Proof of strictly stationary process

  • Context: Graduate 
  • Thread starter Thread starter trenekas
  • Start date Start date
  • Tags Tags
    Process Proof
Click For Summary
SUMMARY

The process defined as Y_n = 1/2*X_n + 1/4*X_{n-1} + 1/8*X_{n-2} is proven to be strictly stationary under the condition that X_n are independent and identically distributed (i.i.d.). The correlation between Y_n and Y_m depends solely on the parameter difference |n-m|, while the distributions of Y_n remain identical due to the identical distributions of X_n. Therefore, the process Y_n satisfies the criteria for strict stationarity.

PREREQUISITES
  • Understanding of strictly stationary processes in probability theory
  • Knowledge of independent and identically distributed (i.i.d.) random variables
  • Familiarity with correlation functions and their properties
  • Basic concepts of random processes and their characteristics
NEXT STEPS
  • Study the properties of stationary processes in depth
  • Learn about correlation functions and their implications in random processes
  • Explore the concept of independence in random variables
  • Investigate examples of strictly stationary processes in applied statistics
USEFUL FOR

Students and researchers in statistics, mathematicians focusing on probability theory, and anyone interested in the analysis of random processes and their properties.

trenekas
Messages
61
Reaction score
0
Hi all. I need to prove or disprove if process Y_n=1/2*X_n+1/4*X_{n-1}+1/8*X_{n-2} are stricly stationary. X_n,n\in R i.i.d.
So almost i have the answer. But don't know if it is correct or not. I have a question of situation when \Gamma_Y(t,s) and |t-s|≤2 for example:
Y_{10}=1/2*X_{10}+1/4*X_{9}+1/8*X_{8}
Y_8=1/2*X_8+1/4*X_{7}+1/8*X_{6}
There are x_8 in both but situations and not sure if Y_{10} and Y_{8} are i.d.
Hope you understand my problem.
Maybe its a little stupid question but I'm just started to learn random processes. :)
Thanks for help.
 
Physics news on Phys.org
The Y's will be correlated as you noticed. The correlation depends only on the parameter difference.
However, the distributions will be identical, since the X's have identical distributions.

Put these together and you get strictly stationary.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K