t ii.d. random samples Xi(- 1,2, , m) have the population mean E (X)- and the po

Question

t ii.d. random samples Xi(- 1,2, , m) have the population mean E (X)- and the population variance Var (X) 2. (a) For the sample mean X, show that E (X)- and Var (X)-2/n (b) Suppose we know that the random samples Xi (i 1,2, ,n) are drawn from normal distribution. Then what is the distribution of the sample mean X? [Note: When finding a distribution, you also need to determine the parameter values. (c) Now suppose we do not know the distribution of the random samples Xi (i 1,2,... ,n). For sufficiently large n, what distribution does X approximately have? 2. Keller Exercise 9.7 and 9.9. 3. Keller Exercise 9.19. 4. Keller Exercise 9.21. V 5. Keller Exercise 9.24. and 9.25

Explanation / Answer

b)

1st method:

Let X=(X1,...,Xn)' be the random vector.

Since X1,..,Xn are i.i.d. N(mu,sigma^2) , hence

X will have a multivariate normal distribution with mean m=(mu,mu,...,mu)' and covariance matrix sigma^2 I

Since Xbar= 1/n(1'X)

where 1'= (1,1,..,1)' (vector of 1's with size n)

1/n(1'X) will have a normal distribution ( Since any linear function of a multinormal distribution is normal).

We now need to evaluate it's mean and variance only.

As you have noted in part a) that E(Xbar)=mu

and Var(Xbar)=sigma^2/n

hence Xbar will follow N(mu,sigma^2/n) distribution.  

2nd method:

You may also use the characteristic function to prove this.

Note that,

Characteristic function of N(mu,sigma^2) distribution is f(t)=exp(itmu - 1/2 t^2 sigma^2)

Hence characteristic function of Xbar is

g(t)=E[exp(it/n(X1+..+Xn)]=(E[exp(itX1/n)])^n

(E[exp(itX1/n)]) = Characteristic function of N(mu,sigma^2) evaluated at t/n = f(t/n)=exp(i(t/n)mu - 1/2 (t^2 /n^2)sigma^2)

Hence

g(t)=(f(t/n))^n= exp(itmu-t^2sigma^2/2n)

Which is the characteristic function of N(mu,sigma^2/n) distribution.

Since Characteristic fucntion characterizes a distribution uniquely hence

Xbar will have this distribution.

3rd method:

If you don't want to prove it from the multinormal point of view or you don't know about characteristic function then you may consider the transformation

(X1,...,Xn) ----> (X1,X2,..Xn-1,X1+.....+Xn)

which is a one-one,continuously differentiable function. and the inverse of this transformation is given by

(y1,..,yn) ......>(y1,y2,...,yn-1,yn-y1-y2-..-yn-1)

which has determinant of the Jacobian matrix = 1.

Hence the joint distribution of (X1,..,Xn-1,X1+...,Xn) would be

g(y1,...,yn)=f(y1,....,yn-1,yn-y1-y2-..-yn-1)

Where f is the joint distribution of

(X1,..,Xn) which is the product of individual N(mu,sigma^2) distribution.

Now all you need to do is integrate out the first (n-1) co-ordinates and then you'll get the desired result.

c)

Now we don't know from which distribution the observations are coming but all we know that

X1,..Xn are i.i.d. observations from a distribution with mean mu , and variance sigma^2.

Then by Lindeberg-Levy Central limit theorem you can say that

sqrt(n)(Xbar-mu) converges to N(0,sigma^2) distribution as n goes to infinity.

So for large n the distribution of Xbar is approximately N(mu,sigma^2/n) distribution.

If you don't know the proof of this theorem I can give you an idea of how to prove it.

Again we'll use characteristic function

Characteristic function of sqrt(n)((Xbar-mu)/sigma) is g(t)

which is same as (f(t/sqrt(n)))^n

where f is the characteristic function of the same family of distribution only with mean 0 instead of mu and variance 1 instead of sigma^2. This is because when we're using ((Xbar-mu)/sigma) we've scaled the variables so that their variance is 1 and we have shifted the origin so that the mean is 1 now.

Recall that if a distribution has finite second moment then the characteristic function is twice continuously differnetiable (with uniformly continuous derivative) with

f'(t)=-iE(X) and f''(t)=-E(X^2)

Now

f(t/sqrt(n))= f(0)+ t/sqrt(n) f'(0) + t^2/2n f''(s) , where s lies in 0 and t/sqrt(n)

f(0)=1 (Characteristic function value is 1 at t=0)

f'(0)=0 since the mean is zero now.

f''(s)=f''(0) + (f''(s)-f''(0))

f''(0)=-1 as variance is 1 and mean is 0 now.

Hence

f(t/sqrt(n))= 1- t^2/2n +t^/2n(f''(s)-f''(0))

f''(s)= - int x^2 exp(isx) d(F(x)) (Recall the formula for derivatives of characteristic function)

(f''(s)-f''(0)) = - int x^2 (exp(isx)-1) d(F(x))  

|(f''(s)-f''(0)) | <= c t/sqrt(n)

for some constant c.

This is because |exp(isx)-1| <=sx

Now the integral is over whole real line and if you simply use this inequlity then it is of no use because we havenot assumed finiteness of any moment greater than 2.

So break the integral into two parts

First is x< n^(1/3) and the next part is x >= n^(1/3)

As n goes to infinity the integral over the 2nd part( i.e. over x> n^(1/3) ) goes to zero because the measure of that set goes to zero.

hence most important is to evaluate the integral over that first part (i.e. if x<n^(1/3) )

|exp(isx)-1| <= |sx|

s lies in 0 and t/sqrt(n) and x< n^(1/3)  

implies |sx| <= t/n^(1/6)

Hence

f(t/sqrt(n))=1-t^2/2n + o(t^2/2n)

Note that this small o notation means   o(t^2/2n) /t^2/2n converges to 0 as n goes to infinity.Simply the last term converges to 0 with rate smaller than 1/n

Now g(t)=(f(t/sqrt(n))^n

=(1-t^2/2n + o(t^2/2n) )^n

converges to e^(-t^2/2) which is the characteristic function of standard normal distribution.

Recall that (1-x/n)^n converges to e^(-x) as n tends to infinity.

It doesnot matter if a sequence of order smaller than 1/n is added to it.

Now convergence of characteristic function implies converges in distribution.

Hence sqrt(n) (Xbar-mu)/sigma converges in distribution to N(0,1)  

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