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I am reading a research paper and they use the expression: "This is a weak signal, i.e. 3-4σ detection over the continuum".

I saw this sentence in a lot of papers, but I can not find what exactly it means.

My interpretation: I only know the 3σ interval from normal distributions, where the 3σ interval contains 99.73 % of all values. So, my interpretation of the above quote is , based on the fact that it is a weak signal, that the signal is only 100%-99.73% = 0.27% above the continuum intensity. Is this right?

Thanks a lot!

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Measurements in Astrophysics are usually done is what is called Signal to Noise Ratio (SNR), r defined as

$$r=S/σ$$where S is the net signal - counts, voltage,whatever
σ is the standard deviation of the noise process.

This determines not just the error on our measurement, but whether we have managed to measure a signal at all. If the true signal is zero, every so often we will see a large value just by chance. For example, if the noise distribution is Gaussian with standard deviation σ then the probability in one experiment of getting $r >2$, i.e. a fake signal with $S= 2σ$, is $1/20$

Here the Signal is 3-4 $\sigma$, it means signal is higher than noise by a factor between 3 to 4 (not 3 subtracted by 4!).

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  • $\begingroup$ Thanks a lot, this was very helpful! Just one follow up quesiton: How did you come up with the 1/20 in the case for r>2? $\endgroup$
    – Welcome_Green
    Commented Mar 12, 2021 at 11:14
  • $\begingroup$ You are welcome! It has to do with Probability of detection. So basically you convert everything in terms of power (because that is what you detect) and then integral the Noise - here Gaussian PDF to find the probability of such a detection. Just google 'signal to noise ratio and probability of detection'. What i just said is not exactly right, as in you have to account for the threshold after which detection (depends on instruments) happens, when integrating the Noise PDF and such factor. That is to say $1/20$ is not that exact, I just meant it to give rough idea of what we are dealing with. $\endgroup$
    – TheImperfectCrazy
    Commented Mar 12, 2021 at 11:37

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