设X1,X2是来自任意总体UND的一个容量为2的-|||-样本, =dfrac (2)(3)(X)_(1)+dfrac (1)(3)(X)_(2) =dfrac (1)(4)(X)_(1)+dfrac (3)(4)(X)_(2),-|||-=dfrac (2)(5)(X)_(1)+dfrac (2)(5)(X)_(2), =dfrac (1)(2)((X)_(1)+(X)_(2)).-|||-(1)确定上面4个E (X)的估计量中,那些是E(X-|||-)的无偏估计量-|||-(2)在无偏估计量中找出最有效的估计量,并说-|||-明原因。

设sim N(mu ,1), X1、X2、X来自总体样本, hat (mu )=dfrac (2)(5)(X)_(1)+dfrac (2)(5)(X)_(2)+

x1，x2是取自总体N(μ，1)(μ未知)的样本。hat (mu )_(1)=dfrac (2)(3)(X)_(1)+dfrac (1)(3)(X)_(2) ;

4.设总体 sim N(mu ,(sigma )^2), x1,x2,x3为来自X的样本 hat (mu )=dfrac (1)(4)(x)_(1)+b(x)_

1.设X1,X2,···,xn来自总体X的样本, (X)=(sigma )^2, overline (X)=dfrac (1)(n)sum _(i=1)^n(X

设X1,X2为来自正态总体N(μ,σ^2 )的样本,下列不是μ的无偏估计量的是-|||-A (mu )_(1)=dfrac (2)(3)(X)_(1)+dfra

[题目]设总体 approx N(mu ,1), (x1,x2,x3)为其样本,-|||-若估计量-|||-mu =dfrac (1)(2)(x)_(1)+df

(8)设X1,X2,···,X9是来自正态总体N(μ,σ^2)的样本, _(1)=dfrac (1)(6)sum _(i=1)^6(X)_(i) , _(2)=

(B) dfrac (1)(2)(X)_(1)+dfrac (1)(2)(X)_(2)-|||-(C) dfrac (1)(2)(X)_(1)+dfrac (1

样本X1,X2,X3来自总体X,若 hat (mu )=dfrac (1)(3)(X)_(1)+-|||-(X)_(2)+dfrac (1)(2)(X)_(3)

已知随机变量X1和X2的概率分布为-|||-X1 -1 0 1 X2 0 1-|||-pi dfrac (1)(4) .dfrac (1)(2) dfrac (