1
$\begingroup$

I have an instrument of which I would like to understand the uncertainty on the measurements taken, so that every time that I perform a single measurement, I can apply the error obtained and therefore compare it to other measurements. In order to estimate the uncertainty, I took 30 measurements one after another.

The variable that I am interested in getting out of this instrument is actually the ratio of two variables, A and B, measured by the instrument at the same moment, so that for each round of measurement, I will have a pair (A$_i$, B$_i$), with i = 1, ... 30. My final variable therefore will be $y = \frac{A}{B}$.

What some people in my lab are doing is to take these 30 measurements, calculate the mean and the standard deviation of the ratio as if the ratio were a direct measurement. That is:

$\bar{y} = \frac{1}{N} \sum \frac{A_i}{B_i}$ and $\sigma_y = \frac{1}{N - 1} \sum (y_i - \bar{y})^2$

My first thought, though, would have been to treat A and B as two separate measurements (as they are), dependent to each other (since coming from the same instrument), and to calculate the standard deviation using the error propagation formula:

$\bar{y} = \frac{\bar{A}}{\bar{B}}$ and $\sigma_y = \sqrt((\frac{\partial y}{\partial A})^2 \sigma_A^2 + (\frac{\partial y}{\partial B})^2 \sigma_B^2 + 2 \frac{\partial y}{\partial A} \frac{\partial y}{\partial B} \sigma_{AB})$

Now my question: What is the theoretical difference between the two approaches? When should one use the first (taken that it is the correct approach), and when to use the second?

$\endgroup$

1 Answer 1

1
$\begingroup$

$\bar{y} = \frac{\bar{A}}{\bar{B}}$ isn't true. Small example with $n = 2, A = (1, 3), B= (1, 2)$ then $\bar y = 1.25$ but $\frac{\bar{A}}{\bar{B}} = 4/3 = 1.33...$. For the general point consider $C = \frac{1}{B}$ and compare the formula for the Covariance of $A$ and $C$.

If you had strong priors about $A,B$ then maybe it would make sense to model them, but if you just are trying to estimate mean and $\sigma$ of $y$ from just your sample, then you aren't gaining anything by not directly looking at $y$.

$\endgroup$
2
  • $\begingroup$ Could you please support this statement with some reference? Because if I look at any book for uncertainty calculation (for example, uscibooks.aip.org/books/…), then it is written that for any indirect measurement one should use the error propagation formula. That is where my confusion comes from. $\endgroup$
    – s.cerioli
    Commented Jan 25 at 8:59
  • $\begingroup$ Maybe I was a little fast on the draw after I saw the error in $\bar y$. I'm not quite an expert on error propagation. We yould test it with a simulation, but I'm not even sure waht you plug in for A,B when you only have a sample. $\endgroup$ Commented Jan 26 at 17:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.