第一章先验分布与后验分布1.1解:令120.1,0.2设A为从产品中随机取出8个,有2个不合格,则22618()0.10.90.1488PAC22628()0.20.80.2936PAC从而有5418.03.02936.07.01488.07.01488.0)()|()()|()()|()|(2211111APAPAPA4582.0)|(1)|(4582.03.02936.07.01488.03.02936.0)()|()()|()()|()|(122211222AAorAPAPAPA1.2解:令121,1.5设X为一卷磁带上的缺陷数,则()XP:3(3)3!ePXR语言求:)4(/)exp(*)3(^gamma1122(3)(3)()(3)()0.0998PXPXPX从而有111222(3)()(3)0.2457(3)(3)()(3)0.7543(3)PXXPXPXXPX1.3解:设A为从产品中随机取出8个,有3个不合格,则3358()(1)PAC(1)由题意知()1,01从而有.10,)1(504)|(504)6,4(/1)6,4(1)6,4()1()1()1()1()1()1()1()()|()()|()|(535310161453105353105338533810AbetaBRBdddCCdAPAPA:语言求(2).10,)1(840)|(840)7,4(/1)7,4(1)7,4()1()1()1()1()1()1(2)1()1(2)1()()|()()|()|(636310171463106363105338533810AbetaBRBdddCCdAPAPA:语言求1.5解:(1)由已知可得.5.125.11,110110/1)()|()()|()|(,2010,101)(5.125.111)|(2112211)|(12,2121,1)|(5.125.11201011111111ddxpxpxxpxpxxxp,,即,时,当(2)由已知可得.6.115.11,1010110/1)()|,,()()|,,(),,|(,2010,101)(6.115.111)|,,(,219.1121,214.1121,211.1121,217.1121215.11212112211)|,,(9.11,4.11,1.11,7.11,5.11,0.12,6,2,1,2121,1)|,,(6.115.112010621621621621621654321621ddxxxpxxxpxxxxxxpxxxpxxxxxxixxxxpi,即,,时,当【原答案:由已知可得()1,0.50.5Pxx1(),10201011.611.51()0.0110mxdv从而有()()()10,11.511.6()Pxxmxvvv】1.6证明:设随机变量()XP:,的先验分布为(,)Ga,其中,为已知,则即得证!),(~),,|()()|,,(),,|(,0,)()(,!!)|,,(121)(121211112111?nxGaxxxexxxpxxxexexexxxpniinnxnnniinxniixnniiniii【原答案:(),0!xePxx1(),0()e因此11(1)()()()xxxPxeee?所以(,1)xGax:】1.7解:(1)由题意可知.1},max{,1)/(1)/(122)()|,,()()|,,(),,|(,10,1)(,,2,1,10,22)|,,(121},max{221},max{21211021212112122111nnxxnnxxnniinnniinnnnniinniininxxddxxdxxxpxxxpxxxnixxxxxxpnn【原答案:由题意可知()1,01因此122()12(1)xxmxdx?因此2()()1(),1()1Pxxxxmxx(实质是新解当n=1的情形)】(2)由题意可知.1},max{,1)/(1)/(13232)()|,,()()|,,(),,|(,10,3)(,,2,1,10,22)|,,(12-21},max{2-22-21},max{22122110212121212122111nnxxnnxxnniinnniinnnnniinniininxxddxxdxxxpxxxpxxxnixxxxxxpnn【原答案:由题意可知12202()36xmxdx?因此()()()1,01()Pxxmx】1.8解:设A为100个产品中3个不合格,则3397100()(1)PAC由题意可知199(202)()(1),01(200)因此3971994296()()()(1)(1)(1)APA?由上可知)297,5(~)|(BeA1.9解:设X为某集团中人的高度,则2(,5)XN:25(,)10XN:2(176.53)51()5pxe由题意可知2(172.72)5.081()5.08e又由于X是的充分统计量,从而有()()()()xxpx?v222(176.53)(172.72)(174.64)55.0821.26eee?因此(174.64,1.26)xNv:1.10证明:设22(,),,Nuu:其中为已知又由于X是的充分统计量,从而有()()()()xxpx?v222222251()()11252()11225252uxxueee因此222251(,)112525uxxNv:又由于21112525所以的后验标准差一定小于151.11解:设X为某人每天早上在车站等候公共汽车的时间,则(0,)XU:.8,861)/(1192192)()|,,()()|,,(),,|(,4,192)(.81)|,,(8,8,5.3,2,1,0,1)|,,(768778774321321321433213213321dddxxxpxxxpxxxxxxpxxxixxxxpi,时,当【原答案:设X为某人每天早上在车站等候公共汽车的时间,则(0,)XU:1(),0pxx当8时,31()pxv43819211()8192mxdv从而有7()()3()()128pxxmxvvv,计算错误】1.12证明:由题意可知1(),0,1,2,...,inpxxinv从而有()()()()xxpx?vv00111nnn?因此的后验分布仍是Pareto分布。1.13解:由题意可知2133164511.15解:(1)设的先验分布为(,)Ga,其中,为已知由题意可知.0,)()|,,(),,|(.0,)()(.,2,1,0,)|()|,,()(121211112111?niiniiixnnnixnnixniinexxxpxxxenixeexpxxxp所以(,)Ga是参数的共轭先验分布。【原答案:设的先验分布为(,)Ga,其中,为已知由题意可知11()(),0,1,2,...,niinxniiipxpxexinv从而有(...