??(x,y)dxdy??xdx?yfxy??(x,y)dyI???xyfxyD0011?10??(x,y)dy??ydfx?(x,y)?yfxy?(x,y)1?yfxy0??fx(x,y)dy,00111110000011??(x,y)dy??xfx?(x,1)dx??xdx?yfx?(x,y)dy于是，I??xdx?yfxy???xf(x,1)10??xdx?yfx(x,y)dy???dy?xfx(x,y)dx00001111
??[?xfx(x,y)dy??dy?fx(x,y)dx]??dy?f(x,y)dx???f(x,y)dxdy?a00000D110111120解：
1)??1,?2,?3101?013?1?0115又??1,?2,?3不能由?1，?2，?3线性表示，?r(?1，?2，?3)?3,于是?1，?2，?3?0，解得a?5?10111?2)(?1,?2,?3,?1，?2，?3)??00312?11513??100210???1?2?1?4?2??3?????010420?于是??2??1?2?2?0?3?001?101????0??0???123???11??1??3???0?05???0111131401221??1??3???0?64???01011311?11201??3?1???r(?1,?2,?3)?321.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???T1?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??2T?3?0?x??3???0???1?x3?0即x解得?1?x1?x3?0?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??11???????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 0 1 -1 0 1/3 1/3 0 1/3 0 1/3 1 0 1/3 1/3 1/3 1/3 2)Z取值为?1、0、11P(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 3)EX?-1 1/3 0 1/3 1 1/3 222,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dlnL??n??(x??)i0i?1n22?2?0得2?41n22?的极大似然估计值???(xi??0)2.ni?1d?2?(x??)i0i?1n22?2(2)因为1?22?(x??)i0i?1n2~?(n).所以E221?2?(x??)i0i?1n2?n??于是E???D?n2?2nE(11?n2?(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(2?(x??)ii?1n2?4?)?2n2?则D(?nn??42?2/?2)?2n)?D(n?