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(Inspired by this question on the photography site)

Say you have N copies of the same signal, each with a layer of noise on top. You average these copies together in an attempt to reduce the effect of noise. How much noise will the average have?

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up vote 6 down vote accepted

Assuming the noise "components" of the $N$ copies are taken independently from the same normal distribution, then noise in the average will also be normally distributed with the same mean and with standard deviation equal to $1/\sqrt{N}$ times the standard deviation of the original noise distribution. See here.

If we assume the mean of the noise distribution is zero, then this amounts to saying that the averaged noise is reduced according to the factor $1/\sqrt{N}$.

This can be generalized in various ways, either by assuming different normal distributions of the noise in the copies or by assuming the noise in the copies is identically distributed and independent but not necessarily a normal distribution. An illustration of the latter is the so-called "random walk" phenomenon.

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If each noise sample is independent,zero-mean and has identical variance, then the noise variance actually $increases $ by a factor of $N$. However the noise variance relative to the squared magnitude of signal would decrease by $1/N$

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The answers you got are correct assuming (as they state) the noise runs are independent. But if not, you can't guarantee that. If you have a noise spike that repeats say 5 sec after the start of the signal, it will not average out. This is the ongoing interplay between random and systematic error.

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Statistics states that if the noise is normally distributed, it is reduced by $\sqrt {N}$ times if summed.

Of course, noise patches being summed must be equally distributed (must have same distribution parameters) and independent.

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