Just to make things clear - this property is not fundamental but important. It is the fundamental difference when it comes to using DCT instead of DFT for spectrum calculation.
Why do we do Cepstral Mean Normalisation
In speaker recognition, we want to remove any channel effects (impulse response of vocal tract, audio path, room, etc.). Providing that the input signal is $x[n]$ and channel impulse response is given by $h[n]$, the recorded signal is a linear convolution of both:
$$y[n] = x[n] \star h[n]$$
By taking the Fourier Transform we get:
$$Y[f] = X[f]\cdot H[f] $$
due to the convolution-multiplication equivalence property of FT - that is why it's so important property of FFT at this step.
The next step in the calculation of cepstrum is taking the logarithm of the spectrum:
$$Y[q] = \log Y[f] = \log \left( X[f] \cdot H[f]\right) = X[q] + H[q]$$
because: $\log(ab) = \log a +\log b $. Variable $q$ is the quefrency. As one might notice, by taking the cepstrum of convolution in the time domain we end up with the addition in the cepstral (quefrency) domain.
What is the Cepstral Mean Normalisation?
Now we know that in the cepstral domain any convolutional distortions are represented by addition. Let's assume that all of them are stationary (which is a strong assumption as a vocal tract and channel response are not changing) and the stationary part of speech is negligible. We can observe that for every i-th frame true is:
$$Y_i[q] = H[q] + X_i[q] $$
By taking the average over all frames we get
$$\dfrac{1}{N}\sum_{i} Y_i[q] = H[q] + \dfrac{1}{N}\sum_{i} X_i[q]$$
Defining the difference:
$$\begin{array}
&R_i[q] &= Y_i[q] - \dfrac{1}{N}\sum_{j} Y_j[q]\\
& = H[q] + X_i[q] - \left(H[q] + \dfrac{1}{N}\sum_{j} X_j[q]\right) \\
& = X_i[q] - \dfrac{1}{N}\sum_{j} X_j[q]\\
\end{array}$$
We now removed channel effects from our signal. Putting all the above equations into simple English:
- Calculate cepstrum
- Subtract the average from each coefficient
- Optionally divide by variance to perform Cepstral Mean Normalisation as opposed to Subtraction.
Is Cepstral Mean Normalisation necessary?
It's not mandatory, especially when you are trying to recognise one speaker in a single environment. In fact, it can even deteriorate your results, as it's prone to errors due to additive noise:
$$y[n] = x[n] \star h[n] + w[n] $$
$$Y[f] = X[f]\cdot H[f] + W[f] $$
$$\log Y[f] = \log \left[X[f]\left(H[f]+\dfrac{W[f]}{X[f]} \right) \right] = \log X[f] +\log \left(H[f]+\color{red}{\dfrac{W[f]}{X[f]}} \right)$$
In poor SNR conditions marked term can overtake the estimation.
Although when CMS is performed, you can usually gain a few extra percent. If you add to that performance gain from derivatives of coefficients then you get a real boost in your recognition rate. The final decision is up to you, especially since there are plenty of other methods used for the improvement of speech recognition systems.