证明S²(x)=1/(n-1)∑[x_i-E(x)]²为var²(x)的无偏估计

需证明E(S²)=var²(x)

∑[x_i-E(x)]²=∑[x_i-1/n∑x_j]²，∑条件为j=1→n

         =1/n²∑[(n-1)x_i-∑x_j]²，∑条件为j=1→n且j≠i

         =1/n²∑[(n-1)²x_i²-2(n-1)∑(x_i x_j)+ ∑x_j²+2∑x_j x_z]，∑条件为j=1→n，z=1→n，且j≠z≠i

E∑[x_i-E(x)]²=1/n²∑[(n-1)² E(x_i²)-2(n-1)∑E (x_ix_j)+ ∑E (x_j²)+2∑E(x_jx_z)]，

知抽样样本相互独立E (x_ix_j)=E(x_i)E(x_j)，且var(x)= E(x²)- E(x)²，且∑有n项，∑有n项，∑有n-1项，∑有(n-1)(n-2)/2项

E∑[x-E(x)]²=1/n²∑[(n-1)²E(x_i²)-2(n-1)(n-1)E(x)²+(n-1)E(x_j²)+(n-1)(n-2)E(x)²]，

           =1/n²∑[(n-1)² var²(x)+ (n-1) var²(x)]，

           =1/n² * n *[(n-1)² var²(x)+ (n-1) var²(x)]

           =(n-1) var²(x)

所以E(S²)=var²(x)