《概率论》计算与证明题
n????F?(y)?1?exp??y???i?
??89
当y?0时F?(y)?0。求导得?的密度函数为,当y?0时p?(y)?0;当y?0时
n??p?(y)?F??(y)???jexp??y??j?.
j?1?j?1?n
30、解:设(0,a)在内任意投两点?1,?2,其坐标分别为x,y,则?1,?2的联合分布密度为
(x,y)?(0,a)?(0,a)?0,?p(x,y)??1。
,(x,y)?(0,a)?(0,a)??a2设??|?1??2|,则?的分布函数为,当z?0时F?(z)?0;当z?a时F?(z)?1;当0?z?a时,
F?(z)?P{|?1??2|?z}??z?x?y?z0?x,y?a2??p(x,y)dxdy?21a2?z?x?y?z0?x,y?a2??dxdy?1S, 2a积分S为平面区域ABCDEF的面积,其值为 a?(a?z)?2az?z,所以
F?(z)?(2az?z2)/a2.
31、证:由独立性得,V?(x,y,z)的概率密度为 p(x,y,z)?12?21(2?)?33e?(x2?y2?z2)
S?x2?y2?z2的分布函数为,当s?0时,
F(s)?Px?y?z?s?2?222?x?y?z?S2??122(2?)3?3e?12?2(x2?y?z2)dxdydz
2作球面坐标变换,x??cos?sin?,y?sin?sin?,z??cos?,则|J|??sin?,
F(s)??2?0d??sin?d??0?a01(2?)3?31??2/?22e1??2/?22??2d?
?2??2
?a01(2?)3?3e??2d?
由此式对s求导可得,当s?0时,S的密度函数为
?s2?F'(s)?f(s)?exp???2?2??. ??2??
32、证:(3.14)式为
2s2
《概率论》计算与证明题 90
p(x)?1?1?2??n??2?1n2x11n?1?x22e,x?0。
令y?当y?0时
x,则x?ny2,x'y?2ny,由p(y)?p[fn?1(y)]|[f?1(y)]'|得,
?的密度函数为,np?/n(y)?(ny)1n221n?12?1?2????2?e1?ny22?2ny?2nyn?1?1?2????2?1n21n2e1?ny22
?与
?仍独立。记T??/?/n,则由商的密度函数公式得T的密度函数为 n?pT(t)??|y|p?(ty)p??(y)dy??y??/n0?12?e1?t2y22?2ny1n21n2n?1e1?ny22?1?2??n??2?dy
???0n1n2?1?2?2??n??2?21n2?(y)21(n?1)?12e1?y2(n?t2)2dy2,
令u?y(n?t),则dy?22du,得 2(n?t)pT(t)?n(n?t)1n21n221?(n?1)2?1?2?2??n??2???u0?11(n?1)?1?u22edu
?n12?1?2?2????2?1n2?1???(n?1)?1?(n?1)2??2 ?(n?t)2?1????2?1(n?1)2?1?1??(n?1)?2?2(n?1)t?2???? ???t?? ?pT(t)??1???n??1?n???n???2?
33、解:U的分布函数为,当t?0时F(t)?0;当t?0时有
《概率论》计算与证明题 91
F(t)?x?y?z?t???p(x,y,z)dxdydz??dx?0tt?x0dy?t?x?y06dz 4(1?x?y)tt?2t22 ???dxdy 33??00(1?t)2(1?x?y?z)tt?t2t11tt2 ????dx?dx?1???3222300t?1(1?t)(1?t)(1?x)(1?t)(1?t)3t2对F(t)求导可得U的密度函数为,当t?0时p(t)?0;当t?0时p(t)?。 4(1?t)
34、证:(U,V)联合分布函数为
1?2(x2?y2)F(u,v)???edxdy
2?x2?y2?ux?vy1当s?0时作变换,s?x?y,t?22x,反函数有两支 y?sx?t?(1?t2)??s?y?2?(1?t)?J?1与?sx??t?(1?t2)? ?s?y??s2?(1?t)?2x2y2x21x??2?2??2(t2?1),|J|??1 2?2y2(1?t)yy考虑到反函数有两支,分别利用两组
????11uv1?2s1??1?2(x2?y2)F(u,v)????????edxdy?2??e?dt 20??2?2(1?t)?x2?y2?ux2?y2?u?2?x?x?v,y?0?v,y?0?y?y?对F(u,v)求导,得(U,V)的联合密度为(其余为0)
1?u1p(u,v)?e2?,22?(1?v)1?u若令pU(u)?e2(u?0),211u?0,0?v?? 1?(1?v2)pV(v)?(???v??),
则U服从指数分布,V服从柯西分布,且p(u,v)?pU(u)?pV(v),所以U,V两随机变量独立。
《概率论》计算与证明题 92
35、证:当x?o时,?与?的密度函数分别为
p?(x)??r1?(r1)xr1?1??xe,p?(x)??r2?(r2)xr2?1e??x;
当x?0时,p?(x)?p?(x)?0。设U????,V??。当s?0或t?0时,(U,V)联合密度为?当s?0,t?0时,作变换s?x?y,t?p(s,t)?0;所以
sxsts,得x?,y?而|J|?,2y(1?t)(1?t)(1?t)p(s,t)??r?r12?(r1)?(r2)xr1?1yr2?1e??(x?y)|J|
r?1r?112s?st??s???s ?????e2?(r1)?(r2)?1?t??1?t?(1?t)12?r?r??r1?r2??(r1?r2)tr1?1r1?r2?1??s???se????r1?r2??(r1)?(r2)???(r1)?(r2)(1?t)
由此知U服从分布服从分布,且U与V相互独立。
36、解:令U????,V????pU(s)pV(t) ??(???),当s?0或t?(0,1)时,U,V联合密度p(s,t)?0;当s?0且t?(0,1)时作变换s?x?y,y?x,则x?st,y?s?st,|J|?s,
(x?y)p(s,t)?e?xe?y|J|?se?(x?y)?se?s?1?pU(s)pV(t)
由此得U服从??分布G(1,2),V服从(0,1)分布,且U与V相互独立。
37、解:
p(x,y)?12??1?2??(x?n)22r(x?a)(y?b)(y?b)2??1??exp???????2222??2(1?r)???1?r1212?????
设Ui????,Vi????;U?U1?a?b,V?V1?a?b。作变换s?x?y?a?b,
t?x?y?a?b则x?a?111(s?t),y?b?(s?t), |J|?。U,V的联合密度函数为 222f(s,t)?p(x,y)|J|
??(s?t)22r(s?t)(s?t)(s?t)2??111????exp???? 2?22??22??1?21?r24??2(1?r)4?4??1212?????
《概率论》计算与证明题 93
?14??1?2?122222222? exp??s????2???t????2???2st(???)?12121212212221?r2?8(1?r)?1?2???????设U,V的边际分布密度函数分别为fU(s),fV(t),欲U与V独立,必须且只需f(s,t)?fU(s)?fV(t),由f(s,t)的表达式可知,这当且仅当?2??1?0时成立。U,V相互独立与Ui,Vi相互独立显然是等价的,所以Ui????,Vi????相互独立的充要条件是?1??2。当?1??2??时,得
22????s21s2, fU(s)?exp??f(t)?exp??V2?2?2??(1?r)2??(1?r)?4(1?r)???4(1?r)??1U~N(0,2(1?r)?2),V~N(0,2(1?r)?2)。
38、解:(1)因为指数中二次项x,y,xy的系数分别为?1,?较知,可设其配方后的形式为
221,?1,所以与(2.22)式(见上题解答)比21?1?(x?s)2?(y?t)2?1?(x?s)(y?t)。
2??2s?t?11??比较系数得 ? ?s?t?7??s2?1t2?st?321?22?此方程组有唯一解s??4,t??3,由此得
p(x,y)???11??exp???x?4)2?(y?3)2?(x?4)(y?3)?? 2?2??????2??11(y?3)1(x?4)(y?3)????2?exp?(x?4)??2? ???? 12121?2???2(1?)?2??1?21???2??2(2)与(2.22)式比较得,a?4,b?3,?1?1,?2?2,r??12。
(3) p1(x)?
?(x?4)2?exp???, p2(y)?2?2??1?(y?3)2?exp???。
4?2??12?11?p(x,y)11??????1??1(4)p(x|y)??exp???x???y?5???,它服从N??y?5,?。
22?p2(y)2????2?????2?
概率论答案 - 李贤平版 - 第三章
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