In my data analysis course we were taught that whenever we perform a measurement, the measured value is to be interpreted as the mean of the distribution for the true value. And the error in measurement is to be interpreted as the standard deviation of the distribution. The full distribution may be taken as Poisson or Exponential of Gaussian depending on the kind of experiment being performed. In general, the Gaussian distribution is preferred because of the Central Limit Theorem.

The book says:

For the experimenter the true value is an unknown parameter of the distribution. The measurement and its error are estimates of the true value and of the standard deviation of the distribution. (Remark that we do not need to know the full error distribution but only its standard deviation.)

-"Introduction to Statistics and Data Analysis for Physicists" by Bohm and Zech Pg. 83

But I have always been very uncomfortable with this interpretation as it does not bound the true value to any interval and allows it to take any value although with diminishing probability. I can think of atleast one context in which this interpretation seems absurd:

Suppose I wish to measure the length of a rod using a metre scale and the marking of, say, 1.414m seems to coincide with the length of the rod. Then, although I cannot conclude that the length is precisely 1.414m because of an error involved due to the limited least count of our measuring instrument which is impossible to get around as lengths are "continuous". I still do know precisely, ignoring systematic errors, that the length is no more than 1.415m and no less than 1.413m.

If I employ the conventional interpretation to this experiment, the true value is supposed to be a number lying anywhere in the distribution but with probability decreasing as the value goes more and more standard deviations away from the mean. But that is absurd since it cannot lie outside $(1.413, 1.415)$.

So, should there be a need to devise a new method of expressing measurements using a range of real numbers within "hard" limits rather than a mean and a standard deviation?

  • $\begingroup$ No, because measuring on the scale "exactly" 1.413 or 1.415 in half of the cases the real length will be smaller than 1.413 or greater than 1.415. $\endgroup$ Mar 9, 2017 at 9:22
  • $\begingroup$ What I am saying is that, as long as I make no procedural errors as an experimenter (no systematic errors) and the length of the rod is constant (as anyone would expect it to be), I would never measure 1.413 or 1.415 or any number outside the range. $\endgroup$ Mar 9, 2017 at 11:38

1 Answer 1


The measurement of the length of the rod is not a good example, because it will drive you to many distractive ideas, blurring the principle itself. Let us select more pure example - measure electrical signal from your apparatus. The signal originates in some exact process that (not a necessary condition, but it makes things easier).

1. First point for each debate is that you assume you know the statistical distribution of your experimental arrangement. It would be a problem by itself, but people mostly assume Gauss distribution and they are mostly right.

2. Then you perform several single measurements - convert your voltage (arising from all the electronic's chain) to one digital value (using an ADC converter).

2a. if you have a converter that is not very good - like 256 bit ADC converter, it probably happens that your measurement will be similar to the length measurement of a rod. Always the same value, no way to extract sigma or a distribution.

2b. with a better precision (12bit, 14bit) you are sensitive to many things and the electrical noise in all your devices in the measurement chain will start to distort the measurements. If you perform several single measurements (that give slightly different values), you can start to compute statistics - and evaluate your sigma and mean.

2c. if you perform many single measurements (100s or 1000s and more), you put your values in a histogram (showing how frequent is each obtained value) you will see a Gaussian shape forming and - voila - this is kind of prove, that your assumption about Gauss distribution of your measurements was ok. And here you can start further analyses of the shape you obtained, but it was already not your question.

You see, why the rod measurement is not a very good example, you easily slip into heat dilatation of rod, the rule all things that are quite complex, correlated but not concerning the merit of the statistical problems.

  • $\begingroup$ Good answer! So are you saying that once you try to increase the precision above a certain level it becomes impossible to keep track of background effects. Hence, we use a distribution to account for them? $\endgroup$ Mar 9, 2017 at 11:33
  • $\begingroup$ Even if that is the case a distribution extending all over the real line seems totally unrealistic to me. $\endgroup$ Mar 9, 2017 at 11:35
  • $\begingroup$ Every case - once you go into very detail - is specific. If we apply to this example - signal + noise : with 12bit ADC, you see approx 1mV changes (typically). This is almost a level of noise of our electronic's chain. If you gonna increase to 16bit or 18bit and use the same setup, you do not win too much in precision. But keeping track of background effects it is a different thing. $\endgroup$
    – jaromrax
    Mar 9, 2017 at 11:51
  • $\begingroup$ real line: ADC converts the voltage to a discreete number - it compares (just a toy model) the tension with other defined voltages and if it is more than $V_A$ and less than $V_{A+1}$, it gives away $V_A$ e.g. You get a discreete number - no real line. $\endgroup$
    – jaromrax
    Mar 9, 2017 at 11:55
  • $\begingroup$ we use a distribution to account for background effects - not exactly. If principally you cannot measure a (precise) value precisely (physical value that is not descreete) - at some precision level - you always get slightly different values in repeated measurements. Use statistics to handle that/or improve your conditions to win one order of magnitude and get to the same situation just later. $\endgroup$
    – jaromrax
    Mar 9, 2017 at 12:03

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