Question about ergodic theorems

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In summary: I'm not sure what to call it. Explanation?In summary, the author is saying that if a stochastic process is not ergodic, then the time average of that process will not be equal to the mean of that process. However, if the stochastic process is ergodic, then the time average of that process will be equal to the mean of that process.
  • #1
Frank Einstein
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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.
 
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  • #2
Frank Einstein said:
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 don't 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
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
I can't seem to understand why ##\int_{-T}^T A dt = A##. Shouldn't this be equal to ##2TA##?
 
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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
Frank Einstein said:
I don't 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?
 
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  • #9
Math_QED said:
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
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:

1. What are ergodic theorems?

Ergodic theorems are mathematical theorems that describe the behavior of a system over time. They are used to analyze the long-term averages of a system and how they relate to its initial conditions.

2. What is the significance of ergodic theorems?

Ergodic theorems are important because they allow us to make predictions about the behavior of complex systems without having to observe them over a long period of time. This is particularly useful in fields such as economics, physics, and biology.

3. How are ergodic theorems used in science?

Ergodic theorems are used in science to study the behavior of systems that are too complex to be understood by traditional methods. They allow scientists to make predictions and draw conclusions about the long-term behavior of these systems.

4. Can ergodic theorems be applied to all systems?

No, ergodic theorems are not applicable to all systems. They are most commonly used for systems that exhibit a certain level of randomness or unpredictability, such as chaotic systems or systems with many interacting variables.

5. What are some real-world examples of ergodic theorems in action?

One example of ergodic theorems in action is in weather forecasting. By studying the long-term behavior of weather patterns, meteorologists can make predictions about future weather conditions. Another example is in the stock market, where ergodic theorems are used to analyze market trends and make predictions about future stock prices.

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