贝叶斯统计茆诗松版大部分课后习题答案Word格式.docx
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1192192,,,,,,,,,,,,,hXpXxxx(,)()()(8,0,,)123347,,,
,,,,192于是,,,,,mXhXdd()(,)7,,88,
的后验分布为
76hX(,)192/68,,,()X,,,,,7,,192mX(),d,7,8,
6,68,,8,,,7()X,,,,,
0,8,,,
1.12样本联合分布为:
1pxx,,,,,(),0n,
,,1,,,,,,/,,00(),,,,0,,,,,0
,,,,,,nn11,,,,,,,,,,,,()()()/1/,max,,,xpxxx,,,,,,,0101n
,,n11/,因此的后验分布的核为,仍表现为Pareto分布密度函数的核,
,,,,nn1,()/,,,,,,,,n11即()x,,,,0,,,,1,
即得证。
1.15
n,,xi,nnnx,,i,1样本的似然函数:
,1()pxee,,,,,
,,,1,,,,()e,,,,,,,
nnx,,,,1(),,,参数的后验分布,,()()()xpxe,,,,,,,
服从伽马分布,,Gannx,.,,,,
,0.0002,,,,
(2)4,20000.,,,,,,,2,0.0001,2,,,
二、1,2,3,5,6,7,8,10,11,12
,tpte(),,,2.2解:
由题意,变量t服从指数分布:
,tni,样本联合分布pTe(),,,
,,,1,,,~(,),0Gae,,E()0.2,,Var()1,,且,,,,,,,(),
由伽玛分布性质知:
,0.2,,,,0.04,0.2,,,,,,,,1,2,,,
又已知n=20,t,3.8
nn
t,,,203.876nt,,,,,,20.04,76.2,所以,,ii,1i,1i由于伽玛分布是指数分布参数的共轭先验分布,而且后验分布
,,,,,tt(),,,,,nn,,,11,,ii()()()tpTeee,,,,,,,,,,,
GantGa(,)(20.04,76.2),,,,,即后验分布为,i
,n20.04,|TE()0.263,,,,,t76.2,,i
1,,,IGantIGa(,)(20.04,76.2),,,,,服从倒伽玛分布,i
,t,i,,||1TT,()()4.002EE,,,,,1,,n,
11Ga(11,4)2.3可以算出的后验分布为,的后验期望估计的后验方差为.,,16
2.5.n,36
,,1,,,,,,/,,002.7的先验分布为:
(),,,,0,,,,,0令,,,max,,,xx,,101n
,,,,nn1,()/,,,,,,,,n11可得后验分布为:
()x,,,,0,,,,1,
(),,,n1则的后验期望估计为:
,,Ex(),,n,,1,2(),,,n1Varx(),后验方差为:
.,2
(1)
(2)nn,,,,,,
n12.8由可以得出,,,xGaIGa~(,),~(,)22,
n12()1n,,1x,2,22pxxex,,,(),0n,()2
,,,,,
(1),,(),0,,e,,,,,(),
(1)的后验分布为:
x,2,n,,,,
(1),22,,,,,,,()()()xpxe,,
nx即为倒伽玛分布的核。
IGa(,),,,,22
nx所以的后验分布为,IGa(,),,,,22
x,,x2,,2
(2)后验均值为Ex(),,,nn22,,,1,,,2
x2(),,2后验方差为Varx(),,nn2
(1)
(2),,,,,,22
(3)样本分布函数为:
nnn,1,,n,xnn2i,,1
(2),,2,,,12ipxpxxe()(),,,,ii,,,,n(/2),,,11ii,,,,所以的后验分布为:
nx,2,i,2ni,1,,,,
(1),22,,,,,,,()()()xpxe,,
n
x,2in,1i(,),,,,IGa即为的核。
22
n1n21(),n,,xni,,,1,2,,,
(1)n,,2,1i2,(xpxxee,,,,,,,,)()()[]*i,n,(),,1i,()2
(dx,,),令0d,
即:
nnnn1,,22,,xxii,,222x,,2()nnn,2i,,,11iin,,,,,,,,,121,,n,n,12i22222,,xee,,,,,[][
(1)*]0,,,i,2n,()22,,,1i,()2
xn,i,1i,2,x,,,i,,12i可得,,,MD22n,,22n,,,1,2
xn,i,1i,2,x,,,i,,12i而由公式得,,,E22n,,22n,,,1,2
因此,倒伽玛分布的这两个估计是不一样的,原因是它不对称。
xNN~(,1),~(3,1),,2.10解:
已知
2设的后验分布为N(,),,,11
可得:
,,22x,,,,0,,122,,,,,0
111,,222,,,10
2,,1243,,2,,由已知得:
,,x,,30n33
333111,,,2,,?
,,,3,1131134,,
[30.51.96,30.51.96],,,,所以的95,的可信区间为:
[2.02,3.98]即为.
22xNIGa~0,,~,,,,,2.11已知,,,,
nn1,,22,可得的后验分布为,,IGax,,,,i,,22,1,,i
n12,,x,i2i,1ˆ后验均值为:
,En,,1,2
2n,,12,,x,i,,2i1,,,2Varx,后验方差为:
,,2nn,,,,,,,,12,,,,,,22,,,,
变换:
n11n,,2,,~,Gax,,,i,,222,,1,,i
n1,,n,,,,22,,,2~2,,x,i,,,,2,,2,,,,1,,i,,
n,,1,,22令:
Pxn,,,220.9,,,,,,,0.1i,,,,2,,1,,i,,
n22,,x,i2,1i,可得的0.9可信上限为.2n,2,,,,0.1
,,1,,,,,,/,,002.12,的先验分布为:
(),,,,0,,,,,0
,,max,,,xx令,,101n
设的可信上限为1,,,,U
U,,,,xd,,1则,,,,1
带入有:
U,,,nn1,,()/1,,,,,,,,nd1,,1
,n,,1,,,,n,,U
1,n1,,,,,,,U1,,,,,
三、10,11,12,133.10解:
依题意
1x,,pxxexp,0,,,,,,,,,,,,
0.01,,,20.01exp,0,,,,,,,,,,,,,,
,,0.01x,,,,3则mxpxdd0.01exp,,,,,,,,,,,,,,,,,,,0,,,,
0.01,0x,,2x,0.01,,
该元件在时间之前失效的概率200:
2002000.01pmxdxdx,,,0.99995,,2,,00x0.01,,,
3.11:
解依题意
xi,,,iipxe,,,,iix!
i
,,,1i,,,,,e,0,,,,,,iii,,,,
xi,,,,,,1iii,,,,,,mxpxdeed,,,,,,,,iiiiiii,,,,,,,,0x!
,,i,,
,,,,x,,ix,i,,,,x1!
,,,,i,,nnn,,,,x,,,i,mxmx,,,,,,,,,,,ix,i,,ii,,11,,,,1!
x,,,,,i,,
3.12解:
超参数和的似然函数为,,
333,,,3,,,,,,,xf35,,,,,,,,,,,,,,,,,,iL,,,,其中,,,,,,338383x,,,,,,ii,1720,,,x1!
13!
5!
1,,,,,,,,,,,,,,,,,,,,,i,,,,
222f,,,,,1234.,,,,,,,,,,,,,,,,
由
L,,,,,,,0,,,,,,L,,,,,,,0,,,,
38,,,,,,,,1,,ff,3ln,,,,,,,,,,,,,,从而有:
38,,,,,,,,ff,3ln,,,,3,,,,
3,利用软件计算,可得,,,1.033599=0.3875996,,83.13证明:
泊松分布的期望和方差分别为
2,.,,,,,,,,,,,,
,1,,,,,,=,0e,,,,,,,,,,,,,
,,,,,,,,,,,ed,,m,,,,,,0,,,,,
x,22,,,,,,,,,EE,,,,,,,,,,,,,,,22,12,,,,,,2,,,,,,,,,,,,,,,,EE,,,,,,,,,,,,,m,,,,22,,,,,,,,,,,,,,,,,,
2,,,,,,,m2,,,,,利用样本矩代替边际分布的矩,列出如下方程:
,x,,,,,,,2,S,,2,,,,
2,,x,,,2,Sx,,,,x,,2,,Sx,,
四、1,4,8,9,10,11,12,15,16
4.4
15,6,7,8,9,10,5,6,7,8,9,10状态集行动集,,,,,,,,,,
2收益函数,,
5,10aa,,,,Qa,,,,,,6,5,,,aa,,,
收益矩阵
aaaaaa123456
252423222120,,1,,,2530292827262,,,
,2530353433323,Q,,,2530354039384,,,,,2530354045445,,,,,253035404550,,6,
3根据定义可知,最优a5行动是,即采摘朵鲜花,,1
4按折中准则:
,,
HQaQamax,1min,,,,,,,,,,,,,,,,,,,,,,,
H,25,,,,1,H246,,,,,,,2
H2312,,,,3,,,,H,,2218,,4,,,,H,,2124,,5,,,
H,,2030,,6,,,
105a当时,选择,每天摘朵鲜花,,1,6
1当时,选择,每天摘朵鲜花,,110a.66,4.8
La,1500,,13
购买8件.4.9
对于行动,其收益函数为a1
100,00.1,,,,
Q,,,30,0.10.2,,,,,
,,,50,0.21,,对于行动,其收益函数为a2Q,,,,,40,01,,
从而可得在和处的损失函数:
aa12
0,