=??2x,0?x?1,
?0,其他.关于Yの边缘概率密度
fY(y)=??????1dx,0?y?2,?yf(x,y)dx=??2
其他.??0,y??1?,0?y?2, =? 2其他.??0,(II) 令FZ(z)?P{Z?z}?P{2X?Y?z}, 1) 当z?0时,FZ(z)?P{2X?Y?z}?0;
2) 当0?z?2时,FZ(z)?P{2X?Y?z} =z?12z; 4 3) 当z?2时,FZ(z)?P{2X?Y?z}?1.
?0,z?0,?1即分布函数为: FZ(z)??z?z2,0?z?2,
?4z?2.?1,1??1?z,0?z?2,故所求の概率密度为:fZ(z)?? 2其他.??0,(23)(本题满分9分)
设X1,X2,?,Xn(n?2)为来自总体N(0,1)の简单随机样本,X为样本均值,记
Yi?Xi?X,i?1,2,?,n.
求:(I) Yiの方差DYi,i?1,2,?,n; (II)Y1与Ynの协方差Cov(Y1,Yn).
【分析】 先将Yi表示为相互独立の随机变量求和,再用方差の性质进行计算即可;求
Y1与Ynの协方差Cov(Y1,Yn),本质上还是数学期望の计算,同样应注意利用数学期望の运
算性质.
【详解】 由题设,知X1,X2,?,Xn(n?2)相互独立,且
EXi?0,DXi?1(i?1,2,?,n),EX?0.
11n(I)DYi?D(Xi?X)?D[(1?)Xi??Xj]
nnj?i121 =(1?)DXi?2nn?DXj?inj
(n?1)21n?1??(n?1)?. =
nn2n2(II) Cov(Y1,Yn)?E[(Y1?EY1)(Yn?EYn)] =E(Y1Yn)?E[(X1?X)(Xn?X)]
2 =E(X1Xn?X1X?XnX?X)
=E(X1Xn)?2E(X1X)?EX
n222 =0?E[X1??X1Xj]?DX?(EX)
nj?22 =?211???. nnn
2005考研数学一真题及答案
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