Question

prove that expected value of sample mean equals population

Answer 1

$Population:x_0,x_1,x_2,x_3,.....,x_N\\A\ sample\ of\ k\ data\ elements:S(i,k)=x_{i+1},x_{i+2},x_{i+3},....,x_{i+k}$

$E(X)=\Sigma_{j=0}^N\ x_j\ * p(x_j)=\mu,\ \ p(x_j)=probability\ of\ x_j\\\\P(S(i,k))=Probability\ of\ sample=\Sigma_{j=i+1}^{i+k}\ p(x_{j})$

$E(S(i,k))=\frac{\Sigma_{j=i+1}^{i+k}\ x_{j}\ * p(x_{j}) }{\Sigma_{j=i+1}^{i+k}\ p(x_{j})}\\\\=\frac{\Sigma_{j=i+1}^{i+k}\ x_{j}\ * p(x_{j}) }{P(S(i,k))}=\mu_{S(i,k)}\\\\E(\mu_{S(i,k)})=\Sigma_{i=0}^{\frac{N}{k}-1}\ E(S(i,k))\ *\ P(S(i,k))\\\\=\Sigma_{i=0}^{\frac{N}{k}-1}\ \Sigma_{j=i+1}^{i+k}\ x_{j}*p(x_j)\\\\=\Sigma_{j=1}^{N}\ x_j*p(x_j)= \mu,\ \ mean\ of\ population,\\\\. \ \ \ \ \ \ \ \ \ as\ \Sigma_{j=1}^{N}p(x_j)=1$