Let Xn bar 1nn sum i1 Xi be the sample mean A Show that the

Let Xn bar = (1/n)(n sum i=1) Xi be the sample mean A. Show that the expected value of X bar sub n is equal to theta, that is E( Xn bar) = theta B. Show that the variance of X bar sub n is equal to 1/12n, that is Var (Xn bar) = 1/12n

Solution

a)where E[X_n]=teta

E[x-bar] = E[ (x1 + x2 + ... + xn) / n ]
= (1/n) (E[x1] + E[x2] + ... + E[xn])
= (1/n) (teta + teta + ... + teta)
= (1/n) (n teta)
= teta

b)
Var(Sum(Xi)) = Sum(Var(Xi)) = Var(X1) + ... + Var(Xn)

Variance[x-bar] = E[ (x-bar)^2 ] - (E[ x-bar ])^2

E[ (x-bar)^2 ] = (1/n) Variance[X]