Question about ergodic theorems

  • #1

Summary:

I am trying to understand an example which doesn't seem to make sense

Main Question or Discussion Point

Hello everyone. I am currently reading the book Probability statistics and random processes for electrical engineering by Alberto Leon Garcia. In page 540, one can find example 9.47, in which is shown how a stationary random process doesn't have to be ergodic by defining a random variable A of zero mean and variance one. Then, a stochastic process is defined as X(t)=A, therefore, mX(t)= E[X(t)]=E[A]=0. However, the integration over a time interval returns

<X(t)>T=∫-TT A dt=A

Which shows that a stationary process isn't necessary ergodic. However, in the very next page the book states that <X(t)>T is an unbiased estimator of m. Am I understanding everything wrong or these two statements contradict each other?

Any answer is appreciated.
Regards.
 

Answers and Replies

  • #2
Math_QED
Science Advisor
Homework Helper
2019 Award
1,349
484
Summary:: I am trying to understand an example which doesn't seem to make sense

Hello everyone. I am currently reading the book Probability statistics and random processes for electrical engineering by Alberto Leon Garcia. In page 540, one can find example 9.47, in which is shown how a stationary random process doesn't have to be ergodic by defining a random variable A of zero mean and variance one. Then, a stochastic process is defined as X(t)=A, therefore, mX(t)= E[X(t)]=E[A]=0. However, the integration over a time interval returns

<X(t)>T=∫-TT A dt=A

Which shows that a stationary process isn't necessary ergodic. However, in the very next page the book states that <X(t)>T is an unbiased estimator of m. Am I understanding everything wrong or these two statements contradict each other?

Any answer is appreciated.
Regards.
Can you add the relevant definitions to your question?

In particular, what is this m you are referring to? What is your definition of ergodic stochastic process?
 
  • #3
First of all. Thanks for taking your time to answer me.

In my question, m refers to the mean. That is why in the example it says that mX(t)=0; the mean of the stochastic process is zero for all times since it's CDF is equal to the one of the random variable A for all times.

Also, <X(t)>T is a time average of a single realization of the stochastic process

In the next page, the author does something similar. He writes

E[<X(t)>]=E[(1/2T)∫-TTX(t)dt]=(1/2T)∫-TTE[X(t)]dt=m and calls <X(t)>T an unbiased estimator of m.

I dont understand it, since in the example it was stated that the time average of a random process can't be used to calculate the mean of said stochastic process.
 
  • #4
Math_QED
Science Advisor
Homework Helper
2019 Award
1,349
484
And what definition for ergodic stochastic process were you using? Then I can try to answer your question.
 
  • #5
An ergodic process is a stochastic process in which a time average converges to the actual value (the mean or the autocorrelation) as the observation interval becomes large. An example of an ergodic process is in this case a independent identically distributed process with finite mean.
 
  • #6
Math_QED
Science Advisor
Homework Helper
2019 Award
1,349
484
I can't seem to understand why ##\int_{-T}^T A dt = A##. Shouldn't this be equal to ##2TA##?
 
  • #7
You are right. I am very sorry. The integral should be multiplied by 1/2T. However, I cannot seem to edit the original post
 
  • #8
Math_QED
Science Advisor
Homework Helper
2019 Award
1,349
484
I dont understand it, since in the example it was stated that the time average of a random process can't be used to calculate the mean of said stochastic process.
Yes, this example shows that you can't do that if your stochastic process is not ergodic.

But isn't what the author says true if the stochastic process is ergodic?

(Also please include definition of unbiased estimator, my background is in formal probability theory and not in statistics).

If I understand you correctly, we have for an ergodic process
$$\lim_{T \to \infty }\langle X(t) \rangle_T = m$$
Is that correct?
 
  • #9
Yes, this example shows that you can't do that if your stochastic process is not ergodic.

But isn't what the author says true if the stochastic process is ergodic?

(Also please include definition of unbiased estimator, my background is in formal probability theory and not in statistics).

If I understand you correctly, we have for an ergodic process
$$\lim_{T \to \infty }\langle X(t) \rangle_T = m$$
Is that correct?
Yes. An ergodic process is defined as an stochastic process in which the parameters can be estimated as time averages.

An unbiased estimator is defined as an estimator which mean is exactly the value of the paramter which is being estimated.

However, that would imply that the sample mean of ∞ repetitions of a mesurement of an interval of longitude 2T would end up as the mean.

I think that that is what the author was trying to say. One time measurement cannot estimate the mean, but a high number of them can. I still don't see the point, since the whole purpose of the ergodic process is to avoid doing that. However, I will se if the author does this in an attempt to justify the second ergodic theorem.

Thanks for the help.
 
  • #10
Stephen Tashi
Science Advisor
6,923
1,182
Taking a hint from:
https://dsp.stackexchange.com/questions/1167/what-is-the-distinction-between-ergodic-and-stationary

It's possible that the book intends ##X(t)## to denote generating a path by selecting a single value of the random variable ##X## and using it for all times ##t## rather that using an independent realization of of ##X## at each time ##t##. In such a situation, averaging one sample path of the process over long intervals of time does not produce the mean of ##X##, instead it produces the particular value of ##X## that happened to be selected in making the single realization of ##X## used to generate the path.
 
Last edited:

Related Threads for: Question about ergodic theorems

Replies
0
Views
2K
  • Last Post
Replies
1
Views
497
  • Last Post
Replies
1
Views
6K
  • Last Post
Replies
2
Views
2K
Replies
1
Views
586
Replies
4
Views
1K
  • Last Post
Replies
1
Views
2K
Replies
1
Views
2K
Top