Are uncertainties higher than measured values realistic?

Whenever I measure a positive quantity (e.g. a volume) there is some uncertainty related to the measurement. The uncertainty will usually be quite low, e.g. lower than 10%, depending on the equipment. However, I have recently seen uncertainties (due to extrapolation) larger than the measurements, which seems counter-intuitive since the quantity is positive.

So my questions are:

• Do uncertainties larger than the measurements make sense?
• Or would it be more sensible to "enforce" an uncertainty (cut-off) no higher than the measurement?

(The word "measurement" might be poorly chosen in this context if we are including extrapolation.)

• It is very bad practice to extrapolate very much beyond your measurements. I suggest you take measurements in the range of the extrapolations, or realize that the farther you extrapolate, the more uncertainty you will have to deal with. – David White Oct 25 '18 at 15:48
• I would suggest Feldman & Cousins's paper for a detailed technical understanding of what's going on here. arxiv.org/abs/physics/9711021 – Sean E. Lake Oct 25 '18 at 18:05
• @DavidWhite I know. But I am receiving time-series data from third-party suppliers, so I need to extrapolate whenever their unstable system does not deliver data. That is why the uncertainty can become large -- which seems unrealistic after "too much" extrapolation. I just tried to formulate the question in a more general manner. – Thomas Oct 26 '18 at 6:46
• I see "uncertainty higher that value" cases every day when I measure the voltage and it turns out to be zero. – Dmitry Grigoryev Oct 26 '18 at 7:48
• @Thomas, if you have a good mathematical model for your supplier's data, extrapolation may not be a terrible thing to do. In any event, the supplier needs to know that extrapolation for some data carries uncertainty with it. – David White Oct 26 '18 at 15:11

Indeed, uncertainties that large don't really make sense.

In reality, we have some probability distribution for the parameter we're describing. Uncertainty is an attempt to describe this distribution by two numbers, usually the mean and standard deviation.

This is only useful if the uncertainties are small, because often you'll end up combining a lot of similarly-sized uncertainties together (e.g. by averaging) and the central limit theorem will kick in, making your final distribution very nearly Gaussian. The mean and standard deviation of this Gaussian only depend on the means and standard deviations of the pieces; all other information is irrelevant.

But if you're looking at just a single quantity, with a very broad distribution, just knowing the standard deviation is just not useful. At that point it's probably better to give a 95% confidence interval instead. Of course the bottom of that interval would never be negative for a physical volume.

• Current uncertainty for the mass of antihydrogen is larger than the measured value. – OrangeDog Oct 26 '18 at 10:16
• @OrangeDog That's different; they're giving a confidence interval there, and they're not including any unphysical values, because the whole point is that the value could be negative. – knzhou Oct 26 '18 at 10:20
• This way of presenting experimental data can also make sense for unphysical values, like the negative values for the neutrino mass from tritium beta decay: link.springer.com/article/10.1140%2Fepjc%2Fs2005-02139-7 – Pieter Oct 27 '18 at 7:14

Uncertainties larger than measured values are common. Especially in measurements where the value is expected to be (close to) zero. For example values for the neutrino mass.

The particle data group lists these as smaller than some value with a 90 % confidence limit. But I have seen papers where $$m^2$$ was given as a negative number, with estimated errors smaller than the value.

For cases where the value may also be negative, symmetric standard deviations larger than the value are no problem at all. Like the difference between the $$g$$-values of the electron and the positron, or the electrical dipole moment of the electron.

• What does it mean for m squared to be negative? I can imagine m being negative, but not m squared. – Joshua Oct 25 '18 at 23:07
• @Joshua Then you need to be more imaginary :) – orlp Oct 26 '18 at 2:57
• @Joshua If an experiment results in a mass squared (relativistic formulas for example), then one can report that result. It could be due to extrapolation (as an the OP's question) or as difference between other measured quantities. – Pieter Oct 26 '18 at 8:03
• @knzhou Neutrino nasses deduced from beta decay give $m^2$. From tritium for example $m^2 = -27 \pm 20$ eV$^2$ was the 1998 value of the Particle Data Group. – Pieter Oct 26 '18 at 9:08
• Current measurements for gravitational interaction of antimatter are ±7500%. – OrangeDog Oct 26 '18 at 10:17

You can't really say whether it "makes sense" or not without the full details of the experiment. But, in many cases, the data is still useful and significant.

An example is that you theorize a value $$x$$ to be about 100. So you design an experiment to measure $$x$$ around 100 $$\pm$$ 10, with the uncertainty being what you can afford, what current technology allows, what the multi-billion dollar particle accelerator can do.

If it turns out that the true value of $$x$$ is actually 1, then you'll probably get an experimental result of $$1 \pm 10$$. This doesn't mean a mistake occurred, or the measurement is invalid, it's simply what happened.

Is the data still useful? Yes! You've now bounded $$x$$ to be less than 11. You should publish that so that future experiments are designed measure the lower range, instead of looking around $$100$$.

You've also shown that there's an issue with the theory and it needs to be modified so that it gives a value that is less than 11, and not 100. This could, for example, support a competing theory, or prompt others to identify mistakes or weaknesses in the existing theory.

Something which has an exponential distribution (so positive) with expected value $$\mu$$ also has standard deviation $$\mu$$ - there are other distributions on positive values where the standard deviation can be many times the size of the expected value

By a vague recollection from the normal distribution, you might naively think there could be roughly a $$95\%$$ chance that a observation of this would be in the range $$\mu \pm 2 \mu$$. This is a wrong approach in general, but in this special case you would turn out to be correct; more precisely this is the probability of an observation below $$3\mu$$, which is $$1-e^{-3} \approx 0.9502$$

The mistake, if any, is thinking that the interval of uncertainty should always be symmetric about the central value

If your extrapolated results have more than 100% uncertainty, which is possible, it just means that either you sample data was unrepresentative of the population, or that your extrapolation is wrong. Depending on what your experiment is, a linear extrapolation might lead to vastly incorrect results.

I'm not sure what you mean by 'enforce', but an uncertainty that high should tell you something is probably wrong. It makes sense to choose a cut-off point for your uncertainty, if that's what you mean by 'enforce', but your first strategy should probably be to look at how you're extrapolating.

An uncertainty greater than the value for a known-positive value makes sense in the context of a non-gaussian credence.

If that first sentence was clear, you can stop reading. If not, let me back up. When we talk about "uncertainty", there isn't some sharp line involved. For any given range, we can express a probability that the value is within that range based on what we know. That probability is the integral over the range of our "credence", or "probability density function" ("pdf" for short). You can think of this function as the probability we assign every possible value, multiplied by infinity in such a way that its integral is one (somewhere, a mathematician flinched).

Very often, our credence is gaussian, in which case we can describe it with two parameters: mean and standard deviation. The mean is also the expected value and the maximum likelihood value, so it makes a great point estimate. We can then call the standard deviation "the uncertainty".

If the credence is not gaussian, which seems to be the case here, then we need to describe it a bit more verbosely.