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Can someone verify my understanding of what the continuous periodogram is/means, and please tell me if I say something wrong:

As I've learned so far, the power spectral density of a wide-sense stationary stochastic process $x(t)$ is given by

$$S_{xx}(f) = \lim_{T\rightarrow\infty}\frac{1}{T}\mathbf{E}\Big\{|X_T(f)|^2\Big\} \tag{1}$$

where $X_T(f)$ is the Fourier transform of a single truncated sample path. The sample path itself is over $\{{-\infty<t<\infty}\}$, but the truncation is over time $T$. Finally, the expectation is over all sample paths in the ensemble.

1) Is it correct to write then that: if in practice we only have access to a single member of the ensemble which is over a finite observation time $T$, then the simplest possible estimator, $\hat{S}_{xx}(f)$, of the true power spectral density, $S_{xx}(f)$, is found by dropping both the $T\rightarrow\infty$ limit and the expectation (because we don't have any other members of the ensemble), and so

$$\hat{S}_{xx}(f) = \frac{1}{T}|X_T(f)|^2 \tag{2}$$

and that this is known as the continuous periodogram?

2) I have seen some resources which say that the periodogram assumes ergodicity - is this true? If so, where and how exactly does this assumption become necessary? As far as I can see, if we simply drop the expectation in (1) (because we only have access to a single realisation), then there should be no reason to need to invoke ergodicity (which would somehow involve replacing the ensemble expectation with a temporal average).

3) Finally, I would like to ask about the variance. I see that the main problem with the discrete periodogram is that it's variance does not decrease as the number of samples is increased. Is there an equivalent problem with the continuous periodogram, and what is the analogy in that case to be able to calculate the asymptotic variance (since we don't have any discrete points to take $N\rightarrow\infty$). Should it be the time window $T$, and one should be able to show that the variance becomes constant as $T\rightarrow\infty$?

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  • $\begingroup$ Surely ergodicity is assumed when you only use one sample path? You're assuming that the single sample path is representative of all sample paths. See section 22.3 of this. $\endgroup$ – Peter K. Apr 9 at 12:04
  • $\begingroup$ But ergodicity hasn't actually been necessary to invoke in order to obtain this definition of periodogram. It is still a valid estimator, but would be a worse estimator if the process was not ergodic - but the equation is still valid? $\endgroup$ – teeeeee Apr 9 at 12:07
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Eq. $(2)$ is indeed the periodogram of the truncated signal where said signal is the sample path $x(t)$ of the random process. There is nothing random about this sample path, which is why dropping the Expectation operator makes sense when one goes from $(1)$ to $(2)$. Dropping the limit is asking everyone to take your word for it that the $T$-second observation interval $\left[-\frac T2, +\frac T2\right]$ is adequate to capture enough of the general behavior of the process. Would increasing $T$ to $T^\prime > T$ make a significant difference in $\hat{S}_{xx}(f)$, at least as far as the central lobe is concerned? We don't know but we trust you to make an informed judgement.

With regard to your query $2)$ as to whether the periodogram assumes ergodicity, the answer is No and Yes. No in the sense that Eq. $(2)$ by itself says nothing about the power spectral density $S_{xx}(f)$ all by itself: the right side of $(2)$ is the periodogram of the observed finite-length segment of the sample path of the process. Yes in the sense that it is you who can be said to be assuming ergodicity when you assert that $\hat{S}_{xx}(f)$ is a good estimate of $S_{xx}(f)$. In general, the averages of the sample paths (e.g. the limit as $T\to \infty$ of the right side of $(2)$) need not have the same properties as the averages over ensembles (e.g. $S_{xx}(f)$), and assuming that they are the same not only means that you are assuming ergodicity (which really deals with limiting or asymptotic values of the averages) but also that $\hat{S}_{xx}(f)$ is pretty close to $\lim_{T\to\infty}\hat{S}_{xx}(f)$, that is, the short observed segment of the sample path is generally representative of the process.

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