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1、第7章1.设总体的概率密度为xx2/(22)f(x)2e,x0,0,x0求参数的矩估计量和极大似然估计量。解:矩估计xx2/(22)x2/(22)E()xf(x)dxxedxxde02021xx2/(22)2x2edxed002令E()2,得的矩估计量ˆ22极大似然估计:设样本观测值为x,x,,x12nnn1x2nx2222iiexi/(2)2nei1x,x0(i1,...,n)似然函
2、数L()2iii1i10,其它1n2n当xi0(i1,...,n)时,lnL2nln2xilnxi2i1i1dlnL2n1n2令3xi0di1ˆ1n2,极大似然估计量ˆ1n2得的极大似然估计值xii2ni12ni1k12.设总体X服从几何分布:P{Xk}(1p)p(0p1),k1,2,3,。求p的极大似然估计量。x1解:总体分布律写成P{Xx}(1p)p(0p1),x1,2,3,似然函数nnxinL(p)(1p)xi1p
3、pn(1p)i1,x1,2,3,...(i1,...,n)ii1nlnLnlnpxinln(1p)i1nxndlnLni1ni1ˆ令0,得p的极大似然估计值pxixdpp1pni13.设总体X的分布律为:X012322pθ2θ(1-θ)θ1-2θ其中01/2为未知参数,利用如下样本值:3,1,3,0,3,1,2,3求θ的矩估计值和最大似然估计值。2解:矩估计:E(X)2(1)23(12)34ˆ3Xˆ1令E(X)34X得θ的矩估计量,代入样本观测值得
4、矩估计值44极大似然估计:似然函数为L()P{X3,X1,X3,X0,X3,X1,X2,X3}12345678P{X3}P{X1}P{X3}1282224[2(1)](12)6244(1)(12)lnLln46ln2ln(1)4ln(12)dlnL6282令0121430,d112ˆ713注意01/2,得的极大似然估计值0.282912n14.设总体~(,2)22的XN,X1,X2,,Xn为其样本,试求常数C使
5、C(Xi1Xi)为i1无偏估计量。n1n1222解:E(C(Xi1Xi))C[E(Xi1)2E(Xi1Xi)E(Xi)]i1i1n122C[D(X)E(X)]2E(X)E(X)[D(X)E(X)]i1i1i1iiii122222C[(n1)(n1)2(n1)(n1)(n1)]22C2(n1)1C2(n1)21115.设总体X~N(,1),X1,X2为其样本,问:估计量ˆ1X1X2,ˆ2X1X2,332211ˆXX中,哪一
6、个是的较有效的估计量?31232212121解:E(ˆ)E(XX)E(X)E(X)E(X)E(X)11212333333111111E(ˆ)E(XX)E(X)E(X)E(X)E(X)212122222221111115E(ˆ)E(XX)E(X)E(X)E(X)E(X)312123232326故只有ˆ,ˆ是的无偏估计量,下面对它们考察有效性1221415D(ˆ)D(XX)D(X)D(X)112123399911111D(ˆ)D(XX)D(X)D(X)2121222
7、442ˆ比较有效。212226.设总体X~B(1,p),X1,X2,Xn为其样本,验证统计量T(X1X2Xn)是n参数p的相合估计量。2证明:X~B(1,p)E(X)p222由X,X,X的独立同分布性可知X,X,X仍然独立同分布,满足独立同分布大数12n12nnn1212定律的条件,而E(T)E(Xi)E(X)p故ni1ni11222pT(XXX)p(n)12nn1222即统计量T(XXX)是参数p的相合估计量12nn注:也可用切比雪夫不等式直接证明nn1212E(T)E(Xi
8、)E(X)p,ni1ni1nnn2121212ppD(T)2D(Xi)2D(X)2(pp)ni1ni1n