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When dealing with quantizers, and in many other communications problems, the interest is usually on the mean-squared distortion or mean-squared error, rather than mean absolute error or anything else. Gallager explains the reason here, probably perfectly if I were to extrapolate from the quality of the rest of his lectures, but I don't get it.

He says that the mean-squared distortion of the quantizer maps nicely to the mean-squared distortion of the waveform we sample&quantize, and this mapping wouldn't be as nice if we used another metric such as mean absolute distortion. Why is that? Shouldn't the mean-absolute difference between the original and the sampled-quantized waveform also map to the mean-absolute difference of the quantizer? What makes mean-square different in this sense?

gtak
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3 Answers3

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Consider a band-limited function $u(t)$ sampled at an appropriate sampling rate $f_s=1/T$ such that it is perfectly represented by its samples $u(kT)$. If those samples are quantized resulting in values $v(kT)$ then it can be shown that the mean-squared error

$$\epsilon=\sum_k\big|u(kT)-v(kT)\big|^2$$

is equal to the mean-squared error between the waveforms

$$\epsilon=\frac{1}{T}\int_{-\infty}^{\infty}\big|u(t)-v(t)\big|^2dt$$

where $v(t)$ is the waveform reconstructed from the quantized samples $v(kT)$. So quantizing in such a way that the mean-squared error between the original and the quantized samples is minimized, guarantees that also the mean-squared error between the corresponding continuous waveforms is minimized. This is not the case with other error measures.

Matt L.
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    @gtak: It's actually pretty remarkable that it holds for the mean-squared error, and there is not obvious reason why it should hold at all. As it turns out, that equivalence can be proved for the mean-squared error but it doesn't hold for any other error measure. – Matt L. Dec 20 '19 at 22:58
  • thanks for the explanation! conceptually, it sounds so obvious that it never occurred to me that it might not hold. but seems it is not that trivial. – gtak Dec 20 '19 at 23:12
  • @MattL. As an attempt for an obvious reason: Wouldn't it turn out in this case that the quantization noise is a uniform distribution while the reconstructed signal would tend toward a Gaussian distribution (given the reconstruction filtering would provide the "law of large numbers"). – Dan Boschen Dec 21 '19 at 15:03
  • @DanBoschen: You have to help me here, I still don't see why from your argument it is obvious that the mean-squared error in the continuous domain should equal the mean-squared error in the discrete domain. – Matt L. Dec 21 '19 at 15:20
  • @MattL Assuming the signal is sampled at 2B so that there is no excess bandwidth, then if you consider the error waveform alone, the continuous waveform will pass through every quantized sample and be interpolated in between. From that it is clear that they would have the same mean squared error, but it you look at the distribution, one will be uniform while the other would approach Gaussian so I thought that would help explain why other error metrics would not be equal. – Dan Boschen Dec 21 '19 at 15:44
  • @DanBoschen: Ok, I kind of see what you mean. But that would assume that the equation $\int |e(t)|^2dt=T\sum_k|e(kT)|^2$ is immediately clear, wouldn't it? – Matt L. Dec 21 '19 at 15:48
  • @MattL. Yes, clear to me from the perspective that when we sample the waveform we don't change the signal energy as long as that energy was all within B (and to avoid confusion we are talking strictly about the sampling process, not quantizing since our "waveform" of interest is now the quantization error). Am I off track? – Dan Boschen Dec 21 '19 at 15:55
  • @DanBoschen: Yes of course, I'm just putting myself in the shoes of an undergrad student for whom that correspondence might not be immediately clear. You need to understand sampling and you need to know Parseval's theorem, at least that's the way I see it. So without that background I thought that that equivalence was not really obvious. – Matt L. Dec 21 '19 at 16:10
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In addition to what's being said on the video, I would also add the followings for a general consideration of errors.

Odd powers of errors (such as $e(n)$, $e^3(n)$) have the drawback that same magnitude positive and negative errors would cancel when added while finding the cumulative error. Even powers of errors, such as $e^2(n)$, $e^4(n)$, would not have this problem.

Another solution to this sign problem would be to use absolute values such as $|e(n)|$; but since the absolute value function is not differentiable at all points, it makes it harder to use in optimizations when the total error is to be minimised by differentiation. The quadratic function, in geeral, is smooth and differentiable.

Furthermore, apart from the sign problem, square of any signal defines its energy in signal processing, hence it helps defining error energy; which is a good indicatior in many aspects of the analysis.

To sum up; it's mathematical tractability and energy relation of the quadratic function that makes it more relevant and easier to handle.

Fat32
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  • thank you for your answer! is it this energy relation that's been referred to in the video that wouldn't hold for absolute errors? is it because we are interested in the difference between the waveforms in terms of energy? – gtak Dec 20 '19 at 22:32
  • Yes we are **more** interested in the energy of errors, hence their square. But it really depends on the application. But I can't really recall why *only* the quadratic error reflects the minimum from samples to waveforms. Nevertheless, it's another important reason. – Fat32 Dec 20 '19 at 23:03
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Nothing is simplest to minimize than energy.

First, energy is a quantity often considered as preserved in physics, chemistry, mechanics, etc. "Preserved" somehow means constant. Constant means still. This model is often assumed to be valid over a short period in time or space. If the square of something is constant, the slope, or derivative, is zero. Very luckily, the derivative of a square is somehow affine:

$$ (ax^2+bx+c)' = 2ax+b$$

and affine systems are easily, and efficiently resolved by linear algebra inversion, since Gauss elimination.

Second, some random classical phenomena obey, through the law of large numbers, Gaussian laws. And Gaussians are very related to squares.

So: in many practical cases, a modelling as square minimization is sound, and lends itself to tractable, offline or online algorithms, often with relatively direct solutions. Cases where more complicated modeling and algorithms, such as with strictly positive data, or outliers, and other physics-based assumptions are less wide-spread, and still under research, as they often required user input (lasso, bridge regression, elastic nets, with tuned hyper-parameters).

However, with recent works on optimization, least-absolute values are become more popular. Spoiler: as least-squares lead to mean, least-absolute value leads to median estimates. Moreover, especially in simulation, awareness is growing on the importance on assessing relative errors, instead of absolute spread. The results logarithmic measures are more complicated to address.

Laurent Duval
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