Why does signal averaging reduces noise levels by more than $\sqrt{n}$? 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 $&gt; 1.41$. 

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?


Raw sweep; 2.3 ms, 56 kHz sample rate, 127 data points


Histogram of amplitude distribution


FFT of the signal
Well, I would say the assumption that your noise is Gaussian is ill fitting.  If the noise is due to machine interference, it probably has some tonal characteristics.  Tones of the same frequency can reinforce or cancel each other out when added.

To get a better idea of what might be going on, you should:

1) Make a histogram of the noise

2) Take an FFT of the noise

The histogram should look like a bell curve and the FFT results should be flat.  If they aren't, your noise is not Gaussian.

Hope this helps.

Ced