Let's consider the case of a measurement $x$ and its correspondent error $\Delta x$ (which are both always positive).

The relative error is defined as:

$$E = \frac{\Delta x}{x}$$

The relative error has many useful application in error propagation and so on, and it is frequently used to determine the quality of a measurement. However, when the value of the quantity to measure tends to zero, the relative error can become very large, although the measurement itself might be very reliable, because for example the quantity to measure is zero within the instrumentation error.

What would you recommend to use under such situations?

I have seen: How to calculate relative error when the true value is zero? but since I do not have access to a measurement and its true value I am not sure how to apply the discussion there to this case, and I could not find any reference in the literature.

This other question: How to calculate relative error? is somewhat related but I do not think it really applies.

I had considered using this:

$$\tilde{E} = \frac{x \cdot \Delta x}{x^2 + \Delta x^2}$$

which behaves numerically the way I would expect and it is a dimensionless quantity.

Alternatively, I had considered using the Relative Percent Difference (RPD), defined in terms of $d_1(a, b)$:

$$d_1(a, b) = 2 \frac{a - b}{|a| + |b|}$$

(where $a$ is a measurement and $b$ its true value), or, more precisely the absolute Relative Percent Difference (aRPD):

$$|d_1(a, b)| = 2 \frac{|a - b|}{|a| + |b|}$$

Using the following substitutions:

  • $|a| = x$,
  • $|b| = x + \Delta x$
  • $|a - b| = \Delta x$

I would obtain:

$$ \tilde{d_1}(x, \Delta x) = \frac{2 \Delta x}{2 x + \Delta x} = \frac{\Delta x}{x + \frac{1}{2}\Delta x} $$

However, I cannot find any reference actually using any of these or at least discussing these topics for my case. Any answer substantiated with a peer-reviewed paper is greatly appreciated.

  • $\begingroup$ I've seen the "relative" difference $|x-y|/(|y|+1)$ used for this. It clearly has the right behaviour of going to zero as its numerator does and it doesn't explode otherwise. It's not a relative error, but the principle here is that if you sometimes get absurd answers you might think of asking a different question. My informant told me that it is a standard idea among numerical analysts; if so, someone else here should find it familiar. $\endgroup$
    – Nick Cox
    Oct 11, 2017 at 18:07
  • $\begingroup$ The main problem with that definition is that is not invariant for change of units of measurement. This property must hold for any "quality index" of a measurement. $\endgroup$
    – norok2
    Oct 11, 2017 at 18:39
  • $\begingroup$ You're right. So the challenge is to find a natural generalisation of 1, or use some other method. $\endgroup$
    – Nick Cox
    Oct 11, 2017 at 19:35
  • $\begingroup$ The above definitions have this property and a benign behavior for small values of $x$, but I cannot find an authoritative discussion of what to do in such case, or some reference where any of these, or something similar was used. $\endgroup$
    – norok2
    Oct 11, 2017 at 19:38

1 Answer 1


Mathematically you know the answer already: relative error goes to infinity as the measurement value goes to zero.

Taking your question from a practical point of view, there is always additive noise in any real dataset as well as multiplicative noise. If you can characterize your additive noise, you can create a noise model that is reasonable all the way down to zero measurement value. But the concept of "relative error" will still not be relevant at low measurement value, since it is only really useful in situations where multiplicative noise dominates. If you disagree with this statement, please provide a counterexample and I will edit my answer accordingly.

  • $\begingroup$ What do you mean by multiplicative noise? If we consider a measurement of something simple, say speed $v$, you have some position $s_{0,1}$ and some time $t_{0,1}$ with their respective additive error (so that e.g. $s_0 = 40 \pm 1 \mathrm{m}$). Now, if you want to know the error on the speed $v=\frac{s_1 - s_0}{t_1 - t_0}$, you will eventually get to the relation: $\frac{\Delta v}{v} = \frac{\Delta t}{t} + \frac{\Delta s}{s}$ and this will hold true for any value of the quantities involved and will let you compute $\Delta v$ (there are some abs.value here and there, but that's not the point). $\endgroup$
    – norok2
    Oct 11, 2017 at 18:50
  • $\begingroup$ The point here however is not to calculate $\Delta v$, but, for example, to exclude some measurement based on how good the measurement was. If you have a purely linear model, and you want to know whether to include the measurement close to $0$, if you reject bad acquisitions based on the relative error, you will never get to use that point for the regression. Yet that point is probably one of the most valuable for the regression. Hence, the need for some other kind "quality measure". $\endgroup$
    – norok2
    Oct 11, 2017 at 18:57
  • $\begingroup$ @norok2 To your first comment, multiplicative noise has a magnitude that scales linearly with the measurement value. So, for instance, Gaussian multiplicative noise has a standard deviation proportional to the measurement value. And in fact, the error in the speed will become infinite if there is additive noise on both the position measurement and the timing measurement and the time interval goes to zero. That is as it should be. $\endgroup$ Oct 11, 2017 at 20:03
  • $\begingroup$ @norok2 to your second comment, agreed and my answer gives a principled way to deal with this issue. You can't and shouldn't reject bad acquisitions based on the relative error alone, unless the multiplicative contribution to the noise totally dominates the additive contribution in your measurement regime. $\endgroup$ Oct 11, 2017 at 20:07
  • $\begingroup$ Yet, how would you define a "quality measure" based on this? $\endgroup$
    – norok2
    Oct 11, 2017 at 20:19

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