求p(|fn(x)-f(x)|≥i)≦1/4ni^2

Question

如题所述

Answer 1

y=1/2时正好 0<x<1/2
P(X<=1/2|Y=1/2)=1
不过不一定任何时候都这麼巧，下面是通用法

fy(y)=∫(0~1-y) 24y(1-x-y) dx
=24y(x-x2/2-xy)|(0~1-y)
=24y((1-y)(1-y)-(1-y)2/2)
=12y(1-y)2

fx|y(x|y)=f(x,y)/fy(y)
=2(1-x-y)/(1-y)2
P(X<=1/2|Y=1/2)
=∫(0~1/2) fx|0.5(x|0.5) dx
=∫(0~1/2) 2(0.5-x)/(1-0.5)2 dx
=∫(0~1/2) 8(0.5-x) dx
=8(0.5x-x2/2)
=1

2）
fx(x)=∫(0~1-x) 24y(1-x-y) dy
=12y2(1-x)-8y3|(0~1-x)
=4(1-x)3

P(X<1/2)=∫(0~1/2) 4(1-x)3 dx
=-(1-x)^4 |(0~1/2)
=15/16

P(X>1/2)=1/16
P(Y<=1/2,X>1/2)=∫(0~1/2)∫(1/2~1-y) 24y(1-x-y) dx dy
=∫(0~1/2) {24y(1-y)x-12x2y|(1/2~1-y)} dy
=∫(0~1/2) 12y(1-y)2-12y(1-y)+3y dy
=∫(0~1/2) 12y(1-2y+y2-1+y+1/4) dy
=∫(0~1/2)y2-y+1/4 dy
=y3/3-y2/2+y/4 (0~1/2)
=1/24-1/8+1/8
=1/24
P(Y<=1/2|X>1/2)=(1/24)/(1/16) =16/24=2/3