Covariance/Independence Proof Help

[jimbodonut]

Homework Statement

Suppose {X1,X2, ...,Xn} is an iid (independent, identically distributed) sample from N(μ, σ^2). Show that Cov(X_bar, Xi − X_bar ) = 0 for all i, and hence conclude that X_bar is independent of Xi − X_bar for every i.

Homework Equations

The Attempt at a Solution

cov(X_bar ̅,Xi-X_bar)
=cov(X_bar ̅,X_i ) - cov(X_bar ̅,X _bar )
=cov((1/n)*∑Xj ,Xi ) - cov((1/n) ∑Xi, (1/n) ∑Xj)
=∑(1/n) cov(Xj,Xi ) - ∑∑(1/n)^2 cov(Xi,Xj )
=(n/n)*σ^2 - (n/n)^2*σ^2
=σ^2 - σ^2=0


not sure if what I am doing is right, and i don't see how showing Cov( X_bar,Xi − X_bar ) = 0 for all i can help to arrive at the suggested conclusion.


Thanks for ur help guys... :)

 

[mighty2000]

Your approach is correct and you have arrived at the correct conclusion. Let me explain why showing Cov(X_bar, Xi-X_bar) = 0 for all i leads to the conclusion that X_bar is independent of Xi-X_bar for every i.

First, let's recall the definition of covariance: Cov(X,Y) = E[(X-E[X])(Y-E[Y])]. In this case, we have X = X_bar and Y = Xi-X_bar. So, Cov(X_bar,Xi-X_bar) = E[(X_bar-E[X_bar])(Xi-X_bar-E[Xi-X_bar])].

Now, we know that X_bar and Xi-X_bar are both random variables, but they are also functions of the same sample {X1,X2,...,Xn}. This means that they are dependent on each other, and their values will change together as the sample changes. However, the formula for covariance shows that if X and Y are independent, then Cov(X,Y) = 0. So, if we can show that X_bar and Xi-X_bar are independent, then we can conclude that Cov(X_bar,Xi-X_bar) = 0 for all i.

To show that X_bar and Xi-X_bar are independent, we need to show that their joint probability distribution can be written as the product of their individual probability distributions. In other words, P(X_bar = x, Xi-X_bar = y) = P(X_bar = x)P(Xi-X_bar = y) for all x and y. This is equivalent to showing that the joint probability density function (PDF) can be written as the product of the individual PDFs.

Using the definition of X_bar and the fact that the sample is iid, we can write the joint PDF of X_bar and Xi-X_bar as:

f(x,y) = fbar(x)fXi-X_bar(y) = (1/σ√(2π))^(n+1) * e^((-1/2σ^2)(∑(xj-μ)^2 + (yi-x_bar)^2))

where fbar(x) and fXi-X_bar(y) are the individual PDFs of X_bar and Xi-X_bar, respectively. Now, if we try to factor this joint PDF, we get:

f(x,y) = (1/σ√(2π))^(n+1) * e^((-1/2σ^2)((x_bar