《概率論與數(shù)理統(tǒng)計(jì)》復(fù)旦大學(xué)出版社第七章習(xí)題答案

1、[鍵入文字]1習(xí)題七習(xí)題七1.設(shè)總體X服從二項(xiàng)分布b（n，p），n已知，X1，X2，…，Xn為來自X的樣本，求參數(shù)p的矩法估計(jì).【解】【解】因此np=1()()EXnpEXAX???X所以p的矩估計(jì)量?Xpn?2.設(shè)總體X的密度函數(shù)f（xθ）=22()00.xx???????????中中X1，X2，…，Xn為其樣本，試求參數(shù)θ的矩法估計(jì).【解】【解】23022022()()d233xxEXxxx???????????????????令E

2、(X)=A1=因此=X3?X所以θ的矩估計(jì)量為^3.X??3.設(shè)總體X的密度函數(shù)為f（x，θ），X1，X2，…，Xn為其樣本，求θ的極大似然估計(jì).（1）f（x，θ）=000.exxx????????（2）f（x，θ）=1010.xx????????中中【解】【解】（1）似然函數(shù)111()eeeniiinnxxnniiiLfx??????????????????1lnlnniigLnx???????由知1ddln0ddniigLnx???

3、??????1?niinx????所以θ的極大似然估計(jì)量為.1?X??[鍵入文字]3(2)似然函數(shù)i=12…8.8811()iiLfx????????????顯然L=L(θ)↓(θ0)，那么時，L=L(θ)最大18maxiix????所以θ的極大似然估計(jì)值=0.9.??因?yàn)镋()=E()≠θ所以=不是θ的無偏計(jì).??18maxiix????18maxiix??6.設(shè)X1，X2，…，Xn是取自總體X的樣本，E（X）=μ，D（X）=σ2，=

4、k2??問k為何值時為σ2的無偏估計(jì).1211()niiiXX?????2??【解】【解】令i=12…n11iiiYXX???則21()()()0()2iiiiEYEXEXDY??????????于是1222211?[()](1)2(1)niiEEkYknEYnk??????????那么當(dāng)即時22?()E???222(1)nk????有1.2(1)kn??7.設(shè)X1，X2是從正態(tài)總體N（μ，σ2）中抽取的樣本11221231221131

5、1???334422XXXXXX?????????試證都是μ的無偏估計(jì)量，并求出每一估計(jì)量的方差.123??????【證明】【證明】（1）11212212121?()()()333333EEXXEXEX?????????????????21213?()()()44EEXEX?????31211?()()()22EEXEX?????所以均是μ的無偏估計(jì)量.123??????(2)22221122145?()()()3399DDXDXX??