Sometimes I see papers that average the log of PSDs (i.e., in dB space), instead of the PSDs themselves. So they are doing a geometric averages. I found a paper that shows that geometric mean has some advantages.

When one does Welch method of PSD estimation, one use arithmetic mean. Would it make any sense to do a geometric mean in that case instead?

When is it (is it ever) appropriate to do a geometric mean between different PSD estimations? What is the use case of one vs the other?

I also understand that it doesn't matter for the median, i.e., the median of the log of PSDs is the same as log of the median of PSDs.


2 Answers 2


This is a hard question to answer, because it really depends on the specific type of data you're analyzing and what you want from it.

I think a good rule of thumb is: if you're analyzing data that you usually represent in the log scale, then you should typically do the averaging on the linear data first.

Consider an example where you have 3 log values: $-36\tt{dB}$, $-24\tt{dB}$ and $0\tt{dB}$. Averaging these values would give you $-20\tt{dB}$. Converting to linear scale first, then taking taking the $\tt{dB}$ value would give you $\approx -4.75 \tt{dB}$. If you're an audio engineer I think you'd agree the second number makes more sense.


I don't really understand what bias is referring to in this paper as it doesn't seem to be what is typically understood to be bias in spectral estimation, and there's no spectral plots to give a visual aid. It is likely referring to what @Jdip is alluding to, which is highly application dependent. Based on some simple experimentation I did in Matlab it primarily seems focused on dynamic range preservation, but that's a very basic analysis and should be taken with a grain of salt.

In estimation theory, there's a well known tradeoff between bias and variance. For spectral estimators, bias is inherently related to resolution. Bias in spectral estimators is typically expressed as

\begin{equation} b(\hat{\phi}(\omega)) = \lim_{N\rightarrow\infty} E[\hat{\phi}_{N}(\omega)-\phi(\omega)] \end{equation}

where $\phi(\omega)$ is the true PSD and $\hat{\phi}_{N}(\omega)$ is the $N$-point estimate of the PSD. The periodogram is an asymptotically unbiased PSD estimate. $N$-point PSD estimators can be viewed as taking $N$ samples of the true PSD spaced at $\frac{f_{s}}{N}$. As you increase $N$, you more precisely sample the true PSD, giving you an unbiased estimate. However, the periodogram is an inconsistent estimator, meaning that the periodogram values will always bounce around the true PSD value with non-zero variance even if $N$ is allowed to increase without bound.

This is the motivation for Periodogram derivative methods, like Welch's method, to reduce the variance. Bartlett's, Welch's, etc, all successfully reduce the variance and can be made consistent estimators, however, they introduce bias. The way I understand it, there are two main forms of bias introduced, averaging bias and resolution bias. Averaging bias results from the time averaging employed to reduce the variance by these methods. Resolution bias is inherent bias in all practical applications due to having a finite $N$. In filterbank methods, the value at a sample $\omega$ is essentially the average value of a bandpass filtered output of the true PSD with approximate bandwidth $[\omega-\frac{f_{s}}{2N},\omega+\frac{f_{s}}{2N})$. This will result in a bias at the point $\omega$. As $N$ increases, the integration window of the filterbank approaches $\delta_{\omega}$, and therefore becomes unbiased. With Periodogram derivative methods like Welch's, they have resolution $M = \frac{N}{k} < N$ where $k$ is some positive integer. Therefore, these methods always have resolution bias.

The paper you linked seems like it attempts to deal with time-averaging biases introduced by Periodogram derivative methods, but there will always be a strict bias in these methods due to resolution bias that cannot be solved by simply increasing $N$. I'm not fully understanding how the paper deals with the averaging bias, but I don't see how it can deal with the resolution bias.

  • $\begingroup$ I didn't read the paper but avg logs or avg magnitudes results in an offset error for AWGN vs true-rms (square of the average is not the same as the average of the squares). Perhaps it has to do with this? If we avg magnitudes, the result is off by 1.05 dB, if we avg dB quantities, it is of by 2.51 dB! (for white noise). $\endgroup$ Commented Feb 28 at 1:29
  • 1
    $\begingroup$ @DanBoschen It probably has something to do with that. I'm not super familiar with this subject, but this is an interesting wiki to explore en.wikipedia.org/wiki/Pythagorean_means . As I understand it, the arithmetic mean, which I believe is the ML estimate, is better suited for non-skewed data with iid draws, so is perfect for white noise. The geometric mean is potentially better in other cases where these assumptions are violated. Either way, it seems a bit unnecessary to go through this process, as accepting bias is unfortunately necessary for Blackman-Tukey type procedures. $\endgroup$
    – Baddioes
    Commented Feb 28 at 3:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.