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例题

概率论

证明独立

X1,X2\displaystyle X_{1}, X_{2} 相互独立，均服从标准正态分布，证明 (X1+X2)2\displaystyle {\left( X_{1}+ X_{2}\right)}^{2} 和 (X1−X2)2\displaystyle {\left( X_{1}- X_{2}\right)}^{2} 独立

由于样本均值和样本方差相互独立，有

X‾=1n∑i=0nXi=12(X1+X2)S2=1n−1∑i=0n(Xi−X‾)2=(X1−X‾)2+(X2−X‾)2=12(X1−X2)2\begin{aligned}\displaystyle \overline{X}&=\frac{1}{n}\sum_{i= 0}^{n} X_{i}=\frac{1}{2}{\left( X_{1}+ X_{2}\right)} \\ \displaystyle S^{2}&=\frac{1}{n- 1}\sum_{i= 0}^{n}{\left( X_{i}-\overline{X}\right)}^{2}={\left( X_{1}-\overline{X}\right)}^{2}+{\left( X_{2}-\overline{X}\right)}^{2} \\ \displaystyle &=\frac{1}{2}{\left( X_{1}- X_{2}\right)}^{2}\end{aligned}

由于 X‾\displaystyle \overline{X} 和 S2\displaystyle S^{2} 独立，即 (X1+X2)\displaystyle {\left( X_{1}+ X_{2}\right)} 和 (X1−X2)2\displaystyle {\left( X_{1}- X_{2}\right)}^{2}，进一步得到 (X1+X2)2\displaystyle {\left( X_{1}+ X_{2}\right)}^{2} 和 (X1−X2)2\displaystyle {\left( X_{1}- X_{2}\right)}^{2} 独立

卡方 2 和参数为 1/2 的指数分布是同分布

设 X,Y\displaystyle X, Y 均服从标准正态分布，相互独立，有

fX(x)=12πe−x22fY(y)=12πe−y22f(x,y)=12πe−x2+y22\begin{aligned}\displaystyle {f}_{X}{\left( x\right)}&=\frac{1}{\sqrt{2\pi} }\text{e}^{-\frac{x^{2} }{2} } \\ \displaystyle {f}_{Y}{\left( y\right)}&=\frac{1}{\sqrt{2\pi} }\text{e}^{-\frac{y^{2} }{2} } \\ \displaystyle f{\left( x, y\right)}&=\frac{1}{2\pi}\text{e}^{-\frac{x^{2}+ y^{2} }{2} }\end{aligned}

令

Z=X2+Y2\displaystyle Z= X^{2}+ Y^{2}

对于 z<0\displaystyle z< 0，有

P(Z<z)=0\displaystyle P{\left( Z< z\right)}= 0

对于 z≥0\displaystyle z\ge 0，有

P(Z≤z)=P(X2+Y2≤z)=∬x2+y2≤z12πe−x2+y22dxdy=∫02πdθ∫0z12πe−r22rdr=∫0ze−r22d(r22)=1−e−z2\begin{aligned}\displaystyle P{\left( Z\le z\right)}&= P{\left( X^{2}+ Y^{2}\le z\right)} \\ \displaystyle &=\underset{x^{2}+ y^{2}\le z}{\iint}\frac{1}{2\pi}\text{e}^{-\frac{x^{2}+ y^{2} }{2} }{\left.\text{d} x\right.}{\left.\text{d} y\right.} \\ \displaystyle &=\int_{0}^{2\pi}\text{d}\theta\int_{0}^{\sqrt{z} }\frac{1}{2\pi}\text{e}^{-\frac{r^{2} }{2} } r\text{d} r \\ \displaystyle &=\int_{0}^{\sqrt{z} }\text{e}^{-\frac{r^{2} }{2} }\text{d}{\left(\frac{r^{2} }{2}\right)} \\ \displaystyle &= 1-\text{e}^{-\frac{z}{2} }\end{aligned}

故

FZ(z)={1−e−z2ifz≥00otherwise\displaystyle F_{Z}{\left( z\right)}={\left\lbrace\begin{matrix*}[l] 1-\text{e}^{-\frac{z}{2} }&\quad\text{if}\quad z\ge 0\\ 0&\quad\text{otherwise}\quad\\\end{matrix*}\right.}

同时，有

fZ(z)={12e−z2z≥00otherwise\displaystyle {f}_{Z}{\left( z\right)}={\left\lbrace\begin{matrix*}[l]\frac{1}{2}\text{e}^{-\frac{z}{2} }& z\ge 0\\ 0&\text{otherwise}\quad\\\end{matrix*}\right.}

因此，χ2(2)\displaystyle \chi^{2}{\left( 2\right)} 和参数为 12\displaystyle \frac{1}{2} 的指数分布同分布。

最大值和最小值的等价转换

max⁡{X1,X2}=X1+X2+∣X1−X2∣2min⁡{X1,X2}=X1+X2−∣X1−X2∣2\begin{aligned}\displaystyle \max{\left\lbrace X_{1}, X_{2}\right\rbrace}&=\frac{X_{1}+ X_{2}+{\left| X_{1}- X_{2}\right|} }{2} \\ \displaystyle \min{\left\lbrace X_{1}, X_{2}\right\rbrace}&=\frac{X_{1}+ X_{2}-{\left| X_{1}- X_{2}\right|} }{2}\end{aligned}

令 X=max⁡{X1,X2},Y=min⁡{X1,X2}\displaystyle X=\max{\left\lbrace X_{1}, X_{2}\right\rbrace}, Y=\min{\left\lbrace X_{1}, X_{2}\right\rbrace}，有

X+Y=X1+X2X−Y=∣X1−X2∣\begin{aligned}\displaystyle X+ Y&= X_{1}+ X_{2} \\ \displaystyle X- Y&={\left| X_{1}- X_{2}\right|}\end{aligned}