Statistics proof, identically distributed RVs and variance

[vj3336]

Homework Statement

Show that for identically distributed, but not necessarily independent random variables with positive pairwise correlation ρ, the variance of their average is ρσ^2 + (1-ρ)σ^2/B.

ρ - pairwise corellation
σ^2 - variance of each variable
B - number of samples

Homework Equations

?

The Attempt at a Solution

I'm totally stuck on this problem, I don't know where to start.
Can you give me some starting hint ?
I know that variance of the average for independent identically distributed random variables is σ^2/B

 

[micromass]

Let's do this for two samples to ease the notation. Do you know how to calculate:

Var(X+Y)

?

 

[vj3336]

I think I could, I would start with this :
Var(X+Y) = Var(X) + Var(Y) = E[(X-μx)^2] - E[(Y-μy)^2]

where μx is the mean of RV X, and μy the mean of RV Y

 

[micromass]

Var(X+Y) = Var(X) + Var(Y)

This is not true! You should have a formula in your course that calculates the variance of a sum...

 

[vj3336]

right, so when X and Y are correlated
Var(X+Y) = Var(X) + Var(Y) + 2ρ*Sqrt(Var(X))*Sqrt(Var(Y))

where ρ is corelattion coefficient between X and Y

 

[micromass]

OK, so what is

Var\left(\frac{X+Y}{2}\right)

Somehow, you need to show that it equals ρσ^2 + (1-ρ)σ^2/2.

 

[vj3336]

The best thing I could do is :
Var(1/2(X+Y))= Var(1/2*X + 1/2*Y)= 1/4σx^2 + 1/4σy^2 + 2*1/4*ρ Sqrt(σx^2)*Sqrt(σy^2)
=1/4σ^2 + 1/4σ^2 +1/2*ρ*σ^2

 

[vj3336]

but maybe I can rearrange this to your form ...I'll try it

 

[micromass]

Well, can you not rewrite that a bit to form ρσ^2 + (1-ρ)σ^2/2? Or you could work out what ρσ^2 + (1-ρ)σ^2/2 is...

 

[vj3336]

yes, I can do it , so:

1/2σ^2 + 1/2*ρ*σ^2 = 1/2*(2ρσ^2-ρσ^2) + 1/2*σ^2 = ρσ^2 - 1/2*ρσ^2 + 1/2*σ^2
= ρσ^2 + 1/2*(σ^2-ρσ^2) = ρσ^2 + 1/2*(1-ρ)σ ^2

So the next step is to try it with n instead of 2 ?, I'll try that now

 

[micromass]

Very good!

The general step should be quite analogous...

 

[vj3336]

The general step is then :
Var(1/n(X1+...+X2)) = 1/n*σ^2 + 2/n^2 *(1/2 * (n-1)n ) ρσ^2
= 1/n*σ^2 + (n-1)/n *ρσ^2
= ρσ^2 + 1/n*σ^2 + -1/n*ρσ^2
= ρσ^2 + 1/n*(1-ρ)σ^2

 

[micromass]

Seems fine! Well done!

 

[vj3336]

Thanks !
Your hints where very helpful pointing me in the right direction.