21.解： 1)
?1??1?????令?1??0?，?2??0?则A?1???1,A?2??2,??1??1?????根据特征值向量的定义，A的特征值为?1??1,?2?1,对应的线性无关的特征向量为?1??1??????1??0?，?2??0??r(A)?2?3,?A?0故?3?0??1??1??????x1????1T?3?0令?3??x2?为矩阵A的相应于?3?0的特征向量?A为实矩阵，所以有?T??023?x??3???0???x1?x3?0即x1?x3?0解得?1??0?????1??1??1??0??2??1??1??2）?1?2?3单位化得：r1?（r1,r2,r3)??0?0?,r2??0?,r3??1?,令Q?2??2????1?0??1???1????2???100???100??001???????TT则QAQ??010?,于是A?Q?010?Q??000??000??000??100???????12012?0??1?,0???22.解：
1)P(X2?Y2)?1?P(X2?Y2)?0,即P(X?0,Y?1)?P(X?0,Y??1)?P(X?1,Y?0)?01P(Y?1)?P(X?0,Y?1)?P(X?1，Y?1)?31?P(X?1,Y?1)?,同理如图：3
Y X -1 0 0 1/3 1 0 1/3 0 1 2)Z取值为?1、0、11/3 1/3 0 1/3 1/3 1/3 1/3 1P(XY??1)?P(X?1,Y??1)?,P(XY?0)?P(X?0,Y?0)?P(X?0,Y?1)3 1?P(X?0,Y??1)?P(X?0,Y?1)?31P(XY?1)?P(X?1,Y?1)?3Z P -1 1/3 0 1/3 1 1/3 2223)EX?,EY?0,EXY?0,DX?,DY?,?XY?0
39323.解：
i?11(1)似然函数L?f(x1)f(x2)?f(xn)?enn(2?)??(xi??0)22?2nn取对数得，lnL??nln2??ln?2?2令dd?2?(x??)i0i?1n22?22?1n22?的极大似然估计值???(xi??0)2.ni?1(2)因为1lnL??n2?2??(x??)i0i?14n2?0得?22?(x??)i0i?1n2~?(n).所以E221?2?(x??)i0i?1n2?n??于是E???D?n2?2nE(11?2n?(x??)i0i?12i0n)??2,?4niD(?02?(x??)i?12)?2?4因为?(x??)i?1?2~?(n),所以有D(0?(x??)i0i?1n
)?2n?2右式?D(?(x??)ii?1n?2)?2n则D(??2?42?4n2?2/?2)?2n)?D(n??n