How can I compute the error on the interquartile range of a sample? By error I mean its std deviation (e.g. error on the mean = $\frac{RMS}{\sqrt{N}}$.

The sample is from a unimodal distribution, similar to normal distribution, but with asymmetric tails.


Bootstrapping would probably be a feasible and convenient option (I'll give a short example in R at the end of this answer). In general, the asymptotic distribution of the IQR is normal (see page 327 of DasGupta (2011): "Probability for Statistics and Machine learning: Fundamentals and Advanced Topics"). Let $f$ be the density, $F$ the CDF and the population quantile function be $F^{-1}(p)$ of a random variable. Further, let $F^{-1}(p) = \xi_{p}$. Then, the following holds asymptotically: $$ \sqrt{n}\left(\mathrm{IQR} - \left(\xi_{\frac{3}{4}}-\xi_{\frac{1}{4}}\right)\right)\xrightarrow{d} \mathrm{N}\left(0, \frac{1}{16}\left[\frac{3}{f^{2}(\xi_{\frac{3}{4}})}+\frac{3}{f^{2}(\xi_{\frac{1}{4}})}-\frac{2}{f(\xi_{\frac{1}{4}})f(\xi_{\frac{3}{4}})}\right]\right) $$

For iid observations of a normal distribution $\mathrm{N}(\mu, \sigma^{2})$, this result simplifies to: $$ \sqrt{n}\left(\mathrm{IQR} - 1.349\sigma\right)\xrightarrow{d} \mathrm{N}\left(0, 2.476\sigma^{2}\right) $$. So asymptotically, the standard deviation is $1.573\sqrt{\frac{\sigma^{2}}{n}}$.


Let's illustrate the bootstrap with an example where the population has an exponential distribution.

# Load packages


# Function used for the bootstrap

iqr.fun <- function(data, indices) {

  d <- data[indices]  
  iqr <- IQR(d)  


# Do the bootstrap with 100000replications

set.seed(612) # for reproducibility

mysamp <- rexp(100, 1.5) # exponential with rate 1.5

res <- boot(data = mysamp, statistic = iqr.fun, R = 100000)


Bootstrap Statistics :
     original      bias    std. error
t1* 0.9985767 -0.01563602   0.1519901

# Confidence intervals


Intervals : 
Level      Normal              Basic         
95%   ( 0.7163,  1.3121 )   ( 0.8015,  1.4591 )  

Level     Percentile            BCa          
95%   ( 0.5381,  1.1956 )   ( 0.5301,  1.1940 )  
Calculations and Intervals on Original Scale

The bootstrap standard error of the IQR is estimated to be $0.152$ and the 95% bias-corrected confidence interval is $\left(0.5301,\;1.1940\right)$. The theoretical IQR of an exponential distribution with $\lambda = 1.5$ is $\frac{\log{(3)}}{\lambda}\approx 0.7324$ which is well within the calculated confidence interval. The theoretical standard error in the exponential case is $2\sqrt{\frac{2}{3}}\sqrt{\frac{1}{n\lambda^{2}}}\approx 0.1089$.

| cite | improve this answer | |

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.