Suppose I have a voltage versus current graph with various "peaks" in current at given voltages. For example, this is what is seen in a Franck Hertz experiment.

I can measure the value of voltage at the center of each peak, $v_i,$ and assign some uncertainty to this $\delta v_i.$

Now I want to consider the spacings between the peaks. $$\Delta v_i \equiv v_{i+1} -v_{i}.$$ The associated uncertainty here is $$\delta(\Delta v_i) = \sqrt{(\delta v_{i+1})^2 + (\delta v_{i})^2}$$ Finally, I want to consider the average spacing between the voltages. $$\bar{\Delta v} = \frac{\sum{ \Delta v_i}}{N}$$ The uncertainty here would be $$\delta(\bar{\Delta v}) = \sqrt{ \Sigma (\delta(\Delta v_i))^2 }.$$

However, I can rewrite the equation for the average, $$\bar{\Delta v} = ((v_2 - v_1) + (v_3 - v_2) + ...)/N = (v_{N+1} - v_1)/N$$ This would imply that the uncertainty for the average is just $$\delta(\bar{\Delta v}) = \sqrt{ (\delta(v_1))^2 +(\delta(v)_{N+1})^2}/N.$$

How to reconcile these different results?

In addition, what is there to be said about using the standard deviation in $\Delta v_i$ as opposed to the error propagation formula? Why is the error propagation formula "better?"

  • $\begingroup$ Shouldn't your average spacing formula read $\overline{\Delta v}=\frac{\sum_i \Delta v_i}{N}$? $\endgroup$
    – Triatticus
    Commented May 2, 2022 at 22:45

2 Answers 2


This is because adding the squares of uncertainties and taking the square root (known as adding in quadrature) only works if the uncertainties are uncorrelated. In this case, the uncertainty of adjacent spacings is correlated: if you are uncertain as to a given peak voltage, then the spacing to both the peak on its left and to its right are affected by this. For example, if the peak were actually slightly further to the right, then the spacing on the right would decrease while that on the left would increase.

The general formula for any possible correlations is

$$\delta f\equiv\sqrt{\langle (f-\langle f\rangle)^2\rangle}=\sqrt{\sum_{i,j}\frac{\partial f}{\partial x_i}\frac{\partial f}{\partial x_j}\langle (x_i-\langle x_i \rangle)(x_j-\langle x_j\rangle)\rangle}$$

where the angle brackets denote an expectation value. Uncorrelated variables would have $\langle(x_i-\langle x_i \rangle)(x_j-\langle x_j\rangle)\rangle=0$ if $i\neq j$ which reduces to the formula you gave. However, in your case, $x_i\equiv\Delta v_{i}=v_{i+1} -v_{i}$, so adjacent spacings are correlated with each other because they both depend on the same $v_i$. Specifically, you would find that $\langle(\Delta v_{i}-\langle\Delta v_{i}\rangle)(\Delta v_{j}-\langle\Delta v_{j}\rangle)\rangle$ does not equal zero for $j=i\pm1$. Plugging it into the general formula with the function

$$f\equiv\bar{\Delta v} = \frac{\sum_i \Delta v_i}{N}$$

would hypothetically give the (correct) result

$$\delta(\bar{\Delta v}) = \frac{\sqrt{ \delta v_1^2 +\delta v_{N+1}^2}}{N}$$

though obviously you arrived at this via a much simpler method.

  • $\begingroup$ I am almost ready to accept this answer. Any comments about my question related to the standard deviation? $\endgroup$
    – Jbag1212
    Commented May 3, 2022 at 2:20
  • $\begingroup$ I'd say error propagation is probably better because it relates directly to the experimental errors $\delta v_i$ which can be estimated from your specific setup e.g. oscilloscope resolution or fluctuations, whereas standard deviation is just an estimator which knows nothing about the actual experiment. $\endgroup$ Commented May 3, 2022 at 8:51

Careful, you can only sum variances when the variables are uncorrelated. This is obviously not the case for the $\Delta v_i$, so your first formula: $$ \delta(\overline{\Delta v}) = \sqrt{ \sum (\delta(\Delta v_i))^2 } $$ is false. This resolves the contradiction, the method using the propagation of uncertainties is valid. Hope this helps and tell me if you need more details.


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