考研数学讲义概率统计

概率统【1

ex,xY01PX：fY01P0,x求ZXY的分布函数FzCov(X,X2Fz(z)P{Zz}P(XYP(Y1)P(XYz|Y1)P(Y0)P(XYz|Y0)P(Y1)P(XYz|Y1[P(Xz)P(0z)P(X3 1 z

1ez z 3

1(exdx1z z 11ez

z3

Cov(X,X2)E(X3)E(X)E(X2E(X)1,E(X2)D(X)E2(X)11E(X3) x3 (x)dx

3x (s) exxs1dx(4)3(3)0(s1)s(s),(1)(3)(21)2(2)1,0x0y【2f(x,y)0①fX(x),fY(②Z2XY的fz③P{Y1|X ④P{Y

|X

2x,0xf(x)

1y,0y，f(y) 0

z:[0,当z0时Fz(z)当z2时Fz(z)当0z2时Fz

z141f(z) P{Y1,X

11(1

P 2 222 4P{X2

20

4 P fY|X(y )dy21dy

11fY|X(y|2)1

f( 1fX(【3

(XY),的概率密度为f(xy①求ZXY②X与Y相互独立，分别服从N

z(1)z0,Fz(z)zz0,Fz(z)

xxex(1y)dy1e

f(x,y)

1(x2y2e

,x,yF(z)P{Xz}

f(x, xyfz(z)

1|y|f(yz,y)dy 1 XY,

x,z

z |f , xey,0x【4】(X,Y),的概率密度为f(x,y) ,求Umax{X,Y}的fu 1(1xy),|x|1,|y|【5】(X,Y),的概率密度为f(x,y) ,X与Y不独X2与Y2相互独

其当u0orv0，P{X2u,Y2

u,vYu,v

v(1 v

uv,0u1,0vudx11(1 1

=u,0u1,v1v1

v

v,u1,0v v

1 1dx14(1 1,vg(u,v)

2G(u,v)

,0u,v1g ,0 g(u) ,g 【1】X~N(0,1).Ymax(X,0),Cov(X,Y) 【2】XaX1b,YcX2dac【3X,Y相互独立且服从正态分布【4

bY,VaX

a2a21【5

X1,X2,...,Xn,X~N(0,2),Y

nn【6

1ex,x22F(x)1,0x ,E(X)

1(e

, 3f(x)

0,求E(X) 0【7】X,Y相互独立，都服从正态分布N(01求D(|2【8【9

X1X2Xn(n2来自总体X~N(0,1)

,2,..n【10 f(y)

e5,y

1(x2y2e2102 P{X2Y且Y2X}【2

设X,Y)服从二维正态分布，其联合概率密度为f

x28YX

服从正态分 服从2分YXY

2服从t分

1((x)2(y)2 e2X 服从F分

( Y ,22

e 10,X与Y独立，X~N(0,22，Y~N(0, 1Y Y X Y()~ ()~

独 【3】X,Y~N(0B).aX+bY44【6】设总体X

(012设X为一随 A.P{1X1}0.79B.P{0X2}0.79C.P{|X1|1}0.21D.P{|X|1}0.21

[EX]2EX2DX1,EXP{0X2}P{1 E(X)|1}1DX10.21设一批零件长度X服从正态分布N(,2 其中 测得