习题三答案
1. 证明：F?x2,y2??F?x1,y2??F?x2,y1??F?x1,y1? ?P?x1＜x?x2,y?y2??P?x1＜x?x2,y?y1?
?P?x1＜x?x2,y1＜y?y2??0由概率的非负性，知上式大于等于零，故得正. 2. 解：①， x2 x1 0 1 2 ０ 1/56 10/56 10/56 １ 5/56 20/56 10/56 PX2?x2j 3/28 15/28 5/14 ??P?X1?x1i? 3/8 5/8 1 ②. P?X1?0,X2?0???1?P{x2?0|x1?0}?P?x1?0?
?1??1??C27??3203205????821?8?56?14 ?21 56P?X1?X2??P?x1?0,x2?0??P?x1?1,x2?1??P?x1x2?0??1?P?x1x2?0??1?P?x1?1,x2?1??P?x1?1,x2?2?
?1?5102613??? 145656283. 解：①由概率的性质
??又
??0??0?x0?????3x?4yf?x,y?dxdy??0kedxdy?1?k1?12 01????x?3x?4yg?x,y?dxdy??0?0k2edxdy?1?k2?21
????0② fX?x??1?3e?3x x＞0 ?41???3x?4ydx?k1e?4y??4e?4y y＞0 fY?y???0k1e321?3xx?3x?4ydx?e?e?7x x＞0 当y＞0时 gX?x???0k2e4???3x?4y01kedy?k1e?3x???当x＞0时 gY?y?????yk2e?3x?4ydx?7e?7y y＞0
4. 解：①
?1?,0?x?2，0?y?1f?x,y???2
?0,其他?
F?x,y???x0?y0?xy?,0?x?2，0?y?11/2dxdy??2
?0,其他?
②
fX?x???x0101121dy? 0?x?2；fY?y???0d?1，0?y?1 222FX?x???x0?x?2?1?y?ydx??2 ；FY?y???01dx?? 2其他?0??00?y?1
其它
③
Py＜x2????12111dxdy?1?dxdy? ?y＜x22?0?x2235.
?2f?x,y???
?00?x?y?1
其它0?x?1其它
fX?x???x1?2??1?x?2dy??
0??2y2dx??
?0
fY?y???y0?y?1
0其它
0,其它??x?y?1?,0?x?2,1?y?2?0?其它?1?2f?x,y??? x6.①F?x,y??? ,0?x?2,y＞20?x?2,1?y?2??2?2?y?1,x＞2,1?y?2?1,x?2,y＞2?②
G?x,y??PX2?x,Y2?y?P?x＜X?x,?y＜Y?y?(xy?x)/2,0?x?4,1?y?4?0，x?0,或y?1??x G(x,y)??，0?x?4,y?42?y?1,x＞4,1?y?4??1,x＞4,y?4?????
讨论如下：
③
13??21P?x＜１,y＞?=??dxdy??10?3/2dxdy?1/4
22?D2?＜X?x2?P?y1＜Y?y2???Fx?x2??Fx?x1??Fy?y2??Fy?y1? 7. 证明：P?x1?? ?Fx,y?x2,y2??Fx,y?x2,y1??Fx,y?x1,y2??Fy?x1,y1? ?P?x1＜X?x2,y1＜Y?y2? 故独立得证. 8. x1|x2?1 ０ 1 3１ 2 3P ９. 解：① Pij?PixPji ②Py?yj????i?1pipji ③Pij?xpipjix?ipipjix
111P?P? P ?01322124421??p?y??1?p?x?0? 故不独立. ② P11410. 解：① P11?11. x y y1 y2 y3 pix x1 1/24 1/8 1/6 1/8 3/8 1/2 1/12 1/4 1/3 p1,p2,?,pn1/4 3/4 1 x2 pj yp11,p12,?,p1np21,p22,?,p2n独立p1,p2,?,pn?pp,?p１２m12. 证明：必要性： 由：
??????
pm1,pm2,?,pmn?秩为1
p1,p2,?,pnp11,p12,?,p1np11,p12,?,p1np21,p22,?,p2n秩为1p21,p22,?,p2n?充分性，若
??????pm1,pm2,?,pmnpm1,pm2,?,pnm?p?x?x2y?y1??p?x?x2y?y2??p?x?x2y?y3???p?x?x2y?yn??p?x?xy?y??p?x?xy?y??p?x?xy?y???p?x?xy?y??1112131n??
...?..........?p?x?xmy?y1??p?x?xmy?y2???p?x?xmy?yn?? 从上式可得x与y独立. 13. 解：① X2?Bx?c?0,
有实根的概率 B2?4c?0,PB2?4C?0
???P?2,1??P?3,1??P?3,2??P?4,4??P?4,2??P?4,1??P?4,3?
?P?5,1??P?5,2??P?5,3??P?5,4??P?5,5??P?5,6??P?6,1??P?6,2??P?6,3??P?6,4??P?6,5???P?6,6?
=
19 361 18② PB?4C?0?P?2,1??P?4,4??2??14. 解：fx1y1?x,y??f?x,y? fy1?y?x＞0 其它k1e?3x?4y?3e?3x?? 当 y?0时， fx1y1?x,y??k1?4y?0e3 当 y＜0时， fx1y1?x|y??0x????,???
?4e?4y,y?0 当 x?0时， fy1|x1(y|x)＝f1(x,y)/fX1(x)??
?0,y?0 当 x＜0时 fy1x1?xy??0??＜y＜??
f2?x,y??3e?3x?3y??② 当 y?0时， fx2y2?x,y??fy2?y??0Y＜0时，fx2x?y x?yy2(x,y)?0 x?(??,??)
?4e?4y0?y?x?f(x,y)?4x??当X≥0时，fy2x2(yx)?2 (1?e)fx2z(x)?0其它?当X＜0时，
fy2x2(yx)＝0,y?(??,??)
215.解 1.由S(D)=4?1/2?7
得X与Y的联合密度函数为f(x,y)???27(x,y)?D
0其它?2. 由于0≤y≤1时，f(x,y)??从而 0≤y≤1时，fy(y)??2/70?x?y?1
其它?0f(x,y)dx??y?10?????2dx?2(y?1)
77
??27又 1≤y≤3时，f(x,y)????0从而 1≤y≤3时，fy(y)?y?1?x?2其它2y?1?????f(x,y)dx??2/7dx?2(3?y); 7又当Y＜0时,Y＞3时，f(x,y)=0,从而fY(y)＝?????f(x,y)dx?0
2y?1)/70?y?1?（?23?y)/71?y?3 综上得： fy(y)??（?0其它???270?y?x?1此外，0≤X≤1时，f(x,y)??
?其它?0从而 0≤X≤1时，fx(x)??????f(x,y)dx??x?102dy?2(x?1)
77
??27当 1≤X≤2时，f(x,y)????0从而 1≤X≤2时，fx(x)?当X＜0或X＞2时，f(x,y)=0 从而fx(x)?x?1?y?x?1其它?????f(x,y)dy??2dy?4
7x?17x?1?????f(x,y)dy?0
2x?1)/70?x?1?（?41?x?2 综上得：fx(x)??7?0其它?16. 证明：
??1Da?x?b又?（x)?y??(x) （１）f(x,y)??
?其它?0