數(shù)理統(tǒng)計(jì)答案整理

1、8、設(shè)總體具有連續(xù)的分布函數(shù)，是來(lái)自總體的樣本，且X()Fx12nXXX?X，定義隨機(jī)變量iEX??1(12)0iiiXYinX??????????試確定統(tǒng)計(jì)量的分布.1niiY??解由題意知，，.(1)iYBp:()1()1iipPXPX??????????12in??從而由二項(xiàng)分布定義知.1()niiYBnp??:19、設(shè)是來(lái)自總體的樣本，試求：12nXXX?[]XUab:1)、的密度函數(shù)2)、的密度函數(shù).(1)X()nX解因?yàn)?，?/p>

2、以的密度函數(shù)與分布函數(shù)分別為[]XUab:X1[]()0[]xabfxbaxab?????????0()1.xaxaFxaxbbaxb??????????????因此所求的(1)1()(1())()nfxnFxfx???11(1)[]0[]nxanxabbabaxab?????????????1()[]()0[]nnnbxxabbaxab???????????()1()(())()nnfxnFxfx??11()[]0[]nxanxabb

3、abaxab????????????所以解得.??22212abXbaM???????????::121233aXMbXM?????????最大似然估計(jì)法：因?yàn)樵摽傮w的密度函數(shù)為X??1a0xbfxababba??????????未知；其他，則該總體決定的似然函數(shù)為??????????1211a120ninniixbinbaLabLabxxxfxab????????????????其他，因?yàn)樗迫环匠?，，顯然似然方????1ln0nLab

4、naba??????????1ln0nLabnbba???????程關(guān)于無(wú)解.這時(shí)利用似然估計(jì)的定義，當(dāng)時(shí)，有ab??12iaxbin????，則，??????12naxxxb??????????????????12111nnnnLabLabxxxbaxx??????顯然當(dāng)，時(shí)，可使似然函數(shù)取最大值，因此的最大似然估計(jì)為，:??21aX?:??2nbX?ab，.:??211miniinaXX????:??21maxininbXX????

5、其中??????228)11xfxx???????2301.x?????解矩估計(jì)法：因?yàn)榭傮w的數(shù)學(xué)期望為X，????222211xxXxx?????????????E所以，得到.2X??:12X??最大似然估計(jì)法：因?yàn)榭傮w的分布密度為X其中??????2211xfxx???????2301.x?????則該總體決定的似然函數(shù)為，??????????22121111innxniiiiLLxxxfxx????????????????其中23