The percentage change between two measurements is usually calculated as $100 \cdot \frac{x_2 - x_1}{x_1}$. If I have confidence intervals $(a_1, b_1)$ for $x_1$ and $(a_2, b_2)$ for $x_2$, is the confidence interval for the percentage change given as $100 \cdot \left ( \frac{a_2 - b_1}{b_1}, \frac{b_2 - a_1}{a_1} \right )$?

(The thought being that $a_2-b_1$ is the smallest difference, while $b_2-a_1$ is the greatest difference.)

Additional questions:

  • Is this valid for confidence intervals for variables other than the (sample) mean?
  • Is this valid for confidence intervals for sample means that were obtained for different sample sizes?
  • Is this valid regardless of the distribution of the (sample) mean?
  • Is this valid regardless of the procedure used to obtain the confidence interval (assuming that all the assumptions of the procedure were met)?

It would be nice to have references backing up every answer.

  • $\begingroup$ You can simplify a bit by asking for the distribution of the ratio of $x_2 / x_1$ and subtracting 1. $\endgroup$
    – Charlie
    Sep 14, 2012 at 18:11

2 Answers 2


No, it is not valid:

  1. It can produce nonsense: for example if $a_1=-1$, $b_1=2$, $a_2=-2$ and $b_2=3$ then your statement would produce a confidence interval for the percentage change starting at -200% and increasing to -400%.

  2. The width of the confidence interval ought to depend on any dependence between the two random variables.

  3. Even if the the random variables are independent and positive, the confidence interval will depend on their distribution.

  • $\begingroup$ +1 Henry good point! To and to his points,a confidence interval is defined not only by its endpoints but also by it coverage (aka confidence level). Say that you start with two 95% confidence intervals for your respective means and you construct a random interval based on the random endpoint from the confidence intervals (the endpoint are random variables when considering confidence levels and constants when evaluated on a particular sample) and you manipulate the endpoint to make them into your percentage change form. $\endgroup$ Sep 14, 2012 at 21:19
  • $\begingroup$ (continued) Furthermore you take the two point estimates for the means and convert to a percent change for the estimates. If that value falls within the interval you will have some sort of confidence interval for the percent change. But what is its confidence level. Not 95% and probably a lot less. $\endgroup$ Sep 14, 2012 at 21:23
  • 1
    $\begingroup$ What can be done that maybe you were thinking of is if you have say a 95% confidence interval [A, B] for a parameter m and you want a 95% confidence interval for f(m) where f is monotone (either nondeceasing or nonincreasing) [F(A), F(B)] will be a 95% confidence interval for f(m) in the nondecreasing case and [F(B), F(A)] will be a 95% confidence interval in the nonincreasing case. $\endgroup$ Sep 14, 2012 at 21:28
  • $\begingroup$ Can't you easily construct a $90\%$ confidence interval for $f(X,Y)$ as the range of values $f$ takes on a $95\%$ confidence interval for $X$ and a $95\%$ confidence interval for $Y$? $\endgroup$ Sep 15, 2012 at 8:09
  • 1
    $\begingroup$ Thanks for disputing my proposed method. The reason I have not accepted this answer is that it does not answer the question proper, which is how to actually compute the confidence interval. $\endgroup$
    – Vegard
    Sep 18, 2012 at 19:20

If we have distributions of $x_1$ and $x_2$; just creates random $x_1$ and $x_2$ for $10000$ times, then generate $(x_2-x_1)/x_1$, within $10000$ number to get CI.


This question is similar in [Lancet paper DOI:https://doi.org/10.1016/S0140-6736(16)31678-6] they use uncertainty intervals : Each of these 9·3 billion estimates was calculated 1000 times to determine uncertainty intervals.


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.