题目
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题目解答
答案
解析
步骤 1：找到Y1和Y2的累积分布函数(CDF)
- Y1的CDF: ${F}_{1}({y}_{1})=P({Y}_{1}\leqslant {y}_{1})=1-P({Y}_{1}\gt {y}_{1})$
- 由于 ${Y}_{1}=min\{ {X}_{1},{X}_{2}\} $ ${Y}_{1}\gt {y}_{1}$ 当且仅当 ${X}_{1}\gt {y}_{1}$ 且 X2>y1:
- P(Y1>y1)=P(X 1>y1)P(X2 >y1)=(1- Fx(y1))^2=e^(-2y1)
- 因此: ${F}_{{Y}_{1}}({y}_{1})=1-{e}^{-2{y}_{1}}$ 对于 ${y}_{1}\gt 0$
- Y2的CDF: ${F}_{{Y}_{2}}({y}_{2})=P({Y}_{2}\leqslant {y}_{2})=P({X}_{1}\leqslant {y}_{2}且{X}_{2}\leqslant {y}_{2})$
- 由于 ${Y}_{2}=max\{ {X}_{1},{X}_{2}\} $ : ${F}_{{l}_{2}}({y}_{2})=P({x}_{1}\leqslant {y}_{2})P({x}_{2}\leqslant {y}_{2})={({F}_{x}{({y}_{2})}^{2}=$ ${(1-{e}^{-{y}_{2}})}^{2}$
- 因此: ${F}_{{Y}_{2}}({y}_{2})={(1-{e}^{-{y}_{2}})}^{2}$ .对于 ${y}_{2}\gt 0$
步骤 2：找到(Y11,Y2)的联合分布函数
- 联合分布函数FY1 ,22(y1,y2)由下式给出: ${F}_{{1}_{1}}{x}_{2}({y}_{1},{y}_{2})=P({Y}_{1}\leqslant {y}_{1}且{Y}_{2}\leqslant {y}_{2})$
- 对于 ${y}_{1}\leqslant {y}_{2}$ : ${F}_{{y}_{2}}({y}_{2})=P({x}_{1}\leqslant {y}_{2}且{x}_{2}\leqslant {y}_{2})={(1-{e}^{-{y}_{2}})}^{2}$
- 对于 ${y}_{1}\gt {y}_{2}$ : ${F}_{{Y}_{1}{Y}_{2}}({y}_{1},{y}_{2})=0$
- 因此,联合分布函数是: ${F}_{{Y}_{1}}{Y}_{2}({y}_{1},{y}_{2})=$ $\left \{ \begin{matrix} 0,\\ {(1-{e}^{-{y}_{2}})}^{2}\end{matrix} \right.$ 如果 $0\lt {y}_{1}\leqslant {y}_{2}$ 如果 ${y}_{1}\gt {y}_{2}$
- Y1的CDF: ${F}_{1}({y}_{1})=P({Y}_{1}\leqslant {y}_{1})=1-P({Y}_{1}\gt {y}_{1})$
- 由于 ${Y}_{1}=min\{ {X}_{1},{X}_{2}\} $ ${Y}_{1}\gt {y}_{1}$ 当且仅当 ${X}_{1}\gt {y}_{1}$ 且 X2>y1:
- P(Y1>y1)=P(X 1>y1)P(X2 >y1)=(1- Fx(y1))^2=e^(-2y1)
- 因此: ${F}_{{Y}_{1}}({y}_{1})=1-{e}^{-2{y}_{1}}$ 对于 ${y}_{1}\gt 0$
- Y2的CDF: ${F}_{{Y}_{2}}({y}_{2})=P({Y}_{2}\leqslant {y}_{2})=P({X}_{1}\leqslant {y}_{2}且{X}_{2}\leqslant {y}_{2})$
- 由于 ${Y}_{2}=max\{ {X}_{1},{X}_{2}\} $ : ${F}_{{l}_{2}}({y}_{2})=P({x}_{1}\leqslant {y}_{2})P({x}_{2}\leqslant {y}_{2})={({F}_{x}{({y}_{2})}^{2}=$ ${(1-{e}^{-{y}_{2}})}^{2}$
- 因此: ${F}_{{Y}_{2}}({y}_{2})={(1-{e}^{-{y}_{2}})}^{2}$ .对于 ${y}_{2}\gt 0$
步骤 2：找到(Y11,Y2)的联合分布函数
- 联合分布函数FY1 ,22(y1,y2)由下式给出: ${F}_{{1}_{1}}{x}_{2}({y}_{1},{y}_{2})=P({Y}_{1}\leqslant {y}_{1}且{Y}_{2}\leqslant {y}_{2})$
- 对于 ${y}_{1}\leqslant {y}_{2}$ : ${F}_{{y}_{2}}({y}_{2})=P({x}_{1}\leqslant {y}_{2}且{x}_{2}\leqslant {y}_{2})={(1-{e}^{-{y}_{2}})}^{2}$
- 对于 ${y}_{1}\gt {y}_{2}$ : ${F}_{{Y}_{1}{Y}_{2}}({y}_{1},{y}_{2})=0$
- 因此,联合分布函数是: ${F}_{{Y}_{1}}{Y}_{2}({y}_{1},{y}_{2})=$ $\left \{ \begin{matrix} 0,\\ {(1-{e}^{-{y}_{2}})}^{2}\end{matrix} \right.$ 如果 $0\lt {y}_{1}\leqslant {y}_{2}$ 如果 ${y}_{1}\gt {y}_{2}$