Efficiency Of The Mean And Median In Location Models

Why Compare the Mean and Median?

In a location model, the parameter describes where a distribution is centered. Two natural estimators of that location are the sample mean and the sample median. Their behavior depends on the shape of the distribution: the mean uses the magnitudes of all observations, while the median is determined by their ordering. This difference can affect both sensitivity to extreme observations and asymptotic precision.

Here, precision is compared through the variance in the limiting distribution of the estimator after multiplication by $\sqrt n$. The comparison is model-dependent: neither estimator is uniformly more efficient across the normal and Laplace families.

Definition: If an estimator $\hat\theta_n$ satisfies $\sqrt n(\hat\theta_n-\theta)\xrightarrow{d}N(0,V)$, its first-order asymptotic variance is $V$. For estimators $A$ and $B$ of the same parameter, define the asymptotic relative efficiency of $A$ to $B$ as $\operatorname{ARE}(A,B)=V_B/V_A$. Thus, a value above one means $A$ has the smaller asymptotic variance.

General First-Order Variances

Let $X_1,\ldots,X_n$ be independent and identically distributed observations from a location model with location parameter $\theta$. For the mean, suppose $E[X_i]=\theta$ and $0<\operatorname{Var}(X_i)=\tau^2<\infty$. The central limit theorem gives

\sqrt n(\bar X-\theta)\xrightarrow{d}N(0,\tau^2), \qquad \bar X=\frac{1}{n}\sum_{i=1}^n X_i.

For the median, suppose the distribution has a unique median $\theta$ and a density $f$ that is continuous and positive at $\theta$. These conditions ensure that probabilities near the center change locally at a nonzero rate. Define the sample median here as the empirical quantile $M_n=X_{(\lceil n/2\rceil)}$, where $X_{(j)}$ is the $j$th order statistic.

Theorem: Under the median conditions just stated, $$ \sqrt n(M_n-\theta)\xrightarrow{d}N\left(0,\frac{1}{4f(\theta)^2}\right). $$ The asymptotic variance depends on the density at the median, not just on a general measure of spread.

To see why the density enters, a small displacement $h$ from $\theta$ changes the probability of falling below the cutoff by approximately $f(\theta)h$. A median is determined by whether roughly half the observations lie below a cutoff. The theorem makes that local relationship precise.

Move the cutoff locally.
For a fixed real number $t$, take $x_n=\theta+t/\sqrt n$. Differentiability of $F$ at $\theta$, with derivative $f(\theta)$, gives $F(x_n)=1/2+f(\theta)t/\sqrt n+o(n^{-1/2})$.

Count observations below it.
The count $N_n(x_n)=\sum_{i=1}^n \mathbf 1\{X_i\le x_n\}$ has a binomial distribution with parameters $n$ and $p_n=F(x_n)$.

Translate the median event.
$M_n\le x_n$ holds exactly when $N_n(x_n)\ge\lceil n/2\rceil$.

The binomial central limit theorem applies because $p_n\to 1/2$. Write $k_n=\lceil n/2\rceil$. Since $k_n-n/2=O(1)$, the standardized threshold is

\frac{k_n-np_n}{\sqrt{np_n(1-p_n)}} = \frac{-f(\theta)t\sqrt n+o(\sqrt n)}{\sqrt n/2+o(\sqrt n)} \longrightarrow -2f(\theta)t.

Consequently,

P\{\sqrt n(M_n-\theta)\le t\} =P\{N_n(x_n)\ge k_n\} \longrightarrow 1-\Phi(-2f(\theta)t) =\Phi(2f(\theta)t),

where $\Phi$ is the standard normal distribution function. The last expression is the distribution function of a centered normal random variable with variance $1/(4f(\theta)^2)$. This proves the stated limit. $\square$

If the sample size is even, another common convention averages the two central order statistics. Under the same conditions, that convention has the same first-order limit; the displayed proof uses one empirical-quantile convention to keep the median event exact.

Normal Location Model

Suppose $X_i\sim N(\theta,\sigma^2)$, with known positive variance $\sigma^2$ for purposes of describing the model. The mean has asymptotic variance $\sigma^2$. The normal density at its median $\theta$ is $f(\theta)=1/(\sqrt{2\pi}\sigma)$. Substitution into the median formula gives

V_{\mathrm{median}} =\frac{1}{4f(\theta)^2} =\frac{1}{4\{1/(\sqrt{2\pi}\sigma)\}^2} =\frac{\pi\sigma^2}{2}.

Worked Example: Comparing Mean and Median Under Normality

For a normal location model with $\sigma^2=9$, calculate both asymptotic variances and the efficiency of the mean relative to the median.

The mean’s asymptotic variance is $V_{\mathrm{mean}}=\sigma^2=9$. The median’s is $V_{\mathrm{median}}=\pi\sigma^2/2=9\pi/2$, approximately $14.14$.

By the stated convention,

\operatorname{ARE}(\mathrm{mean},\mathrm{median}) =\frac{V_{\mathrm{median}}}{V_{\mathrm{mean}}} =\frac{9\pi/2}{9} =\frac{\pi}{2} \approx 1.571.

Thus the mean has the smaller first-order asymptotic variance under this model. Reversing the comparison gives $\operatorname{ARE}(\mathrm{median},\mathrm{mean})=2/\pi\approx0.637$.

Laplace Location Model

Now suppose the observations have a Laplace distribution centered at $\theta$, with scale $b>0$ and density

f(x)=\frac{1}{2b}\exp\left(-\frac{|x-\theta|}{b}\right).

This distribution is symmetric about $\theta$, so its mean and median are both $\theta$. Its density at the median is $1/(2b)$, and hence the median’s asymptotic variance is $1/(4\{1/(2b)\}^2)=b^2$. The mean’s asymptotic variance is the distribution’s variance, $2b^2$. For example, the latter follows by writing $Z=X-\theta$ and evaluating

E[Z^2] =\frac{1}{b}\int_0^\infty x^2 e^{-x/b}\,dx =\frac{1}{b}\,b^3\int_0^\infty u^2e^{-u}\,du =2b^2,

where $u=x/b$ and $\int_0^\infty u^2e^{-u}\,du=2$, by two integrations by parts.

Worked Example: Comparing Mean and Median Under a Laplace Model

For a Laplace location model with $b=2$, find the two asymptotic variances and compare the median with the mean.

The mean’s asymptotic variance is $2b^2=2(2^2)=8$. The density at the median is $1/(2b)=1/4$, so the median’s asymptotic variance is

V_{\mathrm{median}} =\frac{1}{4(1/4)^2} =\frac{1}{1/4} =4=b^2.

Therefore,

\operatorname{ARE}(\mathrm{median},\mathrm{mean}) =\frac{V_{\mathrm{mean}}}{V_{\mathrm{median}}} =\frac{8}{4}=2.

The median has half the first-order asymptotic variance of the mean in this model. Equivalently, the mean’s relative efficiency compared with the median is $4/8=1/2$.

What the Comparison Does—and Does Not—Say

Model	Mean asymptotic variance	Median asymptotic variance	More efficient under the model
Normal $N(\theta,\sigma^2)$	$\sigma^2$	$\pi\sigma^2/2$	Mean
Laplace with scale $b$	$2b^2$	$b^2$	Median

These comparisons are dimensionless: changing the scale changes both variances by the same squared scale factor, which cancels in their ratio. Under normality, the mean’s relative efficiency compared with the median is $\pi/2$; under the Laplace model, the median’s relative efficiency compared with the mean is $2$. The reversal reflects the different shapes of the distributions, not a contradiction.

The asymptotic variance of the mean depends on the second moment. The median’s depends on the amount of probability density at the center. A distribution can therefore favor one estimator over the other. These are comparisons under specified models; they do not establish that one estimator is best under every possible distribution.

Worked Example: Keeping Scale and Efficiency Separate

Suppose one study uses a normal model with $\sigma=2$, while another uses a Laplace model with $b=1.5$. Calculate the two variances in each study and identify the within-model comparison.

For the normal model, $V_{\mathrm{mean}}=\sigma^2=4$, and $V_{\mathrm{median}}=\pi\sigma^2/2=2\pi$, approximately $6.283$. The mean is more efficient, with relative efficiency $(2\pi)/4=\pi/2\approx1.571$.

For the Laplace model, $V_{\mathrm{mean}}=2b^2=2(2.25)=4.5$, and $V_{\mathrm{median}}=b^2=2.25$. The median is more efficient, with relative efficiency $4.5/2.25=2$.

It would be misleading to infer which model has the more efficient estimator by comparing $4$ with $2.25$ across studies: the models have different scales. Relative efficiency compares estimators of the same location parameter under the same distributional model.

Common Pitfall

A frequent mistake is to treat the median’s variance formula as an exact finite-sample variance, or to use the density from the wrong model. The formula $1/\{4f(\theta)^2\}$ is the variance in the limiting distribution of $\sqrt n(M_n-\theta)$; it is an asymptotic variance, not a general exact formula for $n\operatorname{Var}(M_n)$. Also, $f(\theta)$ must be evaluated under the assumed distribution. Using the normal density to assess a Laplace model, for example, produces the wrong comparison.

A second mistake is to state simply that the median is more robust and conclude that it must be more efficient. Robustness to extreme values and first-order asymptotic efficiency are different properties. The calculations here show that the mean is more efficient under the normal model, while the median is more efficient under the Laplace model.

Key takeaway: For a regular location model, the mean’s first-order asymptotic variance is the observation variance, while the median’s is $1/\{4f(\theta)^2\}$. Applying these formulas gives a normal-model advantage for the mean and a Laplace-model advantage for the median.

Check Your Understanding

Use the formulas and the relative-efficiency convention in this tutorial.

For $X_i\sim N(\theta,4)$, find the asymptotic variances of the mean and median.
For a Laplace location model with $b=3$, calculate the median’s asymptotic variance and the mean’s asymptotic variance.
What regularity condition on the density at the population median is used in the median limit result, and why does the density appear in the variance?
If the mean’s asymptotic variance is $5$ and the median’s is $8$, what is $\operatorname{ARE}(\mathrm{mean},\mathrm{median})$, and which estimator is more efficient?
Why does the normal-model comparison not establish that the mean is more efficient for every location distribution?

Model	Mean asymptotic variance	Median asymptotic variance	More efficient under the model
Normal \(N(\theta,\sigma^2)\)	\(\sigma^2\)	\(\pi\sigma^2/2\)	Mean
Laplace with scale \(b\)	\(2b^2\)	\(b^2\)	Median