In: Statistics and Probability
a.
By Central limit theorem, the distribution of X bar is Normal distribution with mean as population mean and variance = 1/n.
Thus, mean = 0
variance = 1/n
Thus, X bar ~ N(0, 1/n)
b.
We know that the linear combination of normal random variable is a normal random variable. Thus, Xj-Xbar follows Normal distribution.
E[Xj-Xbar] = E[Xj] -E[Xbar] = 0 - 0 = 0
Var[Xj-Xbar] = Var[Xj] + Var[Xbar] - 2 Cov(Xj , Xbar) = 1 + 1/n - 2 Cov(Xj , Xbar)
Thus,
mean of Xj-Xbar = 0
Variance of Xj-Xbar = 1 + 1/n - 2 Cov(Xj , Xbar) = 1 + 1/n - 2/n (From part c, Cov(Xj , Xbar) = 1/n)
= 1 - 1/n
c.
Cov(Xj , Xbar) = Cov(Xj , (X1 + X2 + ... + Xj + ... + Xn )/n )
= (1/n) Cov(Xj , X1 + X2 + ... + Xj + ... + Xn )
= (1/n) [ Cov(Xj , X1) + Cov(Xj , X2) + .... + Cov(Xj , Xj) + .... Cov(Xj , Xn) ]
= (1/n) [ 0 + 0 + .... + Var(Xj) + .... 0 ] X1, X2, ..., Xnare mutually independent
= (1/n) * 1
= (1/n)