21 9 解：Se21D??1xdx?lnx|e21?2
(X,Y)在区域D上服从均匀分布，故f(x,y)的概率密度为
?f(x,y)??1?,(x,y)?D ?2?0，其它?1fx)????X（??f(x,y)dy??x1??0dy,1?x?e2
?2?0,其它?e21e2?1?dx?,??1220?y?e?2f???1y111X（x)????f(x,y)dx???2dx?(?1),e?2?y?1?12y?其它?0,?10 解：f(x,y)?0?x?1,0?y?x??3x,?0,其它
fx,y)Y|X(y|x)?f(fx)?0)
X(x) (fX( f???0?x?1X(x)??f(x,y)dy??x2??03xdy?3x,??
?2?0,其它当0 ?3x0?y?xfY|X(y|x)?f(x,y)?3x2,fx)?? X(?2?0,其它?即，f(y|x)?2,0?y?x?1Y|X?? ?x?0,其它11解：f(x,y)??1,0?x?1,|y|?x??0,其它 ?1dx?1?yy?0f?????y,Y(y)????f(x,y)dx??1 ???ydx?1?y,y?0 21 22 当y?0时，fX|Y(x|y)?f(x,y)?1,???1?yfY(x)??0,?1,???1?yfY(x)??0,0?x?1,?x?y?x其它 当y>0时，fX|Y(x|y)?f(x,y)0?x?1,?x?y?x其它 所以，fX|Y(x|y)?f(x,y)?1,???1?|y|fY(x)??0,f(x,y)fY(x)0?|y|?x?1其它 12 解：由fX|Y(x|y)?得 f(x,y)?fX|Y?15yx2,(x|y)fY(y)???0,0?y?1,0?x?y其它1 P{X?0.5}???0.5??????f(x,y)dydx???0.51x15yxdydx?24764 13解：Z=max(X,Y),W=min(X,Y)的所有可能取值如下表 pi (X,Y) max(X,Y) Min(X,Y) 0.05 (0,-1) 0 -1 0.15 (0,0) 0 0 0.2 (0,1) 1 0 0.07 (1,-1) 1 -1 0.11 (1,0) 1 0 0.22 (1,1) 1 1 0.04 (2,-1) 2 -1 0.07 (2,0) 2 0 0.09 (2,1) 2 1 Z=max(X,Y),W=min(X,Y)的分布律为 Z Pk W -1 0 0.53 1 0.31 y?0y?00 0.2 1 0.6 2 0.2 Pj 0.16 ?1?x?e?,f(x)?14 解：X????0,?1?y?e?,x?0f(y)? Y???x?0?0, 由独立性得X，Y的联合概率密度为 22 23 ?1?x?y?e?,f(x,y)???2?0,?x?0,y?0其它 ??x1?x?y1则P{Z=1}=P{X?Y}=??f(x,y)dxdy????00x?y?2edydx?2 P{Z=0}=1-P{Z=1}=0.5 故Z的分布律为 Z 0 1 Pk 0.5 0.5 ?15 解：f(x,y)??1??,x2?y2?1 ??0,其它???1?x22fX(x)????f(x,y)dy??1??2dy???1?x21x2,|x|?1 ????0，其它?2同理，f)??1?y2,y|?1Y(y??| ??0，其它显然，fX(x)?fY(y)，所以X与Y不相互独立. 16 解：(1)f0?x?1X(x)?1,??y)?,0?y?1?0,其它 fY(??1?0,其它 利用卷积公式：fZ(z)??????fX(x)fY(z?x)dx求fZ(z) f(x)f=?1,0?x?1,x?z?1?xXY(z?x)??0,其它 ??z?0dx?z,0?z?1f???1Z(z)????fX(x)fY(z?x)dx??1?z?2 ??z?dx1?2?z?0,其它?(2) f?1,0?x?1X(x)??(y)???0,其它?e?y f,y?0Y ?0,y?0利用卷积公式：fZ(z)??????fX(z?y)fY(y)dy 23 24 ?e?y,fX(z?y)fY(y)???0,y?0,y?z?y?1其它 fZ(z)???????ze?ydy,??0?z?yfX(z?y)fY(y)dy???edy,z?1??0,?0?z?1z?1其它?1?e?z,??z??(e?1)e,?0,?0?z?1z?1其它? 17 解：由定理3.1（p75）知，X+Y~N(1,2) 故P{X?Y?1}?P{X?Y?12???1?12}??(0)?0.5 1212（x)?18解：(1) fX12???f(x,y)dx????0(x?y)e?(x?y)dy?e?x(x?1)(x>0) 同理，fY(y)?e?y(y?1) y>0 显然，fX(x)?fY(y)，所以X与Y不相互独立 (2).利用公式fZ(z)??????fX(x，z?x)dx x?0,z?x?0其它?1?ze?z,??2??0,x?0,z?x其它?1?(x?z?x)e?(x?z?x),fX(x,z?x)??2?0,??? fZ(z)?????z1?z??zedx,fX(x，z?x)dx??02?0,?z?0?12?z?ze,??2z?0?0,?z?0z?0 19解：并联时，系统L的使用寿命Z=max{X,Y} 因X~E(?),Y~E(?),故 ?1?y?1?x??e?,x?0?e,y?0 fY(y)??? fX(x)?????0,x?0??0,y?0x????FX(x)??1?e,??0,y??x?0 F(y)??1?e?,?Y?x?0?0,y?0 y?0zz??????FZ(z)?FX(z)FY(z)??(1?e)(1?e),?0,?z?0 z?0 24 25 ?11?z?1?z???????z111?e??e??(?)e????,fZ(z)???????0,?z?0 z?0串联时，系统L的使用寿命Z=min{X,Y} ?11???????z??????FZ(z)?1?[1?FX(z)][1?FY(z)]??1?e,z?0 ??0,z?0?f???1??11??1???e???????z?z?0Z(z)????, ??????0,z?0 (B)组 1 解：P{X=0}=a+0.4, P{X+Y=1}=P{X=1,Y=0}+P{X=0,Y=1}=a+b P{X=0,X+Y=1}=P{X=0,Y=1}=a 由于{X=0|与{X+Y=1}相互独立, 所以 P{X=0, X+Y=1}=P{X=0} P{X+Y=1} 即 a=(a+0.4)(a+b) (1) 再由归一性知： 0.4+a+b+0.1=1 (2) 解(1),(2)得 a=0.4, b=0.1 2 解: (1) P{X?2Y}???f(x,y)dxdy??1x20?0(2?x?y)dydx?7x?2y24 ??(2) 利用公式fZ(z)??f(x,z?x)dx计算 ??f(x,z?x)??2?z,0?x?1,0?z?x?1??0,其它 ?z(2?z)dx,0?z????0?1（?2?z),0?z?1fz)??f(x,z?x)dx??1??z?1(2?z)dx,1?z?2???(2?z)2Z(,1?z?2????2??0,z?0,z?2?3.解：(1) FY(y)=P{Y?y}=P{X2?y} 当y<0时，fY(y)=0 25