]p \in[0,1]. We may model Omega={H,T},F=2^(Omega)={{H},{T},{H,T},O/}\Omega=\{H, T\}, \mathcal{F}=2^{\Omega}=\{\{H\},\{T\},\{H, T\}, \varnothing\} and P({H})=p\mathbb{P}(\{H\})=p and then set X(H)=1,X(T)=0X(H)=1, X(T)=0. The corresponding distribution P_(X)\mathbb{P}_{X} can then be written as 示例 2.1.12. 考虑一次简单的抛硬币,正面朝上的概率为 p in[0,1]p \in[0,1] 。我们可以建模 Omega={H,T},F=2^(Omega)={{H},{T},{H,T},O/}\Omega=\{H, T\}, \mathcal{F}=2^{\Omega}=\{\{H\},\{T\},\{H, T\}, \varnothing\} 和 P({H})=p\mathbb{P}(\{H\})=p ,然后设置 X(H)=1,X(T)=0X(H)=1, X(T)=0 。相应的分布 P_(X)\mathbb{P}_{X} 可以写成
and we call P_(X)\mathbb{P}_{X} a Bernoulli distribution with success rate pp. Alternatively, we can start with P_(X)\mathbb{P}_{X} as above and use it as a distributional model for the coin toss experiment with random outcome XX directly, without specifying the underlying probability space (Omega,F,P).^(1)(\Omega, F, \mathbb{P}) .{ }^{1} The latter modelling approach is particularly useful in situations, where a random phenomenon is far too complex to allow an explicit construction of the underlying probability space (think of a stock price for example) and is therefore the appropriate one for our statistical purposes. 我们称 P_(X)\mathbb{P}_{X} 为成功率为 pp 的伯努利分布。或者,我们可以像上面一样从 P_(X)\mathbb{P}_{X} 开始,并将其直接用作硬币投掷实验的分布模型,随机结果为 XX ,而不需要指定基础概率空间 (Omega,F,P).^(1)(\Omega, F, \mathbb{P}) .{ }^{1} 。后一种建模方法在随机现象过于复杂以至于无法明确构建基础概率空间的情况下特别有用(例如股票价格),因此在我们的统计目的中是合适的。
证明(草图)。一般来说,可以证明以下是正确的:如果 P,Q\mathrm{P}, \mathrm{Q} 是两个具有底层 sigma\sigma -代数 F\mathcal{F} 的概率测度,该代数由 nn -稳定的集合家族 E\mathcal{E} 生成,即,如果 F=sigma(E)\mathcal{F}=\sigma(\mathcal{E}) 和 A nn B inEA \cap B \in \mathcal{E} 成立,当 A,B inEA, B \in \mathcal{E} 时,则对于所有 B inEB \in \mathcal{E} , P=Q\mathbb{P}=\mathbb{Q} 当且仅当 P(B)=Q(B)\mathrm{P}(B)=\mathbb{Q}(B) 成立。在这里,我们有 B(R)=sigma(E)\mathcal{B}(\mathbb{R})=\sigma(\mathcal{E}) 对于 E={(oo,x]:x inR}\mathcal{E}=\{(\infty, x]: x \in \mathbb{R}\} ,参见引理 1.1.7,并且由于 (-oo,x]nn(-oo,y]=(-oo,x^^y](-\infty, x] \cap(-\infty, y]=(-\infty, x \wedge y] , E\mathcal{E} 是 nn -稳定的。因此, P_(X)=P_(Y)\mathbb{P}_{X}=\mathbb{P}_{Y} 当且仅当
AA x inR:quadubrace(P_(X)((-oo,x])ubrace)_(=F_(X)(x))=ubrace(P_(Y)((-oo,x])ubrace)_(=F_(Y)(x))\forall x \in \mathbb{R}: \quad \underbrace{\mathbb{P}_{X}((-\infty, x])}_{=F_{X}(x)}=\underbrace{\mathbb{P}_{Y}((-\infty, x])}_{=F_{Y}(x)}
命题 2.1.17 (累计分布函数的性质)。一个累计分布函数 F F_(X)F_{X} 具有以下性质:
f(x):=sum_(k=1)^(n)ubrace(alpha_(k)1_(A_(k))(x)ubrace)_(f_(k)(x)),quad x in Ef(x):=\sum_{k=1}^{n} \underbrace{\alpha_{k} \mathbf{1}_{A_{k}}(x)}_{f_{k}(x)}, \quad x \in E
Theorem 2.2.1. For any non-negative, measurable function f:E rarrRf: E \rightarrow \mathbb{R}, there exists an increasing sequence of simple functions (f_(n))_(n inN)\left(f_{n}\right)_{n \in \mathbb{N}} such that we have the pointwise convergence 定理 2.2.1. 对于任何非负的可测函数 f:E rarrRf: E \rightarrow \mathbb{R} ,存在一个简单函数的递增序列 (f_(n))_(n inN)\left(f_{n}\right)_{n \in \mathbb{N}} ,使得我们有逐点收敛。
AA x in E:lim_(n rarr oo)f_(n)(x)=f(x)\forall x \in E: \lim _{n \rightarrow \infty} f_{n}(x)=f(x)
证明。设
那么 f_(n)f_{n} 是简单的,对于任何 x inRx \in \mathbb{R} 明显成立 f_(n)(x)rarr f(x)f_{n}(x) \rightarrow f(x) 作为 n rarr oon \rightarrow \infty 。此外,如果 x inx \inf^(-1)([k2^(-n),(k+1)2^(-n)))f^{-1}\left(\left[k 2^{-n},(k+1) 2^{-n}\right)\right) ,对于某些 k <= n2^(n)-1k \leq n 2^{n}-1 ,那么 f_(n)(x)=k2^(-n) <= f(x)f_{n}(x)=k 2^{-n} \leq f(x) ,类似地, f_(n)(x)=n <= f(x)f_{n}(x)=n \leq f(x) 对于 x inf^(-1)([n,oo))x \in f^{-1}([n, \infty)) 。因此, f_(n) <= ff_{n} \leq f 。最后,由此得出
We call a measurable function f:E rarrRmuf: E \rightarrow \mathbb{R} \mu-integrable if int_(E)|f|dmu < oo\int_{E}|f| \mathrm{d} \mu<\infty and define the Lebesgue integral of an integrable function by 我们称一个可测函数为 f:E rarrRmuf: E \rightarrow \mathbb{R} \mu -可积的,如果 int_(E)|f|dmu < oo\int_{E}|f| \mathrm{d} \mu<\infty ,并通过以下方式定义可积函数的勒贝格积分:
int_(E)fdmu=int_(E)f^(+)dmu-int_(E)f^(-)dmu\int_{E} f \mathrm{~d} \mu=\int_{E} f^{+} \mathrm{d} \mu-\int_{E} f^{-} \mathrm{d} \mu
^(1){ }^{1} Note here that if we start with a distribution P_(X)\mathbb{P}_{X} on (R,B(R))(\mathbb{R}, \mathcal{B}(\mathbb{R})), we can always construct a probability space (Omega,F,P)(\Omega, \mathcal{F}, \mathbb{P}) and a random variable X:Omega rarrRX: \Omega \rightarrow \mathbb{R} such that P_(X)\mathbb{P}_{X} is the distribution of XX. Indeed, we may simply set Omega=R,F=B(R),P=P_(X)\Omega=\mathbb{R}, \mathcal{F}=\mathcal{B}(\mathbb{R}), \mathbb{P}=\mathbb{P}_{X} and X(omega)=omegaX(\omega)=\omega for all omega inR\omega \in \mathbb{R}. This is referred to as the canonical construction. 注意,这里如果我们从分布 P_(X)\mathbb{P}_{X} 开始在 (R,B(R))(\mathbb{R}, \mathcal{B}(\mathbb{R})) 上,我们总是可以构建一个概率空间 (Omega,F,P)(\Omega, \mathcal{F}, \mathbb{P}) 和一个随机变量 X:Omega rarrRX: \Omega \rightarrow \mathbb{R} 使得 P_(X)\mathbb{P}_{X} 是 XX 的分布。实际上,我们可以简单地为所有 omega inR\omega \in \mathbb{R} 设置 Omega=R,F=B(R),P=P_(X)\Omega=\mathbb{R}, \mathcal{F}=\mathcal{B}(\mathbb{R}), \mathbb{P}=\mathbb{P}_{X} 和 X(omega)=omegaX(\omega)=\omega 。这被称为标准构造。