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The spectral exponent \eqn{\theta} is then estimated as the negative slope of the linear regression of the natural log of spectral density as a function of the natural log of frequency.
What reason is there to not just take the slope as it is in the first place? Did you intend for color to have meaning "the higher, the redder"? Or did I get something wrong?
The text was updated successfully, but these errors were encountered:
Hi Ronald, thanks for your comment, and apologies for the late reply.
The model is actually assuming a function of the form $s = 1 / f^\beta$, or $s = f^{-\beta}$ , where $s$ is the spectrum and $f$ the frequency. Empirically we then fit $\textrm{ln}s \sim \textrm{ln}f$ as a linear model, which yields the negative slope. But we're interested in estimating the value of $\beta$ from the first equation above, that's why the absolute value is taken. I hope this makes sense? See this https://esajournals.onlinelibrary.wiley.com/doi/abs/10.1890/02-3122
I might lack some background in spectral analysis but don't we estimate $\beta$ via the slope of the linear regression? Or, well, $-\beta$.
When red noise means more power/energy in lower frequencies (aka "on the left") and white noise an equal amount of variance over all frequencies, shouldn't blue noise then be more power in higher frequencies (aka "on the right")? And if $\beta>=1$ is red(-ish) noise and $\beta=0$ white noise, shouldn't $\beta<0$ be indicate blue noise?
And even though we might not usually see blue noise in nature, couldn't blue noise be mistaken for red noise if we ignore the sign as mentioned in the OP?
Hi Diego,
the doc (
env_stats
) states:And the corresponding code is:
envPred/R/helpers.R
Line 67 in cec9644
Shouldn't it be
instead?
What reason is there to not just take the slope as it is in the first place? Did you intend for color to have meaning "the higher, the redder"? Or did I get something wrong?
The text was updated successfully, but these errors were encountered: