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When studying the "seasonal components" part of time series, I once read the following statement. I do not understand what role does the ergodic theorem play here?

The decomposition of the process is thus $$Y_t = X_t + Z_t + S_t \tag{25.37}$$ where $X_t$ handles the stationary fluctuations, $Z_t$ the long-term trends, and $S_t$ the repeating seasonal component.

If $Z_t = 0$, or equivalently if we have a good estimate of it and can subtract it out, we can find $S_t$ by averaging over multiple cycles of the seasonal trend. Suppose that we know the period of the cycle is $T$, and we can observe $m = n/T$ full cycles. Then $$S_t \approx \frac{1}{m} \sum_{j=0}^{m-1} Y_{t+jT} \tag{25.38}$$ This works because, with $Z_t$ out of the picture, $Y_t = X_t + S_t$, and $S_t$ is periodic, $S_t = S_{t+T}$. Averaging over multiple cycles, the stationary fluctuations tend to cancel out (by the ergodic theorem), but the seasonal component does not.

(Emphasis mine)

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It looks like they're assuming that $E[X_t]=0$. But what that means is that $\int_\Omega X_t(\omega) \mathrm{d}P = 0$. What they would like to be able to say is that $\int_0^\infty X_t \mathrm{d}t = 0$. This is exactly what ergodicity tells you is true, that time averages of a single realization and averages over the space of realizations are equal. By the way a lot of the time in practice this is just assumed to be true for whatever system people are looking at. It was a big deal a few years back when Jonathan Mattingly from Duke was able to actually prove that the stochastically forced 2D Navier Stokes is ergodic for example.

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  • $\begingroup$ ,Thanks. I am not quite familiar with measure theory. In your post, you mentioned sth lik $\int_{\Omega} X_t(\omega)dP$. What does $\Omega$ and $\omega$ mean here? Thanks. $\endgroup$ Commented Feb 12, 2014 at 14:12
  • $\begingroup$ Since $X_t$ is a stochastic process; it needs to be defined on a consisting of $(\Omega,F,P)$ where $\Omega$ is the space or events, a $\sigma$-algebra on that set which essentially means a set of the sets of events you're allowed to ask about, and $P$ is a measure that assigns a probability for each set in $F$. Sometimes it helps me to think about how things would be simulated on a computer to make things firmer in my head when thinking about stochastic processes. If you're simulating a stochastic process, and say it takes 100 random numbers to simulate a path. $\Omega$ would be all the ... $\endgroup$ Commented Feb 12, 2014 at 19:08
  • $\begingroup$ possible sets or 100 random numbers. One realization of the 100 number long random vector corresponds to a $\omega$. $F$ would contain sets like $A=\{\omega :\; v_{i} < 0 \;\text{for all}\; i\}$ if $v_i$ is the $i$th elemnt of the random vector. So the event $A$ would contain all the realizations where every single element in the random vector is negative. And then you can figure out what the probability of that event is which corresponds to the measure $P(A)$. If all the elements of the random vector are independent, $v_i < 0$ has probability $.5$ for each index $i$, then $P(A) = 2^{-100}$. $\endgroup$ Commented Feb 12, 2014 at 19:13

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