I have an electrophysiology signal, which is time-based. It basically measures neural activity in the form of potential differences over time.
The noise in this signal is assumed to be a random background noise caused by the device electronics and is assumed to have a normal (i.e., Gaussian) distribution (based on previous literature) and it is supposed to be stationary. Below I have added some plots that show a raw sweep (2.3 ms of recording per sweep, sample rate= 56k, 127 points were stored), the histogram of the values, and the FFT.
In answer to the answer below: I am fairly confident we can assume normally distributed data, since it's a bell shape. However, The FFT is not flat as answerer suggests it should be when normally distributed. However, we only have a few ms of data, which may complicate this type of analysis.
Assuming a normal distribution, theoretically, doubling the number of sweeps and averaging them should reduce the noise by a factor of the square root of $2$ (about $~1.41$). This kind of signal averaging to reduce random background noise is called ensemble averaging.
Now, I have a recording (without a signal, just background) and I have determined the noise level at $1$, $2$, $4$, $8$ and $16$ averages.
Edit: This was done by re-recording the background noise and determining the average with MATLAB, in the format
B = mean(A1,A2), with
Ai being a vector with a time-based signal.
Noise level was defined as the standard deviation of the sweep. When I determine the factor of improvement between these averages, i.e., 2 vs 1, 4 vs 2, 8 vs 4, and 16 vs 8, I find the factors $1.4$, $2.5$, $2.5$ and $2.3$, respectively. These improvement factors were simply calculated by determining the ratio between SD$n$ / SD$2n$, with $n$ being the number of averaged sweeps.
The factors I find are averages across 31 electrodes measured in a total of N = 13 people. In other words, it's consistent across trials and 13 different devices. I didn't do stats, but I bet the latter three improvements are significantly $> 1.41$.
Why is the noise reduction higher than the theoretical 1.41 when doubling my sweeps from 2 to 4, from 4 to 8, and 8 to 16?