# MAD and RMS SNR relation

Recently I've found a formula relating mean average deviation (MAD) SNR to root mean square (RMS) SNR:

$$\text{SNR}_\text{MAD} = \sqrt{\frac{2}{\pi}} \text{SNR}_\text{RMS}$$

which assumes Gaussian statistics. This is for a definition of MAD something like

$$\text{MovingAverage}[\ |x - \text{MovingAverage}[x]|\ ]$$

for some signal $x$ measured in the time domain. My question is where does the above SNR formula come from? I'm pretty new to signal processing so maybe there's a very basic reference that describes this. Is there some intuition for why MAD SNR is smaller than RMS SNR?

## 1 Answer

The two main measures of scatter in the Gaussian case are the mean absolute deviation:

$$d_n = 1/n\sum |x_i-\overline{x}|$$

and the mean square deviation:

$$s_n = \sqrt{1/n\sum (x_i-\overline{x})^2}$$

If the random Gaussian noise has deviation $\sigma$, $s_n \to \sigma$ and $d_n \to \sqrt{\frac{2}{\pi}}\sigma$. I don't have the exact references right now, you can find them in Robust Statistical Procedures, 1996, P. Huber, or texts by Hampel, or Rousseeuw, on robusts statistics.

The derivation is OK, let us use the zero-mean case: the expected value of the absolute value of the random variable:

$$E ( | x | ) = \int_{- \infty}^{ \infty} | x | p ( x ) d x$$

then you derive:

$$E(|x|) = 2\int_0^{\infty} \frac x {\sigma\sqrt{2\pi}} e^{-\,\frac {x^2} {2\sigma^2} } dx$$

then

$$E(|x|) = \left[-\,\sigma \sqrt{\frac 2 {\pi}} e^{-\,\frac {x^2} {2\sigma^2} }\right]_0^{\infty} = \sigma \sqrt{\frac 2 {\pi}}$$

But this is for the mean absolute deviation, I am not sure of the equivalence with the Median Absolute Deviation.