# Is the sample mean absolute deviation unbiased for a normally distributed population?

Let $$X_1,\ldots,X_n$$ be a random sample from a normally distributed population. Is the sample mean average deviation $$\frac{\sum_{i=1}^n|X_i-\bar{X}|}{n}$$ an unbiased estimator of the population mean average deviation?

• For a normal distribution (but not others) $\mathbb E\left[\frac{\sum_{i=1}^n|X_i-\mu|}{n}\right] =\sqrt{\frac{2}{\pi}} \sigma$ while it seems empirically $\mathbb E\left[\frac{\sum_{i=1}^n|X_i-\bar{X}|}{n}\right] = \sqrt{\frac{n-1}{n}} \sqrt{\frac{2}{\pi}} \sigma$ or close to that. Nov 21, 2023 at 8:11

Clearly not for $$n=1$$ when you always get $$0$$ and (less dramatically) not for larger $$n$$.

$$\frac{\sum_{i=1}^n\left|X_i-\bar{X}\right|}{n}$$ faces the same issue as $$\frac{\sum_{i=1}^n(X_i-\bar{X})^2}{n}$$ in that $$\bar X$$ tends to be closer to the $$X_i$$ than $$\mu$$ is.

For a normal distribution (but not others) $$\mathbb E\left[\frac{\sum_{i=1}^n|X_i-\mu|}{n}\right] =\sqrt{\frac{2}{\pi}} \sigma$$, while it seems empirically $$\mathbb E\left[\frac{\sum_{i=1}^n\left|X_i-\bar{X}\right|}{n}\right]$$ $$= \sqrt{\frac{n-1}{n}} \sqrt{\frac{2}{\pi}} \sigma$$ or close to that.

As an illustration, with a standard normal and sample size $$n=4$$, the expected absolute distance to the sample average seems to be closer to $$\sqrt{\frac3{2\pi}} \approx 0.691$$ than the expected absolute distance to the mean of $$\sqrt{\frac2{\pi}} \approx 0.798$$:

avabsdevnorm <- function(n, mu=0, sigma=1){
X <- rnorm(n, mu, sigma)
meanX <- mean(X)
return(c(mean(abs(X-meanX)), mean(abs(X-mu))))
}
set.seed(2023)
n <- 4
cases <- 10^5
sims <- replicate(cases, avabsdevnorm(n))
c(mean(sims[1,]), mean(sims[2,]))
# 0.6917616 0.7990887


For comparison, for a uniform distribution on $$[a,b]$$, you have $$\mathbb E\left[\frac{\sum_{i=1}^n|X_i-\mu|}{n}\right] =\frac{b-a}{4}$$, while it seems empirically $$\mathbb E\left[\frac{\sum_{i=1}^n\left|X_i-\bar{X}\right|}{n}\right]$$ $$= \left(1-\frac{2}{3n}\right)\frac{b-a}{4}$$ or close to that at least with $$n\ge 2$$. Another simulation of $$U(0,1)$$, again with $$n=4$$, shows the expected absolute distance to the sample average seems to be closer to $$\frac5{24} \approx 0.208$$ than the expected absolute distance to the mean of $$\frac14=0.25$$:

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)
n <- 4
cases <- 10^5
simunif <- replicate(cases, avabsdevunif(n))
c(mean(simunif[1,]), mean(simunif[2,]))
# 0.2081946 0.2498248

• In a sense there seems to be something similar to Bessel's correction to the sample variance for a normal distribution but not for other distributions Nov 21, 2023 at 11:46

Substituting $$x^2/\sigma^2=u$$, $$\int_0^\infty\frac{2x}{\sigma^2}e^{-x^2/2\sigma^2}\,dx=\int_0^\infty e^{-u/2}\,du=[-2e^{-u}]_0^\infty=2.$$ Hence, $$\int_0^\infty xe^{-x^2/2\sigma^2}\,dx=\sigma^2,$$ and therefore $$\int_{-\infty}^\infty|x|e^{-x^2/2\sigma^2}\,dx=2\sigma^2.$$ Then $$\int_{-\infty}^\infty |x|\frac{1}{\sqrt{2\pi}\sigma}e^{-x^2/2\sigma^2}={\sqrt{\frac{2}{\pi}}}\cdot\sigma.$$ Translating by $$\mu$$, $$\int_{-\infty}^\infty|x-\mu|\frac{1}{\sqrt{2\pi}\sigma}e^{-(x-\mu)^2/2\sigma^2}\,dx=\sqrt{\frac{2}{\pi}}\cdot\sigma.$$ That is, if $$X\sim N(\mu,\sigma^2)$$, then the mean absolute deviation (MAD) of $$X$$ is $$E\left[|X-\mu|\right]=\sqrt{\frac{2}{\pi}}\cdot\sigma.$$ Now let $$X_1,\ldots,X_n\sim N(\mu,\sigma^2)$$ be a random sample. Let us compute the mean of the statistic $$\frac{\sum_{i=1}^n|X_i-\bar{X}|}{n}.$$ Note that \begin{align*} X_1-\bar{X}&=X_1-\frac{X_1+\cdots+X_n}{n}\\ &=\frac{(n-1)X_1-X_2-\cdots-X_n}{n}\\ &\sim N\left( 0, \frac{(n-1)^2+(n-1)}{n^2}\sigma^2 \right)=N\left( 0,\frac{n-1}{n}\sigma^2 \right), \end{align*} so $$E\left[|X_1-\bar{X}|\right]=\sqrt{\frac{2}{\pi}}\cdot\sqrt{\frac{n-1}{n}}\cdot\sigma.$$ Similarly, $$E\left[|X_i-\bar{X}|\right]=\sqrt{\frac{2}{\pi}}\cdot\sqrt{\frac{n-1}{n}}\cdot\sigma$$ for all $$i=1,\ldots,n$$, so $$E\left[\frac{\sum_{i=1}^n|X_i-\bar{X}|}{n}\right]=\frac{\sum_{i=1}^n E\left[|X_i-\bar{X}|\right]}{n}=\sqrt{\frac{2}{\pi}}\cdot\sqrt{\frac{n-1}{n}}\cdot\sigma.$$
• An interesting point is why the $\sqrt{\frac{n-1}{n}}$ term appears for a normal distribution but not in general. In general you have $\mathbb E[X_1-\bar X]=0$ and $\textrm{Var}(X_1-\bar X)=\frac{n-1}{n}\sigma^2$ (leading to Bessel's correction) but not $(X_1-\bar X) \sim \sqrt{\frac{n-1}{n}}(X_1-\mu)$ and so not $|X_1-\bar X| \sim \sqrt{\frac{n-1}{n}}|X_1-\mu|$ (though you do with a normal distribution). Nov 23, 2023 at 10:06
• It seems that for uniform distribution, we have $E\left[\frac{\sum_{i=1}^n|X_i-\bar{X}|}{n}\right]=\frac{2}{\sqrt{3}}\cdot\frac{n-1}{n}\cdot\sigma$. So we may as well define the "sample mean average deviation" as $E\left[\frac{\sum_{i=1}^n|X_i-\bar{X}|}{n-1}\right]$, at least for uniform distribution. Nov 25, 2023 at 1:27
• For a uniform distribution, I think you get $\mathbb E\left[\frac{\sum_{i=1}^n|X_i-\bar{X}|}{n}\right]=\frac{\sqrt{3}}{2}\cdot\frac{n-\frac23}{n}\cdot\sigma$ Nov 25, 2023 at 13:31
• as my answer said, my result was empirical, but I illustrated it working for $n=4$ and checked it for other $n$. (It is not correct for the trivial $n=1$) Nov 26, 2023 at 3:04