ECNoise uses noisy numerical stochastic processes
$$
\set{\Delta_h^k f_t(\omega) = \Delta_h^k f(t) + \Delta_h^k \epsilon_t(\omega)}_{t\in I}
$$
built on a continuous, analytic function $f$. Using results from the ECNoise article, we see that
$$
\begin{align*}
\mathbb{E}[\Delta_h^k f_t(\omega)] & = \Delta_h^k f(t)\text{ and}\\
\mathbb{V}[\Delta_h^k f_t(\omega)] &= \frac{1}{\gamma_k} \epsilon_f^2.
\end{align*}
$$
According to the text, a $\sigma_k$ value is a good candidate for the noise level estimate of $f_t$ if the $\Delta_h^k f_t$ samples used to compute it have different signs and if the value of $\sigma_k$ is sufficiently similar to the values of $\sigma_{k+1}, \sigma_{k+2}$. The motivation for this criteria is that the uniform step size $h$ be small enough that the $\Delta_h^k f_t$ are sampling noise (i.e., $\Delta_h^k f_t \approx \Delta_h^k \epsilon_t$).
I understand the check on different signs to mean that if the mean value of the distribution of $\Delta_h^k f_t$ is small enough compared to the width of the distribution, then it is probable that the different $\Delta_h^k f_t$ samples will have different signs. In such a case, we might consider the sampling of different signs as something like a necessary condition for $\Delta_h^k f_t$ sampling noise. This condition is a better check as $k$ decreases since I suspect that the probability of sampling different signs should increase with the number of samples. For example, for $k$ large, we might compute $\sigma_k$ from two samples and should not be surprised at both samples having the same sign even though $\Delta_h^k f_t$ is indeed sampling noise.
The text appears to use the sampling of different signs as a sufficient condition for sampling noise, which might be reasonable if we are sampling $\Delta_h^k f_t$ from the same location $t$ and therefore the same distribution. However, we are averaging many shifted $\Delta_h^k f_t$ samples, which presents an additional difficulty that prohibits this check from being sufficient.
As an example, consider measuring a quadratic function at or near to its minimum. If $h$ is large enough that $f_t$ is nonlinear on our region of sampling, then the shifted $\Delta_h^1 f_t$ samples would have different signs due to the fact that we are sampling in both regions of positive and negative slope. The $\Delta_h^2 f_t$ samples would likely all have the same sign since they are effectively sampling the same positive curvature. However, $h$ is in fact large enough that $\Delta_h^1, \Delta_h^2$ are both measuring signal rather than noise.
In the article and the present implementation of ECNoise, we only check the sign difference for the noise level estimate $\sigma_k$ but not for $\sigma_{k+1}, \sigma_{k+2}$. In order to accept $\sigma_k$ as the estimate, should we also insist that both the $\Delta_h^{k+1} f_t$ samples and $\Delta_h^{k+2} f_t$ samples have different signs? This would be in the spirit of using the sign check as a "necessary heuristic" for all three sets of difference operators detecting noise and to potentially catch more failures where $\sigma_k \approx \sigma_{k+1} \approx \sigma_{k+2}$ despite none of these sampling noise.
ECNoise uses noisy numerical stochastic processes
built on a continuous, analytic function$f$ . Using results from the ECNoise article, we see that
According to the text, a$\sigma_k$ value is a good candidate for the noise level estimate of $f_t$ if the $\Delta_h^k f_t$ samples used to compute it have different signs and if the value of $\sigma_k$ is sufficiently similar to the values of $\sigma_{k+1}, \sigma_{k+2}$ . The motivation for this criteria is that the uniform step size $h$ be small enough that the $\Delta_h^k f_t$ are sampling noise (i.e., $\Delta_h^k f_t \approx \Delta_h^k \epsilon_t$ ).
I understand the check on different signs to mean that if the mean value of the distribution of$\Delta_h^k f_t$ is small enough compared to the width of the distribution, then it is probable that the different $\Delta_h^k f_t$ samples will have different signs. In such a case, we might consider the sampling of different signs as something like a necessary condition for $\Delta_h^k f_t$ sampling noise. This condition is a better check as $k$ decreases since I suspect that the probability of sampling different signs should increase with the number of samples. For example, for $k$ large, we might compute $\sigma_k$ from two samples and should not be surprised at both samples having the same sign even though $\Delta_h^k f_t$ is indeed sampling noise.
The text appears to use the sampling of different signs as a sufficient condition for sampling noise, which might be reasonable if we are sampling$\Delta_h^k f_t$ from the same location $t$ and therefore the same distribution. However, we are averaging many shifted $\Delta_h^k f_t$ samples, which presents an additional difficulty that prohibits this check from being sufficient.
As an example, consider measuring a quadratic function at or near to its minimum. If$h$ is large enough that $f_t$ is nonlinear on our region of sampling, then the shifted $\Delta_h^1 f_t$ samples would have different signs due to the fact that we are sampling in both regions of positive and negative slope. The $\Delta_h^2 f_t$ samples would likely all have the same sign since they are effectively sampling the same positive curvature. However, $h$ is in fact large enough that $\Delta_h^1, \Delta_h^2$ are both measuring signal rather than noise.
In the article and the present implementation of ECNoise, we only check the sign difference for the noise level estimate$\sigma_k$ but not for $\sigma_{k+1}, \sigma_{k+2}$ . In order to accept $\sigma_k$ as the estimate, should we also insist that both the $\Delta_h^{k+1} f_t$ samples and $\Delta_h^{k+2} f_t$ samples have different signs? This would be in the spirit of using the sign check as a "necessary heuristic" for all three sets of difference operators detecting noise and to potentially catch more failures where $\sigma_k \approx \sigma_{k+1} \approx \sigma_{k+2}$ despite none of these sampling noise.