这个题应该还有一个条件,就是个样本观察值相互独立吧!
依题意,有E[ỡ]=σ,令E[Xi]=m,则
E[Xi^2]=D[Xi]+E[Xi]^2=σ^2+m^2
所以,
E[ỡ]=E[k∑(xi+1-xi)*(xi+1-xi)]
=kE{∑[(Xi+1)^2+Xi^2-2(Xi+1)*Xi]}
=k∑E[(Xi+1)^2+Xi^2-2(Xi+1)*Xi]
=k∑(σ^2+m^2+σ^2+m^2-2m*m)
=k∑(2*σ^2)
=2k*(n-1)σ^2
因此有σ^2=2k*(n-1)σ^2,解得:
k=1/[2(n-1)]