Determining stationary and mean-ergodicity

In summary, the conversation discusses the difficulties in setting up and characterizing stationary and ergodicity for certain random processes. It is determined that the process in part (a) is not wide-sense stationary (WSS) or strict-sense stationary (SSS) because it is dependent on time. The process in part (b) is also not WSS, but it is mean-ergodic. Finally, the process in part (c) is both WSS and SSS, and it is suggested that the ensemble and time averages may also be equal, indicating ergodicity. Assistance is requested for calculating the time average in this case.
  • #1
ashah99
60
2
Homework Statement
Please see problem below
Relevant Equations
Autocorrelation function R(m) = E(X_k*X_(k+m))
I am having difficulties setting up and characterizing stationary and ergodicity for a few random processes below. I need to determine whether the random process below is strict-sense stationary (SSS), whether it is wide-sense stationary (WSS), and whether it is ergodic in the mean. All help is much appreciated.

1666884811615.png


1666884839272.png
Approach to (a) :
Since the Poisson RV includes a time duration in its PDF, then X_k would depend on time, so it is not WSS.
To test for SSS, I simply replace k with k - d , so X_(k-d) = e^-|k-d - D|, but since it is not WSS, then it would still dependent on time, so it is not SSS.
I'm not sure how to approach proving if the ensemble and time averages are the same or not. A time average is generally a RV and an ensemble average is generally a function of time. Any help appreciated.

As for (b)
The mean would be constant E(X_k) = (1/3)(1/3) = 1/9, but I'm not sure how to write out the autocorrelation function, but my hutch is that it's dependent of time, since it's equally likely to be 1, 2, or 3 -> random process is not WSS.
To test for SSS, a similar approach with a time delay as with part (a). I suspect the process is not SSS and depends on time k since {U_k} ~ i.i.d Unif(1,2,3)
I think that given the U_k, the ensemble and time average would both come out as 1/9, meaning that the process is mean-ergodic.
 
Physics news on Phys.org
  • #2
Approach to (c):The mean is constant E(X_k) = 0, and the autocorrelation function would be r_x(k) = E[X_k X_(k-d)] = 0, since the variance is 0. Therefore, the process is WSS and SSS.To prove ergodicity, the ensemble and time average would need to be equal. The ensemble average would be 0, but I'm not sure how to calculate the time average. Any help appreciated.
 

FAQ: Determining stationary and mean-ergodicity

What is the difference between stationary and mean-ergodicity?

Stationary refers to a process in which the statistical properties (such as mean and variance) remain constant over time, while mean-ergodicity refers to a process in which the long-term average of a sample is equal to the population mean.

How do you determine if a process is stationary?

There are several methods for determining stationarity, including visual inspection of a time series plot, statistical tests such as the Augmented Dickey-Fuller test, and analyzing autocorrelation and partial autocorrelation functions.

What is the significance of mean-ergodicity in data analysis?

Mean-ergodicity is important because it allows us to make inferences about a population based on a sample. If a process is mean-ergodic, the sample mean can be used as an unbiased estimate of the population mean.

Can a process be stationary but not mean-ergodic?

Yes, it is possible for a process to be stationary but not mean-ergodic. This can occur when the process has a time-varying mean, making the long-term average different from the population mean.

How can we ensure that our data is suitable for determining stationary and mean-ergodicity?

To ensure that our data is suitable for analysis, we should first visually inspect the time series plot and check for any trends or patterns. We should also perform statistical tests and analyze autocorrelation and partial autocorrelation functions to confirm stationarity and mean-ergodicity.

Similar threads

Replies
1
Views
951
Replies
1
Views
839
Replies
10
Views
1K
Replies
8
Views
1K
Replies
5
Views
1K
Replies
2
Views
2K
Replies
2
Views
1K
Replies
5
Views
2K
Back
Top