(?1??2)k?(?1??2) ? ，k=0，1，2，?? ek!由数学归纳法，设对n-1成立，即
?Xi?1n?1i服从参数为
??i?1n?1i的泊松分布，因为Xn服从参数为?n的泊松分布，故
?Xi?1n?1i?Xn服从参数为??i??n的泊松分布，即对n成立.
i?1n?126.
??e??xi,xi?0 i=1,2. ,Z=X1+ X2 fi(xi)??xi?0,?0Z＞0时，
f(Z)??f1(z?t)f2(t)dt???e??(z?t)??e??tdt
00zz ??2e??z ??ze2?z0dt
??z
所以fZ(z)?fx?x(z)??120,z?0 2??z??ze,z?0?27.由题意知X1~N(4,3), X2~N(2,1).且两者相互独立.
X1?X2X－X２，Y=1，且X和Y服从正态分布 22EX1?EX24?2??3 故EX?221111DX?DX1?DX2??3??1?1
4444EX1?EX24?211EY???1 , DY?DX1?DX2?1
2244由X1=X+Y, X2=X-Y得X=故X~N(3,1),Y~N(1,1) 即fX(x)?1e2??(x?3)22 , fY(y)?1e2??(y?1)22
28. 用数学归纳法进行推广，与25题类似. Xi~N(?i,?i2),i=1,2??n.
?xi?1ni~N(??i,??i2)
i?1i?1nn1??,(x,y)?G29.f(x,y)??2
??0(x,y)?G当0＜z?2时，
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FZ(z)?P??X??z????f(x,y)dxdy ?Y?G0???1???zy1?? ????dx?dy????dx?dy
???0??2??zy2? ?(0??11zy01dx)dy?0 21zydy 02z ?
4??当z＞2时，FZ(z)?1?1 z???0?0z?0???1?1故FZ(z)??z0?z?2 fZ(z)???4?4?1?1?1z?2???z?z230.
P?XY?0??0.3?0?0.3?0.2?0.8 P?XY??1??0.1 P?XY?1??0.1
z?00?z?2 z?2XY 0 P -1 1 0.8 0.1 0.1 X 0 P 1 ?0?0.8?(?1)?0.1?1?0.1?0 EXYE(X?Y)?EX?EY
=0×0.6+0.4×1+(-1) ×0.4+0×0.2+1×0.4
=0.4 31 由题意知 fX(x)??所以:
0.6 0.4 Y -1 P 0 1 0.4 0.2 0.4 ?11?x?2,其它?0 fY(y)???11?y?2,其它?0
?11?x?2,1?y?2 f(x,y)??0其它?E?max(X,Y)????????????max(x,y)?1dxdy
?? ?21?221max(x,y)dxdy
2??1211xdxdy??21?2121ydxdy
2 x>y y>x ?
?(?x1xdy)dx??(?ydy)dx
x32
??21(x2?x)dx??211(4?x2)dx 221x32x22x32 ?1?1?2x1?326
8137???2? 33265 ?
3 ?32.证明： P(X=ai,Y=
j)=Pij, i,j=1,2…….
P(X= ai)＝Pi?
P(Y=bj)=P?j?? j=1,2……
E(X?Y)???(ai?bj)?Pij
i?1j?1 ???a?P???biiji?1j?1i?1j?1??ii?j????j?Pij
??ap??bi?1j?1??p?j
?EX?EY
EXY???aibjpij??aipj??bj?pj?EX?EY
i?1j?1i?1j?1???33.
Y 0 1 2
………….. 19 19C1C204020C6020 20C0C204020C60
1182020C19C1C19 4012C20C20C20C401920CP40201840C20CEX?0?1??C220?????19?C20?20?4020C40202020 ………….. C60CCC 206060602020C60C60C60 =8
34.因为可以看成是9重见利试验，EX=np=9?1?（）?
935.参考课本P84的证明过程. 36.
X2 ??825??0 1 2 pi1 XX1 0 1 15521 5628285655535 5614285633
pX2j 31510 282828Cov(x1, x2)=E X1 X2-Ex1 ×Ex2
153535－? 28562815 = －
224 =
?12e?3x?4y37.f(x,y)???0?21e?3x?4yg(x,y)???0所以：
x?0,y?0
其它x?y?0
其它?3e?3x fX1(x)???0?4e?4y fY1(y)???0x?0 x?0y?0 y?0
?21?3x?(e?e?7x)gX2(x)??4?0?y?0 其它x?0其它
?7e?7y gY2(y)???01因为f(x,y)?fX1?fY(y),所以X,Y独立. 故Cov(x1,y1)?EX1Y1?EX1Y1?0
Cov(x2,y2)?EX2Y2?EX2?EY2
? ???0??x??x0xy?21e?3x?4ydydx????021?3x1(e?e?7x)dx? 471?3x?4yxy?21edydx?1? ?0?07811?? ? 4974938. X1?X?Y,DX1?DX?DY?2cov(X,Y) 所以 cov(X1,Y2)?DX1?DX?DY3?1?11??
222因为 X1,Y1 独立。
所以 Cov(X1,Y1)=0
(也可以计算：Cov(X1,Y1)= Cov(X+Y,X-Y) = Cov(X,X-Y)+Cov(Y,X-Y)
= Cov(X,X)-Cov(X,Y)+Cov(y,x)-Cov(Y,y)
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=1?39.
11＋?1=0) 22?1? f(x,y)?????0 fX(x)?x2?y2?1其它
?1?x21?1?x2?dy?21?x2? fY(y)??1?y21?1?y2?dx?21?y2?
所以： f(x,y)?fx(x)fy1y 所以：X,Y不独立。 ?x,y?40.
cov(x,y)EXY?EX?EY=……… ?DXDYDXDY?XY?cov(X,Y)EXY?EX?EY??DXDYDXDY44225 ?66112?2257541.
①?rA,rB?cov(rA,rB)61??
DrADrB16?92 ②rP?xrA?(1?x)?rB
Drp?x2DA?(1?x)2DB?2x(1?x)DA ?16x?9(1?x)?2x(1?x)?4?3? ?13x?6x?9
222DB??AB
1 23232 ?13(x?)?13?2?9
1313当x?39108?时，Drp最小为9?
131313min(DrA,DrB)?min(16,9)?9
当13x?6x?0时，即０?x?42.解
①设投资组合的收益率为rp,则 rp?xrA?(1?x)rB
26时，Drp?min(DrA,DrB) 13Dp?x2DA?(1?x)２DB?2x(1?x)DADB??AB
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