Proving the Property of Covariance Function: (r(n)-r(m))^2≤2r(0)(r(0-r(n-m)))

Click For Summary

Homework Help Overview

The discussion revolves around proving a property of the covariance function, specifically the inequality \((r(n)-r(m))^2≤2r(0)(r(0)-r(n-m))\). Participants are exploring the definitions and properties of covariance functions in the context of stationary processes.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the expansion of the inequality and the properties of the covariance function. There is an attempt to relate the inequality to known properties, such as \(2r(0)^2≥r(n)^2+r(m)^2\). Questions arise regarding the definitions of \(r(n)\) and its implications in the context of the problem.

Discussion Status

Some participants have provided definitions and properties of the covariance function, while others have pointed out potential issues with the problem statement, suggesting that there may have been a typographical error. The discussion is ongoing, with various interpretations being explored.

Contextual Notes

There is a mention of the covariance function being defined for a stationary process, and properties such as \(r(0)≥0\), \(r(h)=r(-h)\), and \(|r(h)|≤r(0)\) are noted. The original poster expresses uncertainty about their approach and seeks further clarification.

trenekas
Messages
61
Reaction score
0
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 don't 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.
 
Physics news on Phys.org
trenekas said:
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 don't 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.

What is the function ##r(n)##; that is, how is it defined, and what is its relation to anything?
 
Ray Vickson said:
What is the function ##r(n)##; that is, how is it defined, and what is its relation to anything?
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 :(
 
trenekas said:
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 :(

OK, so you have a stationary process in which
[tex]\text{Cov}(X_{n+k},X_n) = r(k), k=0,1,2, \ldots .[/tex]
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
[tex]\text{Cov}(X_k,X_{k-(-r(n-m))}),[/tex]
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:
Oh my God. I made a mistake :) In the first line it should be ##r(0)*(r(0)−r(n−m))##
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
20
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
20
Views
4K