证明:根据定义:P{1<min(x,y)<=2}=P{min(x,y)<=2}-P{min(x,y)<1}
而:P{1<min(x,y)}=1-P{min(x,y)>=1}=1-P{x>=1,y>=1};(遇到具体题目的进一步转化)
同样的:P{min(x,y)<=2}=1-P{min(x,y)>2}=1-P{x>2,y>2};
所以:P{1<min(x,y)<=2}
=P{min(x,y)<=2}-P{min(x,y)<1}
=1-P{min(x,y)>2}-[1-P{min(x,y)>=1}]
=P{min(x,y)>=1}-P{min(x,y)>2}
若X,Y是连续型随机变量,则:P{min(x,y)>=1}-P{min(x,y)>2}=P{min(x,y)>1}-P{min(x,y)>2}。得证
刚好复习到这个问题,化成P{min(x,y)>1}-P{min(x,y)>2}的意义在于能够计算概率
p{min(X,Y)<2}=1-P{min(X,Y)>2},详见浙大概率论课本81页
同理可得p{min(X,Y)<1}=1-P{min(X,Y)>1},将此式和上式代入下面方程
根据概率论定义P{1<min(x,y)<=2}=F(2)-F(1)=P{min(x,y)<=2}-P{min(x,y)<1}
化简后即为P{min(x,y)>1}-P{min(x,y)>2}
对于P{min(X,Y)>2}来说,要保证X,Y都大于2,因此P{min(X,Y)>2}=p{X>2,Y>2}
题目一般给出的是XY独立同分布,假设题目给出的是xy服从E(1)(指数分布)分布
则最后的计算是[1-(1-e^-1)]^2-[1-(1-e^-2)]^2=e^-2-e^-4