概率密度f(x,y)=24y(1-x-y),求P{X<=1/2|Y=1/2}

Question

概率密度f(x,y)=24y(1-x-y)，x>0,y>0,x+y<1
0 其他
求P{X<=1/2|Y=1/2}，P{Y<=1/2|X>1/2}

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