Asymptotic relative efficiency

Section of Large-sample estimation, part of Statistical inference

Type	Section
Taught in	Large-sample estimation
Part	Statistical inference
Questions using it	1

Suppose T_{1n} and T_{2n} both estimate g(\theta), and

\sqrt n\,\bigl(T_{in}-g(\theta)\bigr)\xrightarrow{d}\operatorname{N}(0,\tau_i^2), \qquad i=1,2.

For large n this is the working approximation T_{in}\approx\operatorname{N}(g(\theta),\tau_i^2/n). The asymptotic relative efficiency of T_2 with respect to T_1 is

\operatorname{ARE}(T_2;T_1)=\frac{\tau_1^2}{\tau_2^2}.

If the ratio is 1/2, then T_2 needs about twice as many observations as T_1 to reach the same limiting variance. The comparison does not require either estimator to be unbiased at finite n. It requires the bias, if any, to be small enough that \sqrt n times the bias tends to zero. Otherwise the limit is not centred at g(\theta).

The sample mean and the sample median of a normal sample are the standard pair. For X_i independent \operatorname{N}(\mu,\sigma^2),

\sqrt n\,(\overline X-\mu)\sim\operatorname{N}(0,\sigma^2)

exactly. The sample median \widetilde X satisfies

\sqrt n\,(\widetilde X-\mu)\xrightarrow{d}\operatorname{N}\Bigl(0,\frac{\sigma^2}{4\phi(0)^2}\Bigr)=\operatorname{N}\Bigl(0,\frac{\pi\sigma^2}{2}\Bigr),

where \phi is the standard normal density, so \phi(0)=1/\sqrt{2\pi} and 1/(4\phi(0)^2)=\pi/2. The asymptotic relative efficiency of the median with respect to the mean is

\operatorname{ARE}(\widetilde X;\overline X)=\frac{\sigma^2}{\pi\sigma^2/2}=\frac{2}{\pi}\approx 0.637.

The median needs about \pi/2\approx 1.57 times as many normal observations to match the variance of the mean. That is the price of a procedure that barely changes if one observation is replaced by an arbitrary outlier. The mean is more efficient in this model and more fragile in a model with heavy tails. The efficiency number is not a reason to discard the median. It is the exchange rate between the two procedures when the normal model is right.

set.seed(10)
n <- 400; R <- 8000
z <- matrix(rnorm(n * R), nrow = n)
v.mean <- var(colMeans(z))
v.med <- var(apply(z, 2, median))
v.mean / v.med          # near 2/pi = 0.637
2 / pi

Asymptotic optimality is not unique in the way a UMVU estimator is unique. If \sqrt n\,(T_n-g(\theta)) has the smallest possible limiting variance and \sqrt n\,R_n\to 0 in probability, then T_n+R_n has the same limit. Adding 1/n to the sample mean does not change the \sqrt n approximation. Finite-sample unbiasedness cares about that 1/n. The asymptotic comparison does not.

Used in

Taught in Large-sample estimation
Questions that use it (1) — How many more observations the median needs
All 2 questions on Large-sample estimation

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