003-2004学年第一学期概率论与数理统计(A)期末考试试卷答案 Px=1,Y=l}=P=1,n=1}=P=lpn=339 P(⑦)=0 P{X=1,Y=3}=P()=0 P{X=2,Y=}=P=1,n=2}+P{=2,n=l} =P2=1P{=2+P{=2P{=1}=1×1+1x 2 3333 P{x=2,y=2=Pk=2,n=2=Pk=2h=2}=1×1=1 P{X=2,y=3}=P( =0 P{X=3,Y=1}=P=3,n=1}+P{=1,n=3} =P=3y=+P=1)P=3}=1×1+1×1=2 333 P{X=3,Y=2}=P=3,n=2}+P=2,n=3 =P=3P=2}P=2P{v=3} 2 33339 P{x=3,y=3}=P{=3m=3}=P=3=3}=1×=1 因此二维随机变量(X,Y)的联合分布律及X的边缘分布律为 pi 0 2 2 (2)E(X)=1×+2×2+3×= 2 E(X)=1×2+2×=+3×=+4x+6×=+9×=4 9 12.设总体X的数学期望为μ,方差为σ>0,现从中分别抽取容量为n与n2的两个独立样本,这 两个样本的样本均值分别为x与X2证明:对于满足a+b=1的任何常数a及b,Y=ax1+b2是的 第9页共10页2003-2004 学年第一学期概率论与数理统计(A)期末考试试卷答案 第 9 页 共 10 页 9 1 3 1 3 1 P X =1, Y =1 = P =1, =1 = P =1 P =1 = = ; PX =1, Y = 2= P()= 0 ; PX =1, Y = 3= P()= 0 ; PX = 2, Y =1= P =1, = 2+ P = 2, =1 9 2 3 1 3 1 3 1 3 1 = P =1 P = 2 + P = 2 P =1 = + = ; 9 1 3 1 3 1 P X = 2, Y = 2 = P = 2, = 2 = P = 2 P = 2 = = ; PX = 2, Y = 3= P()= 0 ; PX = 3, Y =1= P = 3, =1+ P =1, = 3 9 2 3 1 3 1 3 1 3 1 = P = 3 P =1 + P =1 P = 3 = + = ; PX = 3, Y = 2= P = 3, = 2+ P = 2, = 3 9 2 3 1 3 1 3 1 3 1 = P = 3 P = 2 + P = 2 P = 3 = + = ; 9 1 3 1 3 1 P X = 3, Y = 3 = P = 3, = 3 = P = 3 P = 3 = = ; 因此二维随机变量 (X, Y ) 的联合分布律及 X 的边缘分布律为 Y X 1 2 3 i p 1 9 1 0 0 9 1 2 9 2 9 1 0 9 3 3 9 2 9 2 9 1 9 5 j p 9 5 9 3 9 1 ⑵ ( ) 9 22 9 5 3 9 3 2 9 1 E X =1 + + = , ( ) 9 14 9 1 3 9 3 2 9 5 E Y =1 + + = , ( ) 4 9 1 9 9 2 6 9 1 4 9 2 3 9 2 2 9 1 E XY =1 + + + + + = . 12.设总体 X 的数学期望为 ,方差为 0 2 ,现从中分别抽取容量为 1 n 与 2 n 的两个独立样本,这 两个样本的样本均值分别为 X1 与 X 2 .证明:对于满足 a + b =1 的任何常数 a 及 b ,Y = aX1 + bX2 是 的