概率论与数理统计(湖南大学教材《大学数学4(第三版)》)笔记(二)(持续更新中)

前导章节

第四节|分布函数

定义
  • X X X是一个随机变量, x x x是任意实数,称 F ( x ) = P { X ≤ x } F(x) = P \{ X \leq x \} F(x)=P{ Xx} X X X的分布函数
  • P { x 1 < X ≤ x 2 } = P { X ≤ x 2 } − P { X ≤ x 1 } = F ( x 2 ) − F ( x 1 ) P \{ x_{1} < X \leq x_{2} \} = P \{ X \leq x_{2} \} - P \{ X \leq x_{1} \} = F(x_{2}) - F(x_{1}) P{ x1<Xx2}=P{ Xx2}P{ Xx1}=F(x2)F(x1)
性质
  • 0 ≤ F ( x ) ≤ 1 , x ∈ R 0 \leq F(x) \leq 1 , x \in R 0F(x)1,xR

  • F ( − ∞ ) = lim ⁡ x → − ∞ F ( x ) = 0 F(- \infty) = \lim\limits_{x \rightarrow - \infty}{F(x)} = 0 F()=xlimF(x)=0 F ( + ∞ ) = lim ⁡ x → + ∞ F ( x ) = 1 F(+ \infty) = \lim\limits_{x \rightarrow + \infty}{F(x)} = 1 F(+)=x+limF(x)=1

  • F ( x ) F(x) F(x)是单调不减的函数,即 F ( x 1 ) ≤ F ( x 2 ) , x 1 < x 2 F(x_{1}) \leq F(x_{2}) , x_{1} < x_{2} F(x1)F(x2),x1<x2

    • 事实上, F ( x 2 ) − F ( x 1 ) = P { x 1 < X ≤ x 2 } ≥ 0 F(x_{2}) - F(x_{1}) = P \{ x_{1} < X \leq x_{2} \} \geq 0 F(x2)F(x1)=P{ x1<Xx2}0,故 F ( x 1 ) ≤ F ( x 2 ) F(x_{1}) \leq F(x_{2}) F(x1)F(x2)
  • F ( x ) F(x) F(x)右连续,即 F ( x ) = F ( x + 0 ) F(x) = F(x + 0) F(x)=F(x+0)

  • 满足上述 4 4 4个性质的函数也一定是某个随机变量的分布函数

离散型随机变量的分布函数
  • F ( x ) F(x) F(x)是离散型随机变量的分布函数,则 F ( x ) = P { X ≤ x } = P { ⋃ x k ≤ x ( X = x k ) } = ∑ x k ≤ x P { X = x k } = ∑ x k ≤ x p k F(x) = P \{ X \leq x \} = P \{ \bigcup\limits_{x_{k} \leq x}{(X = x_{k})} \} = \sum\limits_{x_{k} \leq x}{P \{ X = x_{k} \}} = \sum\limits_{x_{k} \leq x}{p_{k}} F(x)=P{ Xx}=P{ xkx(X=xk)}=xkxP{ X=xk}=xkxpk,其中和式是对于所有 x k ≤ x x_{k} \leq x xkx的指标 k k k求和

  • p k = P { X = x k } = F ( x k ) − F ( x k − 0 ) p_{k} = P \{ X = x_{k} \} = F(x_{k}) - F(x_{k} - 0) pk=P{ X=xk}=F(xk)F(xk0)

  • 若已知 X X X的分布函数 F ( x ) F(x) F(x),则 P { a ≤ X ≤ b } = P { a < X ≤ b } + P { X = a } = F ( b ) − F ( a ) + P { X = a } P \{ a \leq X \leq b \} = P \{ a < X \leq b \} + P \{ X = a \} = F(b) - F(a) + P \{ X = a \} P{ aXb}=P{ a<Xb}+P{ X=a}=F(b)F(a)+P{ X=a}

连续型随机变量的分布函数
  • X X X是连续型随机变量,其概率密度为 f ( x ) f(x) f(x),则 F ( X ) = P { X ≤ x } = ∫ − ∞ x f ( t ) d t F(X) = P \{ X \leq x \} = \displaystyle\int_{- \infty}^{x}{f(t) dt} F(X)=P{ Xx}=xf(t)dt,即分布函数是概率密度函数的可变积分上限的定积分
  • f ( x ) f(x) f(x)的连续点,有 d F ( x ) d x = f ( x ) \cfrac{d F(x)}{dx} = f(x) dxdF(x)=f(x),即概率密度函数是分布函数的导数
例题 1 1 1
  • 问题:证明:若 X ∼ N ( μ , σ 2 ) X \sim N (\mu , \sigma^{2}) XN(μ,σ2),则 Y = X − μ σ ∼ N ( 0 , 1 ) Y = \cfrac{X - \mu}{\sigma} \sim N (0 , 1) Y=σXμN(0,1)

  • 解答

    • P { a ≤ Y ≤ b } = P { a ≤ X − μ σ ≤ b } = P { μ + σ a ≤ X ≤ μ + σ b } = ∫ μ + σ a μ + σ b 1 σ 2 π e − ( x − μ ) 2 2 σ 2 d x P \{ a \leq Y \leq b \} = P \{ a \leq \cfrac{X - \mu}{\sigma} \leq b \} = P \{ \mu + \sigma a \leq X \leq \mu + \sigma b \} = \displaystyle\int_{\mu + \sigma a}^{\mu + \sigma b}{\cfrac{1}{\sigma \sqrt{2 \pi}} e^{- \frac{(x - \mu)^{2}}{2 \sigma^{2}}} dx} P{ aYb}=P{ aσXμb}=P{ μ+σaXμ+σb}=μ+σaμ+σbσ2π 1e2σ2(xμ)2dx

    • t = x − μ σ t = \cfrac{x - \mu}{\sigma} t=σxμ,得 P { a ≤ Y ≤ b } = ∫ a b 1 2 π e − t 2 2 d t P \{ a \leq Y \leq b \} = \displaystyle\int_{a}^{b}{\cfrac{1}{\sqrt{2 \pi}} e^{- \frac{t^{2}}{2}} dt} P{ aYb}=ab2π 1e2t2dt,则 Y ∼ N ( 0 , 1 ) Y \sim N (0 , 1) YN(0,1)

正态随机变量
  • X ∼ N ( μ , σ 2 ) X \sim N (\mu , \sigma^{2}) XN(μ,σ2),则 Y = a X + b ∼ N ( a μ + b , ∣ a ∣ 2 σ 2 ) Y = aX + b \sim N (a \mu + b , | a |^{2} \sigma^{2}) Y=aX+bN(aμ+b,a2σ2),正态随机变量的线性函数仍为正态随机变量

  • 对于标准正态分布,用 Φ ( x ) \Phi (x) Φ(x)表示分布函数

  • φ ( x ) \varphi (x) φ(x)是偶函数,即 φ ( − x ) = φ ( x ) \varphi (- x) = \varphi (x) φ(x)=φ(x),于是 Φ ( − x ) = 1 − Φ ( x ) \Phi (- x) = 1 - \Phi (x) Φ(x)=1Φ(x)

    • Φ ( − x ) = ∫ − ∞ − x φ ( t ) d t = ∫ x + ∞ φ ( u ) d u ( 令 t = − u ) = ∫ − ∞ + ∞ φ ( u )
评论 1
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值