概率论与数理统计 求解答过程
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发布时间：2023-09-02 14:40
共1个回答
热心网友
时间：2024-06-29 08:57
fx(x)=∫(x~1) 2 dy
=2(1-x)
fy(y)=∫(0~y) 2 dx
=2y
2)
这里不太复杂，正好的方形，一般还是需要作图或不等式来分析
P(0<X<1/2,1/2<Y<3/4)
=∫(0~1/2)∫(1/2~3/4) 2 dydx
=2*1/4*1/2
=1/4
P(1/2<Y<3/4)
=∫(1/2~3/4) fy(y) dy
=y^2 (1/2~3/4)
=9/16-4/16
=5/16
P(0<X<1/2|1/2<Y<3/4)
=(1/4)/(5/16)
=4/5
3)
不独立，因为fx(x)*fy(y)不等於f(x,y)
4)
Fz(z)
P(Z<z)
=P(X+Y<z)
=∫(0~z/2)∫(x~z-x) 2 dydx
=2∫(0~z/2) (z-2x) dx
=2(zx-x^2) (0~z/2)
=2z^2/4
=z^2/2 (0<z<=1)
(1<z<2)时
P(Z<z)
=1-P(Z>=z)
=1-∫(z/2~1)∫(z-y~y) 2 dxdy
=1-2∫(z/2~1) (2y-z) dy
=1-2(y^2-zy) (z/2~1)
=1-2(1-z^2/4-z(1-z/2))
=1-2(1+z^2/4-z)
=2z-z^2/2-1 (1<z<2)
fz(z)=F'z(z)
=z (0<z<=1)
=2-z (1<z<2)
=0 其他z
