"Weighted Median": statistical properties and connection to the sample variance

Question

Suppose we only have the following information:

• A set of sample means : $\bar{x_1}$, $\bar{x_2}$ ... $\bar{x_k}$
• The sample size used to calculated each sample mean: $n_1$, $n_2$ ....$n_k$
• The population size from which each sample mean was taken: $N_1$, $N_2$ ....$N_k$
• The sample variance of each mean: $Var(\bar{x_1})$, $Var(\bar{x_2})$ ... $Var(\bar{x_k})$
• We do NOT have the raw data that were used to calculate the sample means

Given this information, I have seen that this information can be combined together to produce a weighted mean estimator:

1) Count Weighted Mean:

$$\bar{x_W} = \frac{\sum_{i=1}^{k} w_i \bar{x_i}}{\sum_{i=1}^{k} w_i}$$ $$w_i = n_i$$

$$Var(\bar{x_W}) = \frac{\sum_{i=1}^{k} w_i^2 Var(\bar{x_i})}{(\sum_{i=1}^{k} w_i)^2}$$

2) Variance Weighted Mean:

$$\bar{x_W} = \frac{\sum_{i=1}^{k} w_i \bar{x_i}}{\sum_{i=1}^{k} w_i}$$

$$w_i = \frac{1}{Var(\bar{x_i})}$$

$$Var(\bar{x_W}) = \frac{\sum_{i=1}^{k} w_i^2 Var(\bar{x_i})}{(\sum_{i=1}^{k} w_i)^2}$$

Note that in this case, it can be shown (using Lagrange Multipliers) that this weighting scheme (i.e. inverse variance) will produce the estimate with the lowest overall variance (https://en.wikipedia.org/wiki/Inverse-variance_weighting)

My Question: Suppose we are given the exact same information (i.e. sample size, population size, sample variance, no raw data) - but this time, we are provided with the sample medians instead of the sample means.

Is it still common practice to calculate "weighted median" estimators just as we did above? Or does this "ruin" the purpose of the median by "contaminating" all the medians with each other (i.e. influencing) - thus no longer a "true median"? In general, I am not sure if it is a good idea to combine medians like this together: does this combined estimator still have desired mathematical properties (e.g. asymptotic normality, consistency).

I am not sure how exactly the variance of this weighted median estimator will be calculated. I am considering using the weighted bootstrap in this situation (i.e. the probability of selecting an individual median is proportional to its weight) - but I am not sure if this is a statistically valid approach (e.g. Understanding the Weighted Bootstrap)

References:

• The variance of the weighted median and optimal weights

Answer 1

No - your approach does not work without further assumptions.

For example in the counting case, if I told you that

• Group A had a weight of $10\%$ and median $3$
• Group B had a weight of $90\%$ and median $8$

all you can say is that the overall median is somewhere in $[3,8]$.

If the actual observations were as follows, the overall median would be $3$. Change the $42:48$ in group B to $32:58$ and the overall median becomes $8$. Intermediate overall medians are also possible.

Value  Frequency  Group
  2       42        B
  3       10        A
  8       48        B

Answer 2

This was an early answer reacting to the original title which began: Does the "Weighted Median" Exist?

How common it is overall I can't easily say, but applications are easy to mention: e.g. median income as calculated from incomes and weights.

Let's keep an example calculation trivial:

Income bins 1, 2, 3, 4 with weights 0.3, 0.4, 0.2, 0.1.

So cumulative weights are 0.3, 0.7, 0.9 and 1 and cumulative weight 0.5, which defines the position of the median, falls within bin 2, which is reported as the median.

Interpolation can be as fancy as you like, e.g. linear interpolation between bin limits, and so on.

In short, order the values, and scale weights to sum to 1. Then cumulative weight 0.5 defines the position of the median, the precise recipe depending on local tradition and personal taste.

The recipe can be extended to weighted quantiles.

Various common areas:

Data are released for geographic areas. We might even try to get a weighted median from area means and their weights.

Data are released only binned in some other way, e.g. to respect confidentiality, etc.

Incomes are top-coded, so there are no precise details released on the very richest people. In this case, means are out of the question except with strong assumptions, but medians might be tractable (and interesting and useful any way).

Answer 3

Does the "Weighted Median" Exist?

Obviously it exists, you can take a median of medians. Using weighted medians is just an extension.

The more relevant question is whether it is useful to use a median of medians and to use weights.

Is it still common practice to calculate "weighted median" estimators just as we did above? Or does this "ruin" the purpose of the median by "contaminating" all the medians with each other (i.e. influencing) - thus no longer a "true median"?

If the sample medians are drawn from populations with the same distribution then the median of medians can be biased (just like a sample median can be biased), but it is consistent. The quantile of the sample medians will follow a beta distribution which approaches the 0.5 point when the sample sizes increase$^{*}$.

Using weights can improve the properties of the median of medians statistic if the different medians do not follow the same beta distribution.

Below is an example where we simulated the median of medians by using two times ten samples from different beta distributions (representing a sample of twenty medians each sampled based on different sample sizes) and combine them with different weights. It shows that there is an optimal weight.

set.seed(1)

sample = function(n=10, weights = c(1,2)) {
   x = rbeta(n,3,3)
   y = rbeta(n,10,10)
   standard = median(c(x,y))
   weighted = median(c(rep(x,weights[1]),rep(y,weights[2])))
   return(c(standard, weighted))
}


weights = matrix(c(1,2,
                   1,3,
                   1,4,
                   1,5,
                   2,3,
                   2,5,
                   3,4,        
                   3,5,
                   4,5),9, byrow = 1)
 

standard = rep(NA,9)
weighted = rep(NA,9)
k = 1:9

for (ki in k) {
  s = replicate(3*10^4,sample(10, weights = weights[ki,]))
  standard[ki] = var(s[1,])
  weighted[ki] = var(s[2,])
}

plot(weights[,1]/weights[,2],standard, ylim = range(c(standard,weighted)), xlab = "ratio of weights", ylab = "variance of statistic")
points(weights[,1]/weights[,2],weighted,col = 2) 

legend(0.2, 0.00145, c("standard","weighted"), col = c(1,2))

Whether this approach has been applied somewhere in practice or literature I do not know. But the above shows that in principle the weighted median of medians exists and can make sense.

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$^*$ For continuous variables, the sample distribution of the median is $F_X^{-1}(U)$ with $U$ a beta distribution. So the sampling of the sample median is similar to sampling from that beta distribution and then applying the quantile function $F^{-1}$. See for more: Is the median of a very large number of medians always equal to the true median and Central limit theorem for sample medians