# Why is the total noise variance less than the sum of individual noise variances?

I have three random variables:

• $$Y$$: my data
• $$Y_n$$: my data corrupted by additive white Gaussian noise (AWGN)
• $$Y_{nc}$$: my noisy data corrupted by a non-linear transformation $$\mathcal{C}$$.

I have computed the histograms of the noise terms:

• $$N = Y_n - Y$$: the AWGN
• $$N_c = Y_{nc} - Y_n$$: the noise added by the transformation $$\mathcal{C}$$.
• $$N_{tot} = Y_{nc} - Y$$: the total noise (AWGN + transformation noise)

I have found that the histograms of $$N$$ and $$N_c$$ are Gaussian with variances $$\sigma_{N}^2$$ and $$\sigma_{N_c}^2$$, respectively, and that $$N$$ and $$N_c$$ are uncorrelated based on their covariance analysis. Furthermore, I have found that $$N_{tot}$$ also has a Gaussian histogram with variance $$\sigma_{N_{tot}}^2$$.

However, I am surprised to find that $$\sigma_{N_{tot}}^2 < \sigma_{N}^2 + \sigma_{N_c}^2$$, and even more surprisingly, I have $$\sigma_{N_{tot}}^2 < \sigma_{N}^2$$. For example, I have computed these values from an example signal:

• $$\sigma_{N}^2 = 3.97277792$$
• $$\sigma_{N_c}^2 = 0.26705326$$
• $$\sigma_{N_{tot}}^2 = 3.71623896$$

This phenomenom always happens even if I change the input data.

Covariance matrix of $$N$$ and $$N_c$$ is

$$\begin{bmatrix} 3.97277792 & 0.25886258 \\ 0.25886258 & 0.26705326 \end{bmatrix}$$

Can someone help me understand why this is happening? I was pretty sure that the result should have been $$\sigma_{N_{tot}}^2 = \sigma_{N}^2 + \sigma_{N_{c}}^2$$. I am actually closer to $$\sigma_{N_{tot}}^2 = \sigma_{N}^2 - \sigma_{N_{c}}^2$$

• Since you are applying a nonlinear transformation to a noisy signal, there is not a simple relation between the noise on the input side of the nonlinearity and the noise on the output side of the nonlinearity. Furthermore, there are cross-terms (involving both signal and noise) that need to be accounted for. Thus, there are too many unknowns here. Can you reveal the nonlinearity in more detail, or is that proprietary information? Mar 2, 2023 at 18:54

I was pretty sure that the result should have been $$\sigma_{N_{tot}}^2 = \sigma_{N}^2 + \sigma_{N_{c}}^2$$.