analogy between mean and median [X validated]

A recent question on X validated was looking rather naïvely for the median being equal to the mean, as earlier questions there, but this made me realise a certain symmetry in the two notions, namely that the median ς is solution of the balance condition

namely the areas under the density are equal, while the mean μ is solution of a parallel balance condition

which is an equality of the same kind, i.e., of areas under the cdf or its complement…. Courtesy of the general representation

The general problem of having the mean and median equal is not particularly interesting as it is a matter of parameterisation. For instance, using the cdf transform turns the random variate X into a Uniform F(X), with mean and median equal to ½.

sampling the mean

A challenge found on the board of the coffee room at CEREMADE, Université Paris Dauphine:

When sampling with replacement three numbers in {0,1,…,N}, what is the probability that their average is (at least) one of the three?

With a (code-golfed!) brute force solution of

mean(!apply((a<-matrix(sample(0:n,3e6,rep=T),3)),2,mean)-apply(a,2,median))

producing a graph pretty close to 3N/2(N+1)² (which coincides with a back-of-the-envelope computation):

a discovery that mean can be impacted by extreme values

A surprising editorial in Nature about the misleading uses of impact factors, since as means they are heavily impacted by extreme values. With the realisation that the mean is not the median for skewed distributions…

To be fair(er), Nature published a subsequent paper this week about publishing additional metrics like the two-year median.

an ABC experiment

 

In a cross-validated forum exchange, I used the code below to illustrate the working of an ABC algorithm:

#normal data with 100 observations
n=100
x=rnorm(n)
#observed summaries
sumx=c(median(x),mad(x))

#normal x gamma prior
priori=function(N){
 return(cbind(rnorm(N,sd=10),
  1/sqrt(rgamma(N,shape=2,scale=5))))
}

ABC=function(N,alpha=.05){

  prior=priori(N) #reference table

  #pseudo-data
  summ=matrix(0,N,2)
  for (i in 1:N){
   xi=rnorm(n)*prior[i,2]+prior[i,1]
   summ[i,]=c(median(xi),mad(xi)) #summaries
   }

  #normalisation factor for the distance
  mads=c(mad(summ[,1]),mad(summ[,2]))
  #distance
  dist=(abs(sumx[1]-summ[,1])/mads[1])+
   (abs(sumx[2]-summ[,2])/mads[2])
  #selection
  posterior=prior[dist<quantile(dist,alpha),]}

Hence I used the median and the mad as my summary statistics. And the outcome is rather surprising, for two reasons: the first one is that the posterior on the mean μ is much wider than when using the mean and the variance as summary statistics. This is not completely surprising in that the latter are sufficient, while the former are not. Still, the (-10,10) range on the mean is way larger… The second reason for surprise is that the true posterior distribution cannot be derived since the joint density of med and mad is unavailable.

After thinking about this for a while, I went back to my workbench to check the difference with using mean and variance. To my greater surprise, I found hardly any difference! Using the almost exact ABC with 10⁶ simulations and a 5% subsampling rate returns exactly the same outcome. (The first row above is for the sufficient statistics (mean,standard deviation) while the second row is for the (median,mad) pair.) Playing with the distance does not help. The genuine posterior output is quite different, as exposed on the last row of the above, using a basic Gibbs sampler since the posterior is not truly conjugate.