I have a time series which presents two different patterns during time. These patterns are related to two different events that happen during the experiment. I can manually select the temporal interval related to these two events, so we can consider the two portions of signal separately.
I'm looking for a (simple) method to test wether the two portions of signal significantly differ. Given that their pdf don't follow any known distribution, I think that Chebyshev's inequality could help in this task.
I try to explain my idea. Given that no more than 1/k^2 of the distribution's values can be more than k standard deviations away from the mean, I calculate the two-tailed confidence interval for the first distribution (using k=3.2) and then count how many values lie outside this interval in the second distribution. If more than 10% (k=3.2) of the values in the second distribution lie outside the confidence interval calculated for the first distribution, I conclude that the second portion of signal is significantly different from the first one. Here's an image which should help to understand my idea.
There is an example of the signal. The blue line represent the mean of the first portion of signal (Phase A); the dotted lines represent the limits of the confidence interval I calculate on the first portion; the cyan area contains observations of the second portion of signal (Phase B) which lie outside the confidence interval identified on the first portion.
My question is: do you think that this method could represent a valid application of Chebyshev Theorem?
- If not, could you explain why it can not be used in this way? Could you provide alternatives to demonstrate that the two portions of signal significantly differ?
- If yes, could you suggest previous works in which such a method was used before?
SOME USEFUL HYPOTHESIS ABOUT THE SIGNAL:
- Each of the two considered portions counts more than 100 samples, so the original formulation of Chebyshev's inequality should be appropriate, according with Saw (1984).
- You can consider the first portion of the signal (the one from which I estimate the confidence interval) stationary, because the original signal showed in the image is cyclostationary and I can "stationarize" it using an appropriate theorem (Hurd 1974).
Thank you very much for your precious help!!!