神马作文网概率论问题,求教
来源：学生作业帮 编辑：神马作文网作业帮 分类：数学作业 时间：2024/11/12 10:19:49
概率论问题,求教
E(T1)=E(2/n ΣXi)=(2/n)(n*E(X))=2E(X)=sita
设Z=max(X1...Xn)
Fz(z)=(z/sita)^n
fz(z)=nz^(n-1)/(sita^n)
E(Z)=∫(0~sita) (nz^n)/sita^n dz
=(n/n+1) sita
E(T2)=(n+1)/nE(z)=sita
D(T1)=(4/n²)nD(X)=(4/n)D(X)=sita²/3n
E(Z²)=∫(0~sita) (nz^(n+1))/sita^n dz
=(n/(n+2))sita^(n+2)/sita^n
=(n/(n+2))sita²
D(Z)=(n/(n+2)-n²/(n+1)²)sita²
D(T2)=(n+1)²/n²(n/(n+2)-n²/(n+1)²)sita²
=(n(n+1)²/(n+2)-n²)/n²sita²
n(n+1)²-n²(n+2)=n³+2n²+n-n³-2n²=n
D(T2)=sita²/n
T1方差较小,更有效
设Z=max(X1...Xn)
Fz(z)=(z/sita)^n
fz(z)=nz^(n-1)/(sita^n)
E(Z)=∫(0~sita) (nz^n)/sita^n dz
=(n/n+1) sita
E(T2)=(n+1)/nE(z)=sita
D(T1)=(4/n²)nD(X)=(4/n)D(X)=sita²/3n
E(Z²)=∫(0~sita) (nz^(n+1))/sita^n dz
=(n/(n+2))sita^(n+2)/sita^n
=(n/(n+2))sita²
D(Z)=(n/(n+2)-n²/(n+1)²)sita²
D(T2)=(n+1)²/n²(n/(n+2)-n²/(n+1)²)sita²
=(n(n+1)²/(n+2)-n²)/n²sita²
n(n+1)²-n²(n+2)=n³+2n²+n-n³-2n²=n
D(T2)=sita²/n
T1方差较小,更有效