【概率论基础进阶】多维随机变量及其分布-二维均匀分布和二维正态分布

二维均匀分布

定义：如果二维连续型随机变量$(X,Y)$的概率密度为
f ( x , y ) = { 1 A ( x , y ) ∈ G 0 其他 f(x,y)=\left\{

\right. f(x,y)=⎩ ⎨ ⎧​​A1​0​(x,y)∈G其他​
其中$A$是平面有界区域$G$的面积，则称$(X,Y)$服从区域$G$上的均匀分布

设$(X,Y)$在$G$上服从均匀分布，$D$是$G$中的一个部分区域，记它们的面积分别为$S_D$和$S_G$，则 P { ( X , Y ) ∈ D } = S D S G

P{(X,Y)∈D}=SG​SD​​​

例1：设二维连续型随机变量$(X,Y)$在区域$D$上服从均匀分布，其中
$$D=\left\{(x,y)|x^2+y^2\le 1\right\}$$
求

• $X$的边缘密度$f_X(x)$
• 条件概率密度$f_{Y|X}(y|x)$

区域$D$是半径为$1$的单位圆，其面积应为$\pi$，因此$(X,Y)$的联合密度
f ( x , y ) = { 1 π x 2 + y 2 ≤ 1 0 其他 f(x,y)=\left\{

\right. f(x,y)=⎩ ⎨ ⎧​​π1​0​x2+y2≤1其他​
有
f X ( x ) = ∫ − ∞ + ∞ f ( x , y ) d y = { ∫ − 1 − x 2 1 − x 2 1 π = 2 π 1 − x 2 − 1 ≤ x ≤ 1 0 其他 f_{X}(x)=\int_{-\infty}^{+\infty}f(x,y)dy=\left\{

∫−1−x21−x21π=2π1−x2−1≤x≤10其他" role="presentation" style="position: relative;">∫1−x2√−1−x2√1π=2π1−x2−−−−−√0−1≤x≤1其他$$\begin{array}{rlr}& {\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\frac{1}{\pi }=\frac{2}{\pi }\sqrt{1-x^2}& -1\le x\le 1\\ & 0& 其他\end{array}$$

\right. fX​(x)=∫−∞+∞​f(x,y)dy=⎩ ⎨ ⎧​​∫−1−x2 ​1−x2 ​​π1​=π2​1−x2 ​0​−1≤x≤1其他​

注意这里$-1\le x\le 1$的范围是根据$x^2+y^2\le 1$得到的，因此必须有等号

当$-1<x<1$时
f Y ∣ X ( y ∣ x ) = { 1 2 1 − x 2 − 1 − x 2 ≤ y ≤ 1 − x 2 0 其他 f_{Y|X}(y|x)=\left\{

\right. fY∣X​(y∣x)=⎩ ⎨ ⎧​​21−x2 ​1​0​−1−x2 ​≤y≤1−x2其他​

而这里$-1<x<1$，是由于条件概率的分母边缘概率，即$f_X(x)>0$，得到的，因此必须没有等号，有等号的时候可以代入边缘概率$f_X(1)=f_X(-1)=0$

二维正态分布

定义：如果二维连续型随机变量$(X,Y)$的概率密度为
f ( x , y ) = 1 2 π σ 1 σ 2 1 − ρ 2 exp ⁡ ( − 1 2 ( 1 − ρ 2 ) [ ( x − μ 1 ) 2 σ 1 2 − 2 ρ ( x − μ 1 ) ( y − μ 2 ) σ 1 σ 2 + ( y − μ 2 ) 2 σ 2 2 ] ) − ∞ < x < + ∞ , − ∞ < y < + ∞

f(x,y)​=2πσ1​σ2​1−ρ2 ​1​exp(−2(1−ρ2)1​[σ12​(x−μ1​)2​−σ1​σ2​2ρ(x−μ1​)(y−μ2​)​+σ22​(y−μ2​)2​])−∞<x<+∞,−∞<y<+∞​
其中${\mu }_1,{\mu }_2,{\sigma }_1>0,{\sigma }_2>0,-1<\rho <1$，均为常数，$\exp (x)$表示$e^x$，则称$(X,Y)$服从参数为${\mu }_1,{\mu }_2,{\sigma }_1,{\sigma }_2,\rho$的二维正态分布，记作$(X,Y)\sim N({\mu }_1;{\mu }_2;{\sigma }_1;{\sigma }_2;\rho )$

性质

设$(X,Y)\sim N({\mu }_1;{\mu }_2;{\sigma }_1^2;{\sigma }_2^2;\rho )$，则

• $X\sim N({\mu }_1,{\sigma }_1^2),Y\sim N({\mu }_2,{\sigma }_2^2)$
• $X$与$Y$相互独立的充分必要条件是$\rho =0$

如果$(X,Y)$二维正态分布可保证$X$与$Y$均正态，反之则不能成立，即已知$X$与$Y$均正态，并不能保证$(X,Y)$正态

• $aX+bY\sim N(a{\mu }_1+b{\mu }_2,a^2{\sigma }_1^2+2ab{\sigma }_1{\sigma }_2\rho +b^2{\sigma }_2^2)$
• 当 ∣ a b c d ∣ ≠ 0
  |abcd|" role="presentation" style="position: relative;">∣∣∣acbd∣∣∣$$\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|$$
  \ne 0 ∣ ∣​ac​bd​∣ ∣​=0时，$(aX+bY,cX+dY)$也一定为二维正态

在今后的数理统计中，常有
随机变量$X_1,X_2,\cdots ,X_n$相互独立，且$X_i\sim N(\mu ,{\sigma }^2)(i=1,2,\cdots ,n)$，则有
$$\sum _{i=1}^n{c}_i{X}_i\sim N(\sum _{i=1}^n{c}_i\mu ,\sum _{j=1}^n{c}_j^2{\sigma }^2)$$
随机变量$X_1,X_2,\cdots ,X_n$相互独立，且$X_i\sim N({\mu }_i,{\sigma }_i^2)(i=1,2,\cdots ,n)$，则有
$$\sum _{i=1}^n{c}_i{X}_i\sim N(\sum _{i=1}^n{c}_i{\mu }_i,\sum _{j=1}^n{c}_j^2{\sigma }_j^2)$$

例2：设二维随机变量$(X,Y)$的概率密度为
$$f(x,y)=\frac{1+\sin x\sin y}{2\pi }{e}^{-\frac{1}{2}(x^2+y^2)},-\infty <x<+\infty ,-\infty <y<+\infty$$
则关于$X$的边缘概率密度$f_X(x)=()$

f X ( x ) = ∫ − ∞ + ∞ 1 + sin ⁡ x sin ⁡ y 2 π e − 1 2 ( x 2 + y 2 ) d y = 1 2 π e − 1 2 x 2 ( ∫ − ∞ + ∞ e − 1 2 y 2 d y + ∫ − ∞ + ∞ sin ⁡ x sin ⁡ y e − 1 2 y 2 d y ) = 1 2 π e − 1 2 x 2 ( ∫ − ∞ + ∞ e − 1 2 y 2 d y + 0 ) 注意 e x 2 常用两种积分积不出来 , 这里单独讨论

fX​(x)​=∫−∞+∞​2π1+sinxsiny​e−21​(x2+y2)dy=2π1​e−21​x2(∫−∞+∞​e−21​y2dy+∫−∞+∞​sinxsinye−21​y2dy)=2π1​e−21​x2(∫−∞+∞​e−21​y2dy+0)注意ex2常用两种积分积不出来,这里单独讨论​
对于 I = ∫ − ∞ + ∞ e − 1 2 y 2 d y

I=∫−∞+∞e−12y2dy" role="presentation" style="position: relative;">I=∫+∞−∞e−12y2dy$$\begin{array}{r}I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-\frac{1}{2}{y}^2}dy\end{array}$$

I=∫−∞+∞​e−21​y2dy​，有
I 2 = ∫ − ∞ + ∞ e − 1 2 y 2 d y ∫ − ∞ + ∞ e − 1 2 x 2 d x D = { ( x , y ) ∣ x ∈ R , y ∈ R } = ∬ D e − 1 2 ( x 2 + y 2 ) d x d y D = { ( r , θ ) ∣ 0 ≤ r < + ∞ , 0 ≤ θ ≤ 2 π } = ∫ 0 2 π d θ ∫ 0 + ∞ e − 1 2 r 2 r d r = − ∫ 0 2 π d θ ∫ 0 + ∞ e − 1 2 r 2 d ( − 1 2 r 2 ) = − ∫ 0 2 π d θ ⋅ ( − 1 ) = 2 π

I2=∫−∞+∞e−12y2dy∫−∞+∞e−12x2dxD={(x,y)|x∈R,y∈R}=∬De−12(x2+y2)dxdyD={(r,θ)|0≤r<+∞,0≤θ≤2π}=∫02πdθ∫0+∞e−12r2rdr=−∫02πdθ∫0+∞e−12r2d(−12r2)=−∫02πdθ⋅(−1)=2π" role="presentation" style="position: relative;">I2=∫+∞−∞e−12y2dy∫+∞−∞e−12x2dxD={(x,y)|x∈R,y∈R}=∬De−12(x2+y2)dxdyD={(r,θ)|0≤r<+∞,0≤θ≤2π}=∫2π0dθ∫+∞0e−12r2rdr=−∫2π0dθ∫+∞0e−12r2d(−12r2)=−∫2π0dθ⋅(−1)=2π$$\begin{array}{rl}{I}^2& ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-\frac{1}{2}{y}^2}dy{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-\frac{1}{2}{x}^2}dxD=\left\{(x,y)|x\in R,y\in R\right\}\\ & =\underset{D}{\iint }{e}^{-\frac{1}{2}(x^2+y^2)}dxdyD=\left\{(r,\theta )|0\le r<+\mathrm{\infty },0\le \theta \le 2\pi \right\}\\ & ={\int }_0^{2\pi }d\theta {\int }_0^{+\mathrm{\infty }}{e}^{-\frac{1}{2}{r}^2}rdr\\ & =-{\int }_0^{2\pi }d\theta {\int }_0^{+\mathrm{\infty }}{e}^{-\frac{1}{2}{r}^2}d\left(-\frac{1}{2}{r}^2\right)\\ & =-{\int }_0^{2\pi }d\theta \cdot (-1)\\ & =2\pi \end{array}$$

I2​=∫−∞+∞​e−21​y2dy∫−∞+∞​e−21​x2dxD={(x,y)∣x∈R,y∈R}=D∬​e−21​(x2+y2)dxdyD={(r,θ)∣0≤r<+∞,0≤θ≤2π}=∫02π​dθ∫0+∞​e−21​r2rdr=−∫02π​dθ∫0+∞​e−21​r2d(−21​r2)=−∫02π​dθ⋅(−1)=2π​
因此有 I = 2 π

I=2π" role="presentation" style="position: relative;">I=2π−−√$$\begin{array}{r}I=\sqrt{2\pi }\end{array}$$

I=2π ​​，带回原式
$$f_X(x)=\frac{1}{\sqrt{2\pi }}{e}^{\frac{-x^2}{2}}$$

对于本题，显然$X\sim N(0,1),Y\sim N(0,1)$，即$X$与$Y$都服从标准正态，但$(X,Y)$不是二维正态分布

延伸两个公式
∫ 0 + ∞ e − x 2 d x = π 2

∫0+∞e−x2dx=π2" role="presentation" style="position: relative;">∫+∞0e−x2dx=π−−√2$$\begin{array}{r}{\int }_0^{+\mathrm{\infty }}{e}^{-x^2}dx=\frac{\sqrt{\pi }}{2}\end{array}$$

∫0+∞​e−x2dx=2π ​​​
证明：
I 2 = ∫ 0 + ∞ e − x 2 d x ⋅ ∫ 0 + ∞ e − y 2 d y D = { ( x , y ) ∣ x ≥ 0 , y ≥ 0 } = ∬ D e − ( x 2 + y 2 ) d x d y D = { ( r , θ ) ∣ 0 ≤ e < + ∞ , 0 ≤ θ ≤ π 2 } = ∫ 0 π 2 d θ ∫ 0 + ∞ r e − r 2 d r = π 4 I = π 2

I2=∫0+∞e−x2dx⋅∫0+∞e−y2dyD={(x,y)|x≥0,y≥0}=∬De−(x2+y2)dxdyD={(r,θ)|0≤e<+∞,0≤θ≤π2}=∫0π2dθ∫0+∞re−r2dr=π4I=π2" role="presentation" style="position: relative;">I2I=∫+∞0e−x2dx⋅∫+∞0e−y2dyD={(x,y)|x≥0,y≥0}=∬De−(x2+y2)dxdyD={(r,θ)|0≤e<+∞,0≤θ≤π2}=∫π20dθ∫+∞0re−r2dr=π4=π−−√2$$\begin{array}{rl}{I}^2& ={\int }_0^{+\mathrm{\infty }}{e}^{-x^2}dx\cdot {\int }_0^{+\mathrm{\infty }}{e}^{-y^{2}dy}D=\left\{(x,y)|x\ge 0,y\ge 0\right\}\\ & =\underset{D}{\iint }{e}^{-(x^2+y^2)}dxdyD=\left\{(r,\theta )|0\le e<+\mathrm{\infty },0\le \theta \le \frac{\pi }{2}\right\}\\ & ={\int }_0^{\frac{\pi }{2}}d\theta {\int }_0^{+\mathrm{\infty }}r{e}^{-r^2}dr\\ & =\frac{\pi }{4}\\ I& =\frac{\sqrt{\pi }}{2}\end{array}$$

I2I​=∫0+∞​e−x2dx⋅∫0+∞​e−y2dyD={(x,y)∣x≥0,y≥0}=D∬​e−(x2+y2)dxdyD={(r,θ)∣0≤e<+∞,0≤θ≤2π​}=∫02π​​dθ∫0+∞​re−r2dr=4π​=2π ​​​

作者：熊骏、曾祥洲
链接：反常积分∫∞0ex2dx的几种计算方法 - 道客巴巴 (doc88.com)

上面算 ∫ − ∞ + ∞ e − 1 2 y 2 d y

∫−∞+∞e−12y2dy" role="presentation" style="position: relative;">∫+∞−∞e−12y2dy$$\begin{array}{r}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-\frac{1}{2}{y}^2}dy\end{array}$$

∫−∞+∞​e−21​y2dy​，也可以套该式
∫ − ∞ + ∞ e − 1 2 y 2 d y = 2 ∫ 0 + ∞ e − ( y 2 ) 2 d y = 2 2 ∫ 0 + ∞ e − ( y 2 ) 2 d ( y 2 ) = 2 2 ⋅ π 2 = 2 π

∫−∞+∞e−12y2dy=2∫0+∞e−(y2)2dy=22∫0+∞e−(y2)2d(y2)=22⋅π2=2π" role="presentation" style="position: relative;">∫+∞−∞e−12y2dy=2∫+∞0e−(y2√)2dy=22–√∫+∞0e−(y2√)2d(y2–√)=22–√⋅π−−√2=2π−−√$$\begin{array}{rl}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-\frac{1}{2}{y}^2}dy& =2{\int }_0^{+\mathrm{\infty }}{e}^{-(\frac{y}{\sqrt{2}}{)}^2}dy\\ & =2\sqrt{2}{\int }_0^{+\mathrm{\infty }}{e}^{-(\frac{y}{\sqrt{2}}{)}^2}d\left(\frac{y}{\sqrt{2}}\right)\\ & =2\sqrt{2}\cdot \frac{\sqrt{\pi }}{2}\\ & =\sqrt{2\pi }\end{array}$$

∫−∞+∞​e−21​y2dy​=2∫0+∞​e−(2 ​y​)2dy=22 ​∫0+∞​e−(2 ​y​)2d(2 ​y​)=22 ​⋅2π ​​=2π ​​
还有一个公式
$${\int }_0^{+\infty }{x}^n{e}^{-x}dx=n!$$
其实就是分布积分法，然后代入上下限就行
∫ x n e − x d x = − x n e − x + n ∫ x n − 1 e − x d x = − x n e − x − n x n − 1 e − x + n ( n − 1 ) ∫ x n − 2 e − x d x 将分部积分步骤重复 n 次 = − x n e − x − n x n − 1 e − x − n ( n − 1 ) ( n − 2 ) e − x − ⋯ − n ( n − 1 ) ( n − 2 ) ⋯ 2 x e − x + n ! ∫ e − x d x = − x n e − x − n x n − 1 e − x − n ( n − 1 ) ( n − 2 ) e − x − ⋯ − n ( n − 1 ) ( n − 2 ) ⋯ 2 x e − x − n ! e − x + C 设 f ( x ) = x n = − e − x [ f ( x ) + f ′ ( x ) + ⋯ + f ( n ) ( x ) ] + C

∫xne−xdx=−xne−x+n∫xn−1e−xdx=−xne−x−nxn−1e−x+n(n−1)∫xn−2e−xdx将分部积分步骤重复n次=−xne−x−nxn−1e−x−n(n−1)(n−2)e−x−⋯−n(n−1)(n−2)⋯2xe−x+n!∫e−xdx=−xne−x−nxn−1e−x−n(n−1)(n−2)e−x−⋯−n(n−1)(n−2)⋯2xe−x−n!e−x+C设f(x)=xn=−e−x[f(x)+f′(x)+⋯+f(n)(x)]+C" role="presentation" style="position: relative;">∫xne−xdx=−xne−x+n∫xn−1e−xdx=−xne−x−nxn−1e−x+n(n−1)∫xn−2e−xdx将分部积分步骤重复n次=−xne−x−nxn−1e−x−n(n−1)(n−2)e−x−⋯−n(n−1)(n−2)⋯2xe−x+n!∫e−xdx=−xne−x−nxn−1e−x−n(n−1)(n−2)e−x−⋯−n(n−1)(n−2)⋯2xe−x−n!e−x+C设f(x)=xn=−e−x[f(x)+f′(x)+⋯+f(n)(x)]+C$$\begin{array}{rl}{\int }_{}^{}{x}^n{e}^{-x}dx& =-x^n{e}^{-x}+n{\int }_{}^{}{x}^{n-1}{e}^{-x}dx\\ & =-x^n{e}^{-x}-n{x}^{n-1}{e}^{-x}+n(n-1){\int }_{}^{}{x}^{n-2}{e}^{-x}dx\\ & 将分部积分步骤重复n次\\ & =-x^n{e}^{-x}-n{x}^{n-1}{e}^{-x}-n(n-1)(n-2)e^{-x}-\cdots -n(n-1)(n-2)\cdots 2x{e}^{-x}+n!{\int }_{}^{}{e}^{-x}dx\\ & =-x^n{e}^{-x}-n{x}^{n-1}{e}^{-x}-n(n-1)(n-2)e^{-x}-\cdots -n(n-1)(n-2)\cdots 2x{e}^{-x}-n!e^{-x}+C\\ & 设f(x)=x^n\\ & =-e^{-x}[f(x)+f^{\prime }(x)+\cdots +f^{(n)}(x)]+C\end{array}$$

∫​xne−xdx​=−xne−x+n∫​xn−1e−xdx=−xne−x−nxn−1e−x+n(n−1)∫​xn−2e−xdx将分部积分步骤重复n次=−xne−x−nxn−1e−x−n(n−1)(n−2)e−x−⋯−n(n−1)(n−2)⋯2xe−x+n!∫​e−xdx=−xne−x−nxn−1e−x−n(n−1)(n−2)e−x−⋯−n(n−1)(n−2)⋯2xe−x−n!e−x+C设f(x)=xn=−e−x[f(x)+f′(x)+⋯+f(n)(x)]+C​
代入上下限
∫ 0 + ∞ x n e − x d x = lim ⁡ u → + ∞ x n e − x d x = lim ⁡ u → + ∞ [ − e − x ( x n + n x n − 1 + n ( n − 1 ) x n − 2 + ⋯ + n ! x + n ! ) ] ∣ 0 u = lim ⁡ u → + ∞ − e − u ( u n + n u n − 1 + n ( n − 1 ) u n − 2 + ⋯ + n ! u + n ! ) + n !

∫0+∞xne−xdx=limu→+∞xne−xdx=limu→+∞[−e−x(xn+nxn−1+n(n−1)xn−2+⋯+n!x+n!)]|0u=limu→+∞−e−u(un+nun−1+n(n−1)un−2+⋯+n!u+n!)+n!" role="presentation" style="position: relative;">∫+∞0xne−xdx=limu→+∞xne−xdx=limu→+∞[−e−x(xn+nxn−1+n(n−1)xn−2+⋯+n!x+n!)]∣∣∣∣u0=limu→+∞−e−u(un+nun−1+n(n−1)un−2+⋯+n!u+n!)+n!$$\begin{array}{rl}{\int }_0^{+\mathrm{\infty }}{x}^n{e}^{-x}dx& =\underset{u\to +\mathrm{\infty }}{lim}{x}^n{e}^{-x}dx\\ & =\underset{u\to +\mathrm{\infty }}{lim}[-e^{-x}(x^n+n{x}^{n-1}+n(n-1)x^{n-2}+\cdots +n!x+n!)]{|}_0^u\\ & =\underset{u\to +\mathrm{\infty }}{lim}-e^{-u}(u^n+n{u}^{n-1}+n(n-1)u^{n-2}+\cdots +n!u+n!)+n!\end{array}$$

∫0+∞​xne−xdx​=u→+∞lim​xne−xdx=u→+∞lim​[−e−x(xn+nxn−1+n(n−1)xn−2+⋯+n!x+n!)]∣ ∣​0u​=u→+∞lim​−e−u(un+nun−1+n(n−1)un−2+⋯+n!u+n!)+n!​
由于$\forall n\in N\underset{x\to +\infty }{\lim }{x}^n{e}^{-x}=0$，故
$${\int }_0^{+\infty }{x}^n{e}^{-x}dx=n!$$
作者：乌里扬诺夫丶
链接：x^(n)e(-x)和x^(n)e(x)型积分公式 - 知乎 (zhihu.com)

CSDN话题挑战赛第2期
参赛话题：学习笔记