I'm calculating S


  • S = (A - B)/A
  • A = 778 ± 30 (value ± percent error)
  • B = 388 ± 12 (value ± percent error)

I want to report the error in S as S ± error.

I have checked several online resources. I found many equations of how the error of propagation is derived but I couldn't find a specific equation that help me estimate the error in S based on the errors of A and B?

I'll highly appreciate any suggestion what is the equation I should use to estimate the error in S based on its two components A and B?

UPDATE In this webpage about propagation of errors. https://www.lhup.edu/~dsimanek/scenario/errorman/propagat.htm

In 3.5 Examples

Example no (2) below is similar to my case but not the same. First, my question is about 2 variables only A and B. In addition, there is subtraction in the nominator in my case not summation. Yet, they have not estimated the correlation between the three parameters G , H and Z.

A quantity Q is calculated from the law:

Q = (G+H)/Z,

and the data is:

G = 20 ± 0.5

H = 16 ± 0.5

z = 106 ± 1.0

  • 1
    $\begingroup$ Can you assume A, B are independent, do you know there correlation, or can you estimate the correlation from the same data used to estimate A, B? You'll need to do one of these in order to answer the question. $\endgroup$ – Lucas Roberts Jul 16 '17 at 3:09
  • $\begingroup$ @user3164100 Thanks Lucas. I appreciate your time and help. Could you please check my update? I have found a link that shows the estimation of error similar to my case but it is not exactly the same. They have not estimated the correlation. I want to estimate the error similar to the way they did. But I didn't manage to find which equation I should be using in my case. $\endgroup$ – shiny Jul 16 '17 at 6:01

You should be able to get at this by using the delta method. I've made a post here: estimation of population ratio using delta method about the delta method on a ratio which you can use the argument to give you exactly what you need. The final result is:

$$1-\frac{B}{A}=\frac{A-B}{A} \sim N\left(1-\frac{\mu_B}{\mu_A}, \frac{\sigma^2}{n}\right)$$

Here $\sigma^2 = \frac{\sigma_B^2}{n\mu_B^2} - 2\frac{\sigma_{BA}}{\mu_B^2}+\frac{\sigma^2_A\mu_B^2}{n\mu_B^4}$. As I mentioned in the comment you'll either need to estimate the covariance $\sigma_{AB}$ (equivalently the correlation $\rho$) or assume the covariance is 0.

From the data you provided in this post and assuming a $1.96$ multiplier to get 95% confidence intervals. I get the following for the parameter estimates:

$\mu_A =778$, $\mu_B=388$, $\sigma^2_A=15.3061^2$,$\sigma^2_B=6.1224489^2$. Then $1-\mu_B/\mu_A=0.5076$ and $\sigma^2/n =0.0004376/n$. You can then again use $1.96$ as a multiplier on the standard deviation $\sigma$ to get a 95% confidence interval for this quantity. Here in my calculation for $\sigma^2$ I'm assuming the covariance is 0.

  • $\begingroup$ The formula for $\sigma^2$ is fishy. Suppose, for instance, that the correlation between $A$ and $B$ were $1/2$ and $\sigma_A=\sigma_B$. Then $\sigma^2 \lt 0$ for $n \gt 2$! $\endgroup$ – whuber Jul 17 '17 at 19:51
  • $\begingroup$ @whuber Thanks for catching the typo! the $n$ should've been a $2$. $\endgroup$ – Lucas Roberts Jul 17 '17 at 23:17
  • $\begingroup$ I still don't believe your formula, because if we consider cases with very large values of $|\mu_B|$, your $\sigma^2$ grows arbitrarily small, but clearly it needs to increase as $|\mu_B|$ increases. You will be able to defend yourself against algebraic mistakes, typos, and other errors, by clearly explaining how you derive $\sigma^2$: if your derivation method is legitimate, then readers can figure out how to correct incidental errors of exposition. But without any such explanation, your formula is either correct or not--and so far we have no reason to trust it. $\endgroup$ – whuber Jul 18 '17 at 12:17
  • $\begingroup$ @whuber, the derivation is given in the link in my answer. This is a delta method approximation. Please take a look at that link to determine if you think there is an error in the derivation. Here to get to the result stated. I apply the delta method to the ratio B/A. Then convert means and variances using standard results from probability theory. $\endgroup$ – Lucas Roberts Jul 18 '17 at 14:48
  • $\begingroup$ I don't think we need to refer to the link: I have pointed out why this result must be erroneous. In fact, it doesn't even make sense, because you haven't explained what "$n$" refers to. There is no such value in the original question. $\endgroup$ – whuber Jul 18 '17 at 15:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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