题型四、关于二维正态分布(★★)
解题思路——利用二维正态分布的性质来求解，主要以小题为主.【例3.11】 随机变量(X ,Y )服从二维正态分布，且 X 与 Y 不相关，
f ( x)， f ( y)分别表示 X ，
X Y
Y 的概率密度，则在Y = y的条件下， X 的
条件概率密度 f (x | y)为( A ).
X|Y
f ( x)
(A) f ( x) (B) f ( y) (C) f (x) f ( y) (D) X
X Y X Y
f ( y)
Y
X) Y #X Y *
- (x v =
. .
,
E
X & Y &
f(x) fxIN FxM)
fix
(x(Y) · fx)
= = 54 E*
=
fx
my) fxm)【例3.12】 设随机变量 X 和 Y 相互独立，且均服从标准正态分布，则
下列正确的是( D ).
1 1
(A) P{X + Y 0} = (B) P{X − Y 0} =
4 4
1 1
(C) P{max{X ,Y } 0} = (D) P{min{X ,Y } 0} =
4 4
N(01) YeWol) #X X **E
XN
- -
EFH
(A) X-N(0 2) PXY0)
~ X+ , =
,
I
(B) x y N(0 2) PSX-X03
- - . =
,
(c) Pimax3x . <30) = 1 - PlmaxSX . X70) = 1- 19x0 X207
,
3/
**
PSXcoS P(Xco) I
= 1 - . = 1 - = * +
(D) Pimm(x-1307 19x0 4207 99X303 P(X 0) Ex* =
= , = . , =题型五、随机变量函数的概率分布(★★★★★)
(一)离散型随机变量函数的分布
解题思路——离散型随机变量(X ,Y )的函数Z = g(X ,Y )仍为离散型随
机变量，关键是根据 X ,Y 的分布来求出Z = g(X ,Y )的分布律，包括确
定Z 的所有取值，以及算出所有取值的概率.
& ZTIE = 91 , 4.) , 91 , 42 ... - 9(xi . Y ; ) - - -
② PSZ qIi TiPT PEX =Xi X = Y ; ) = Pij
: = . = .
⑤ E
:【例3.13】 假设随机变量 X , X , X , X 相互独立，且同分布
1 2 3 4
X X
1 2
P X = 0 = 0.6, P X = 1 = 0.4(i = 1,2,3,4),求行列式 X = 的概
i i
X X
3 4
率分布.
T
X
XXP-
:
-1
0
1
: = , ,
0, 1 0 1
.
1)
P(x 1) P(X XzXs PSX 0) PSXzX3 13
= - = Xx = 0 = = Xy = . =
, , ,
E PSX , Xx = 07 = P(X = 0 U Xy = 0) = PSX = 07 + PSXx = 07 - PEX, = 07 . PSXEO
= 0 6+ 0 . 6 - 0 . 6x0 . 6 = 0 . 84
.
P(X-X3 = 1) = PEXz = 1 . X3 = 1) = P(Xz = 1 - P(X3 = 7 = 0 . 4x0 . 4 = 0 . 16
~ P(X -17 0 84x0 16 = 0 1344
= = . . .psx = o) = P(X , X4 =0 , Xa X3 = 07 + P9X , X4=, XzX3 = 1)
PEX 03 PSXzXs 07 PEX.Xy 1) P(XzX3 11
= X4= . = + = . =
,
= 0 . 84x0 . 84 +p . 16x0 . 16
7312
= 0
.
P(x 1) P(XXx 1 XzX3 07 PSX Xy 1) PSXzX3 01
= = = . = = , = . =
16 x084 1344
= a = 0 .
.
X I
+ o
↑ aB44 0 7312 0 1344
. .(二)连续型随机变量函数的分布
解题思路：已知连续型随机变量(X ,Y )的联合概率密度 f (x, y)，求函
数Z = g(X ,Y )的概率密度函数，常用如下两个方法.
方法 1——分布函数法(通用方法)：已知 f (x, y), 求Z = g(X ,Y )的密
度 f (z)，应采用分布函数法：
Z
第一步 画出 f (x, y) 0的区域D;
falzl
第二步 求Z 的取值范围，设为[a,b];
Fo
⑧
Al11(/ ⑧
?
a D第三步 求Z 的分布函数F (z)：
Z
(1)当z a时, F (z)=0; 当
Z
z b
falzl
Fo
时, F (z)=1;
⑧
- Z 11/6 ⑧
-
?
(2)当a z b时, a D
&
-
F (z) = P Z z = P g(X ,Y ) z = f (x, y)dxdy;
Z
g(X,Y )z
第四步 f (z) = F (z).
Z Z注意，如果Z = max(X ,Y )，则
F (z) = P{Z z} = P{max(X ,Y ) z} = P{X z,Y z}；
Z
如果Z = min(X ,Y )，则
F (z) = P{Z z} = 1 − P{Z z} = 1 − P{min(X ,Y ) z} = 1 − P{X z,Y z}
Z
.方法 2——卷积公式：
+ +
(1)如果Z = X Y ，则 f (z) = f (x, z x)dx = f (z y, y)dy.
Z
− −
z-ax
(2)如果Z = aX + bY ,(a 0,b 0),则 X
=
1 z − ax 1 z − by
+ +
f (z) = f x, dx = f , y dy.
Z
− | b | b − | a | a
1 z 1 z
+ +
(3)如果Z = XY ，则 f (z) = f x, dx = f , y dy.
z
− | x | x − | y | y
Y
+
(4)如果Z = ，则 f (z) = | x | f (x, zx)dx ；
Z
X −
X
+
如果Z = ，则 f (z) = | y | f ( yz, y)dy.
Z
Y −【例3.14】 设随机变量 X 和 Y 的联合分布是在以点 ( 0 , 1 ) , ( 1 , 0 ) , ( 1 , 1 ) 为顶
点的三角形区域上服从均匀分布，求随机变量 Z = X + Y 的密度函数.
↑
35 =: (X X) ~U(D) (0 1) Cill
- - .
a ↳
D
fNY)
(xY) ED
-
: = 52 .
Te
o S 3
fz(z)
to
Cl 0
O o .
XR [ 1 2]
:
Z
=
X+
.
11
'z
2
I
Fz(z) P(z = z) zel-c c)
= +
/AF Fell
D Z = =0
② z2#J Ez(z) I
=
.El
③ < z22
(z+. 1)
↑
↑ x Cill
(0 1)
Fz(z) = P(z = z) = P(x + Y = z) = P(y = - X + z) . a /1 ;
y
=-XTz
P(X) z)
1 X+
= - -
S 3
Cl 0
11 fixmaxay
(x) .
1-
=
- xzay
=
Y >- X+Z
2
= Sxx I (2 -z,
2
in
-1-
=
1
-
(2
-
z)
I
STE z
fall Eziz (2(2 -z) 12/2
-
= =
12
·
.fNM) 0 =x =, 1x = y = (1 2)
35 = = = 54 , = Xo .
xz
Z
E S
2
Co 1) z = 1
.
f(x 0x11 1 = z = +X !
92
z - x) = , .
.
E
Y
i
E
[fIxz-X)dX
fall
: =
falzl
Ez 1 z) 2#J = 0 ;
① =
,
,
Af
② E 1222
1
fzlzl I
= 2 dX = 2(2 - z)
.
zt【例3.15】
设随机变量
X和Y
的联合密度函数为
6x2 y,0 x, y 1 X
f (x, y) = ，求随机变量Z = 的密度函数 f (z).
Z
0,其它 Y
zX
y
=
fz(z)
+o N
O X
-
(z
,
1)
15- : z = [0 + o) 1211492 I 1
,
-
Y
Ez(z P(z z) zel-c o
= .+
=
,
S
O I
D Ez = 0 At Ezz = 0
,
⑧ Ezo Af
,
Fz(zl P(zzz) P(y = z) PSXEX)
= =
=
31
casel Eo < Z = 1 At
. ,
I
/ = I
Ez(z)
ax 6xy dy
Ez3
=
=
x
o
ZE
EZ)1 Af O< <
case2
.
N
3
X F
1lax/Loyay
Fzlzl
= I
T
=
5 illI
S
Ezizs -z
falzl 0 < z =
: = = O I
I
6
zx
5E3
E
O+
3 + 962 E
=: 4 (z4 4) 0 = z = 0:44 N
=
, , ,
<
ty
z
E
=
G
his
(141
fall fizy
1) any
: =
*
,
fz(z)
① Ez #J
= = o
,
C
Fall 62yay
EocZAt ! gz
②
= =
fall
Jegzytay
EzxAJ
③
= =
,
【例3.16】 设二维随机变量(X ,Y )在矩形G = (x, y) | 0 x 2,0 y 1
上服从均匀分布，求随机变量S Z = XY 的概率密度 f (s). Z
Z
y =
fz(z) +o X y(z 1)
33 XXTRI [0 2] O ⑧ X ,
z
- : = : , 1 -
#11111 'Ill
, z
,
f(xy) E
Fzll PEZ z) & 2
= = H
=
is
O
O 232H] Fzlzl
① ZoAJ Fall = =
#J
③ ocz2
Ezlzl PEzzz) PSXX z) P(X=* ) 1 - PX >* )
= = = = =
1- (ax)
Jay E(H(n) (nz)
= = -
=
&
falzl Faiz (In2-Int)
ocz2
: &
= =
E
OZ X (2 2)
I .
35
=:
f(x
El
1 ax ·
<X=2 0 = z =X
,
, = ,
&
&
Et
Ex
0 Z
12
fall [ fix l ax 8 2 X
: =
fall
*
Ezo 22,2 At = o
①
Af
02
①
fzlzl ( 2 EluX = (n2-Inz
2x
ax
= =↑
【例3.17】 设随机变量(X ,Y )的概率密度为
x + y, 0 x 1,0 y 1,
f (x, y) = ，试求(1)M = max{X ,Y } 概率密
0, 其他,
↑
Y
I =X
度；
P(M z) PimaxiX X7 = z)
1 ) Fu(z) = = = . + x+ y
(zz
: /
1111
= MX =z Y = z) zt( - x , +b) S
.
O I
EZ20AJ Fulzl 0 @EZYIAT Fulzl I
① = ; =
(Eax)* z
8 Eo#T Emil
= (xylay =
,
00
full Fulzs 3
I ocza/
: = = ,
.
o(2) N = min{X ,Y }概率密度．
"
***
Fr(z) PEN = z) Plmin(X-X) = z)
= =
1- PSmin(x-y) z) P(xz Y z) ·
< = 1 - + (ziz)
= ,
S
O I
EZ20#f Fulzl 1-1
& = = 0
,
Ezx1Af Fr(z) 1
② = 1 -o =
,
EOLZ4AF
③
,
(Lax() z2 z
(x yay
Fr(z 1 + = z + -
-
=
22-3z2
Frz
frzl + oczal
9 H
: = = ,
#1
O(三)离散与连续混合型随机变量函数的分布
解题思路——设Z = g(X ,Y ),其中 X 是离散型随机变量， Y 是连续型随
机变量，则Z 是连续型随机变量，且其概率密度 f (z)应用分布函数法
Z
求解，在求解F (z)时应用全概率公式展开计算.
Z
Fz(zl Plz = z) P99(X) = z) X0 1 2
=
= . .
= P(x =0 , 9 (0 , x) = z7 + P9x = 1 , 9(1 , Y1= 2)
+ P(X = 2 9(2 4) = 2)
, ,【例3.18】 设随机变量 X 与 Y 相互独立, X 的概率分布为
1 1, 0 y 1,
P{X = i} = (i = −1,0,1),Y 的密度为 f ( y) = 记
3 Y 0, 其它.
1
Z = X + Y .(1)求P Z X = 0 ;
2
= 0) PSz = E X= 0) Px + X = & X = 0 )
1 )
P(z
x= =
, ,
=
P(X ol P(X 0)
= =
P9XiE X 0 P(X= ) Pol
=
, .
=
=
-
PSX 01 (x 07
= + =(2)求Z 的概率密度.
Fz(zl P(z z) PEXTX z) zE70 O)
= = = , +
=
,
z)
P(x = 1 x+ y = z) + P(X = 0 x+ Y = z) + p(x = 1 , x+y=
= , ,
P(x -1 Y = z+) + P(x =0 Y = z) + P(x = 1 Y = z - 1)
= = , , ,
= P(x = +11 . P(X = zn) + P(x = 01 . PSx = z7 + P(x = 17 PX = -17
Fx (z + ) 5 Fx(zl + 5 Ex (2 + ) 75zfig
+
=
,
5 (fy(z fx(z) fy(z 1)
fz(z) +)
+ + -
= =fall y (fy(z + 1) + fx(z) + fx(z + ) fx M) = [
= =
AJ
14 EP - = z = 0
DE022 +
5
fz( 5(( 01
+ o+ =
= .
5
fz(z * x(0 0)
② E0 : :Af = + 1 + =
.
fz(zl -(0 5
③ EOEHAT EP1 = 2 = 2 #5 = + o + 1) =
, ,
fzlzl
EIR
0
④ =
fal 5
+= zz2
=
:
,
.
o
