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????x102dy,1?x?e2??
??0,其它?e21e2???2dx??112,0?y?e?2f（x)??????f(x,y)dx??1???y112dx?12(1Xy?1),e?2?y?1?其它?0,?10 解：f(x,y)???3x,0?x?1,0?y?x?0,其它
fx,y)Y|X(y|x)?f(f(x) (fX(x)?0) X?x3x f(x)??????f(x,y)dy??2??03xdy?2,0?x?1X
??0,其它当0 ?3xfx)?f(x,y)?3x2,0?y?xY|X(y|f??X(x)?2 ?0,其它?即，f?2,0?y?x?1Y|X(y|x)???x ?0,其它11解：f(x,y)???1,0?x?1,|y|?x?0,其它 ?1f(y)????f(x,y)dx???dx?1?y,y?0Y??y?????1ydx?1?y,y?0 21 22 ?1,0?x?1,?x?y?xf(x,y)???1?y当y?0时，fX|Y(x|y)? fY(x)?其它?0,?1,0?x?1,?x?y?xf(x,y)???1?y当y>0时，fX|Y(x|y)? fY(x)?其它?0,?1,0?|y|?x?1f(x,y)???1?|y|所以，fX|Y(x|y)? fY(x)?其它?0,12 解：由fX|Y(x|y)?f(x,y)得 fY(x)?15yx2,0?y?1,0?x?yf(x,y)?fX|Y(x|y)fY(y)?? 其它?0,P{X?0.5}????0.5?????f(x,y)dydx??10.5x2?15yxdydx?147 6413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 0 0.2 1 0.6 2 0.2 Pj 0.16 xy?1???1????14 解：fX(x)???e,x?0 fY(y)???e,??x?0?0,?0,y?0 y?0由独立性得X，Y的联合概率密度为 22 23 ?y?1?x??,x?0,y?0 f(x,y)???2e?其它?0,x?y则P{Z=1}=P{X?Y}= x,y)dxdy????x1??x??f(?y?00?2edydx?12 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?x2fX(x)?????f(x,y)dy??122????2?1?x2dy?1?x,|x|?1???0，? 其它?同理，f)??2??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)f?1,0?x?1?1,0?y?1X(x)???0,其它 fY(y)???0,其它 利用卷积公式：fZ(z)??????fX(x)fY(z?x)dx求fZ(z) fx)f?1,0?x?1,x?z?1?xX(Y(z?x)=??0,其它 ?z??0dx?z,0?z?1f???1Z(z)????fX(x)fY(z?x)dx????z?dx1?2?z1?z?2 ?0,其它?(2) fx)???1,0?x?1?e?y,y?0X(?0,其它 fY(y)???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?ze?ydy,?z??00?z?1?1?e,0?z?1????z??(e?1)e?z,z?1? fZ(z)??fX(z?y)fY(y)dy???e?ydy,z?1z?1????0,其它其它0,???17 解：由定理3.1（p75）知，X+Y~N(1,2) 故P{X?Y?1}?P{X?Y?11?1?}??(0)?0.5 22f(x,y)dx????0（x)?18解：(1) fX同理，fY(y)??????11(x?y)e?(x?y)dy?e?x(x?1)(x>0) 221?ye(y?1) y>0 2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 ?1?1?(x?z?x)e?(x?z?x),x?0,z?x?0?ze?z,x?0,z?x fX(x,z?x)??2??2??0,其它其它??0,fZ(z)???????z1?z1??zedx,z?0??z2e?z,z?0??2 fX(x，z?x)dx??02?0,z?0?z?0?0,?19解：并联时，系统L的使用寿命Z=max{X,Y} 因X~E(?),Y~E(?),故 yx?1???1???e,y?0?e,x?0 fY(y)??? fX(x)????0,y?0?0,x?0??yx????????FX(x)??1?e,x?0 FY(y)??1?e,??x?0?0,?0,y?0 y?0zz??????FZ(z)?FX(z)FY(z)??(1?e)(1?e),z?0 ?0,z?0? 24 25 ?11??1?z1?z??????z?e??e??(1?1)e????,z?0fZ(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??Z(z)????1?1?11?????????????z??????e,z?0 ?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 1x2 解: (1) P{X?2Y}?f(x,y)dxdy?x???2y??200(2?x?y)dydx?724 ??(2) 利用公式fZ(z)?)dx计算 ??f(x,z?x?f(x,z?x)???2?z,0?x?1,0?z?x?1?0,其它 ?z????0(2?z)dx,0?z?1（?2?z),0?z?1fz)??f(x,z?x)dx????1(2?z)dx,1?z?2??Z(?(2?z)2,1?z????z?12?0,z?2??0,z?2?3.解：(1) FY(y)=P{Y?y}=P{X2?y} 当y<0时，fY(y)=0 25