[E((X{i}-\overline{X})^{2})=(\mu{i}-\overline{\mu})^{2}+\frac{n-1}{n}\sigma^{2}]

$E((X_{i}-\overline{X})^{2})=(\mu_{i}-\overline{\mu})^{2}+\frac{n-1}{n}\sigma^{2}$

Cov(X,aY) = a*Cov(X,Y) #

📃

\[\overline{\mu}=\frac{1}{n}\sum_{i=1}^{n}\mu_{i}\]

$X_i$ oberoende

$E(X_i)=\mu_i$

$\textrm{Var}(X_i)=\sigma^2$

Bevis

\[E\left((X_{i}-\overline{X})^{2}\right)=\left(E(X_{i}-\overline{X})\right)^{2}+\textrm{Var}\,(X_{i}-\overline{X})\]

$E(U^{2})=\left(E(U)\right)^{2}+\textrm{Var}\,U$

$(\mu_{i}-\overline{\mu})^{2}$

$\textrm{Var}\,X_{i}+\textrm{Var}\,\overline{X}-2\textrm{Cov}(X_{i},\overline{X})$

$\sigma_i$

\[\frac{1}{n^{2}}\sum_{i=1}^{n}\sigma^{2}\]

eftersom $X_i$ oberoende, så summan av ind. varianserna

\[=\frac{1}{n}\sigma^{2}\]

\[=-2\frac{1}{n}\sum_{j=1}^{n}\textrm{Cov}(X_{i},X_{j})\]

Cov är ju noll förutom när j=i
Cov(Xi,Xi)=Var(Xi)=$\sigma^2$

\[=\sigma_{i}+\frac{1}{n}\sigma^{2}-2\frac{1}{n}\sigma^{2}=\frac{n-1}{n}\sigma^{2}\]

Cov(a, b + c) = Cov(a,b) + Cov(a,c) #

Cov(X,aY) = a*Cov(X,Y) #