# Difference between absolute deviation to population median and sample mean

I have independent variables $X_i\in[0;1]$ and suppose they are uniformly distributed. If you want to minimize the total absolute deviation to a fixed number, how much can you gain from using the sample mean over the population median?

Therefore I am looking for

$f(n)=E_{\{x\}}\left(\sum^n|x_i-0.5|-\sum^n\left|x_i-\frac{\sum^n x_i}{n}\right|\right)$

Hope this is the correct notation :) The function depends on the number of variables used and the expectation is over all possible values of these $n$ variables.

Apparently $f(1)=0.25$ but it seemed that $f(n>1)\approx 0.17$. It would be interesting to know the analytic expression for that. Any idea?

PS: Are there easier or more interesting results if you use the squared deviation, or other distributions, or the mean of population quantity?

• Why sample mean rather than sample median (which actually minimizes the sum of absolute deviations from the sample values)? Commented Apr 2, 2014 at 21:26
• Because in my particular case I had only the possibility to determine the total sum of the $x_i$. And without knowing the $x_i$ I was trying to minimize the scoring function which was the absolute deviation from a fixed number I provide.
– Gere
Commented Apr 3, 2014 at 9:34

The population mean and median of a uniform distribution on $$[0,1]$$ are both $$\frac12$$ as it is symmetric about $$\frac12$$.

Let's write $$\bar X$$ for $$\frac1n \sum X_i$$. We can say:

• $$\mathbb E\left[|X_i-\frac12|\right]=\frac14$$ so $$\mathbb E\left[\frac1n\sum|X_i-\frac12|\right]=\frac14$$ and $$\mathbb E\left[\sum|X_i-\frac12|\right]=\frac n4$$.
• For $$n=1$$, $$X_1=\bar X$$ so $$\mathbb E\left[|X_1-\bar X|\right]=0$$ and $$\mathbb E\left[\sum|X_i-\frac12|\right] - \mathbb E\left[\sum|X_i-\bar X|\right] =\mathbb E\left[|X_1-\frac12|\right] - \mathbb E\left[|X_1-\bar X|\right] =\frac14$$.
• For $$n>1$$, $$\mathbb E\left[|X_i-\bar X|\right]=\mathbb E\left[\frac1n\sum|X_i-\bar X|\right]= \frac14\left(1-\frac{2}{3n}\right)$$ for $$n>1$$
• so $$\mathbb E\left[\sum|X_i-\bar X|\right]= \frac n4-\frac{1}{6}$$
• and thus $$\mathbb E\left[\sum|X_i-\frac12|\right] - \mathbb E\left[\sum|X_i-\bar X|\right] =\frac16$$ for $$n>1$$, much as you found.

In another question, ashpool stated in comments that the $$\left(1-\frac{2}{3n}\right)$$ term can be proved at least for $$n=2,3,4$$. As an illustration, see the following simulation in R (the average absolute deviation from the sample mean is in black and from the population mean in grey, with the simulations as points and $$\frac14\left(1-\frac{2}{3n}\right)$$ and $$\frac14$$ as lines):

avabsdevunif <- function(n, low=0, high=1){
X <- runif(n, low, high)
meanX <- mean(X)
return(c(mean(abs(X-meanX)), mean(abs(X-(high-low)/2))))
}
set.seed(2023)
cases <- 10^5
avabsdev <- matrix(nrow=2, ncol=10)
for (n in 1:10){
simunif <- replicate(cases, avabsdevunif(n))
avabsdev[,n] <- c(mean(simunif[1,]), mean(simunif[2,]))
}
plot(1:10, avabsdev[1,], ylim=c(0,0.26))
curve((1-2/(3*x))/4, from=1, to=10, add=TRUE)
points(1:10, avabsdev[2,], pch=15, col="lightgrey")
abline(h=1/4, col="lightgrey")


A normal distribution $$N(\mu, \sigma^2)$$ would show a different result,

• with $$\mathbb E\left[\frac1n\sum|X_i-\frac12|\right]=\sqrt{\frac{2}{\pi}} \sigma$$
• and $$\mathbb E\left[\frac1n\sum|X_i-\bar X|\right]=\sqrt{\frac{n-1}{n}} \sqrt{\frac{2}{\pi}} \sigma$$,
• so $$\mathbb E\left[\sum|X_i-\mu|\right] - \mathbb E\left[\sum|X_i-\bar X|\right] = (n-\sqrt{n(n-1)})\sqrt{\frac{2}{\pi}} \sigma$$
• which gradually reduces towards $$\frac{\sigma}{\sqrt{2\pi}}$$ as $$n$$ increases.