# Covariance function property

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1. Oct 25, 2014

### trenekas

Hi all. My task is to prove the property of covariance function:

$(r(n)-r(m))^2≤2r(0)(r(0-r(n-m)))$

My solution:

$1) (r(n)-r(m))^2=r(n)^2-2r(n)r(m)+r(m)^2$
$2) 2r(0)(r(0)-r(n-m)))=2r(0)^2-2r(0)r(n-m)$

From covariance function properties I know that $2r(0)^2≥r(n)^2+r(m)^2$
So now I just need to prove that $r(0)r(n-m)≤r(n)r(m)$
But dont know how to do that. Any thoughts? I'm not sure, maybe my way of solution is bad and I need to find other one. Any help would be appreciate.

2. Oct 25, 2014

### Ray Vickson

What is the function $r(n)$; that is, how is it defined, and what is its relation to anything?

3. Oct 26, 2014

### trenekas

If I understand you I'll try to answer.
$r(n)$ is covariance function of stationary process. $r(n)$ show the covariance between two random variables and $n$ is operator of distance.
$r(n)=EX_nX_0-EX_nEX_0$
Because process X is stationary, then $EX_nX_0-EX_nEX_0=EX_{n+h}X_h-EX_{n+h}EX_h$
And some properties:
$r(0)≥0$
$r(h)=r(-h)$
$|r(h)|≤r(0)$
But it does not help me to solve the problem :(

4. Oct 26, 2014

### Ray Vickson

OK, so you have a stationary process in which
$$\text{Cov}(X_{n+k},X_n) = r(k), k=0,1,2, \ldots .$$
In that case, the notation in the question still does not make sense: the RHS involves $r(0-r(n-m))= -r(n-m)$, and I cannot fathom this. If we stick to your definition, this means
$$\text{Cov}(X_k,X_{k-(-r(n-m))}),$$
and there is no reason for a time like $t = k -(-r(n-m)) = k + r(n-m)$ to be integer-valued.

Did your problem statement have a typo in it? Did you really mean $r(0) - r(n-m)?$

Last edited: Oct 26, 2014
5. Oct 26, 2014

### trenekas

Oh my God. I made a mistake :) In the first line it should be $r(0)*(r(0)−r(n−m))$