What Relative Efficiency Compares
Two estimators can both be consistent and asymptotically normal while having different levels of precision. Their asymptotic relative efficiency summarizes that difference using their limiting variances. The comparison is useful only when its convention is stated: one common definition reports the variance of the comparator divided by the variance of the estimator being evaluated, while another reports the reverse.
Throughout, suppose \(T_{A,n}\) and \(T_{B,n}\) estimate the same scalar parameter \(\theta\). Assume, for the comparisons below, that \(\sqrt n(T_{A,n}-\theta)\xrightarrow{d}N(0,V_A)\) and \(\sqrt n(T_{B,n}-\theta)\xrightarrow{d}N(0,V_B)\), where \(0<V_A,V_B<\infty\). The quantities \(V_A\) and \(V_B\) are called the estimators’ asymptotic variances under this normalization. In particular, they are the variances of the corresponding asymptotic normal distributions.
This definition compares the estimators under a common sample-size normalization and for the same target. A value of \(\operatorname{ARE}(A:B)=2\), for example, means that \(A\)'s asymptotic variance constant is half of \(B\)'s. It does not mean that \(A\)'s actual variance is necessarily half of \(B\)'s at every finite sample size.
How the Convention Determines the Interpretation
The reciprocal conventions encode the same comparison. The efficiency ratio is larger than one when \(A\) has the smaller limiting variance; the asymptotic-variance ratio is then less than one. Stating only “the relative efficiency is two” is therefore incomplete unless the numerator and denominator are specified.
To verify the result, substitute the definition: \[ \operatorname{ARE}(B:A)=\frac{V_A}{V_B} =\left(\frac{V_B}{V_A}\right)^{-1} =\frac{1}{\operatorname{ARE}(A:B)}. \] The asymptotic-variance ratio is \(V_A/V_B\), which is the same reciprocal. Thus, the direction of a comparison matters; switching the estimators reverses the ratio. \(\square\)
A useful interpretation comes from the asymptotic normal approximations. At sample size \(n\), these approximations have variances \(V_A/n\) and \(V_B/n\). If sample sizes can differ, equating the two approximate variances gives \(V_A/n_A=V_B/n_B\), and hence \(n_B/n_A=V_B/V_A=\operatorname{ARE}(A:B)\). So, at first order, this ratio is the number of observations required by \(B\) divided by the number required by \(A\) to attain the same approximate variance. This is an interpretation of the normal approximations, not a claim that actual finite-sample variances equal those expressions.
Worked Example: Full-Sample and Half-Sample Estimators of a Normal Mean
Let \(X_1,\ldots,X_n\) be independent observations from \(N(\mu,\sigma^2)\), with \(\sigma^2\) known, and let \(n\) be even. Compare the full-sample mean \(T_{A,n}=\bar X_n\) with the mean of the first half, \(T_{B,n}=\bar X_{n/2}^{(1)}\).
The full-sample mean has distribution \(N(\mu,\sigma^2/n)\). Therefore \(\sqrt n(\bar X_n-\mu)\sim N(0,\sigma^2)\), giving \(V_A=\sigma^2\). The first-half mean is based on \(n/2\) observations, so its distribution is \(N(\mu,\sigma^2/(n/2))=N(\mu,2\sigma^2/n)\). Thus \(\sqrt n(T_{B,n}-\mu)\sim N(0,2\sigma^2)\), giving \(V_B=2\sigma^2\).
Under the efficiency convention, \(\operatorname{ARE}(A:B)=V_B/V_A=2\sigma^2/\sigma^2=2\). The full-sample mean has half the asymptotic variance constant of the half-sample mean. Under the asymptotic-variance-ratio convention, \(V_A/V_B=1/2\). Here these are also exact finite-sample variance comparisons, because the distributions and variances were calculated exactly.
Limiting Variance Is Not Automatically Actual Variance
A limiting distribution describes probabilities of values in the limit; it does not, by itself, control the contribution of rare, very large values to a second moment. Consequently, from \(\sqrt n(T_n-\theta)\xrightarrow{d}N(0,V)\) alone, one cannot conclude that \(\operatorname{Var}(T_n)\) is approximately \(V/n\). The quantity \(V/n\) is the variance of the asymptotic normal approximation. Additional control of second moments is needed to infer convergence of actual variances.
Worked Example: A Normal Limit With No Corresponding Variance Limit
Let \(Y_n\) have a mixture distribution: with probability \(1-1/n\), draw \(Y_n\) from \(N(0,1)\); with probability \(1/(2n)\) each, set \(Y_n=n\) or \(Y_n=-n\). The probability assigned to the two exceptional values tends to zero. For any bounded continuous function \(g\), \(E[g(Y_n)]=(1-1/n)E[g(Z)]+(g(n)+g(-n))/(2n)\), where \(Z\sim N(0,1)\). The second term tends to zero by boundedness, so \(Y_n\xrightarrow{d}N(0,1)\).
The distribution is symmetric, so \(E[Y_n]=0\). Its variance is \(\operatorname{Var}(Y_n)=E[Y_n^2]=(1-1/n)\cdot 1+(1/(2n))n^2+(1/(2n))n^2=n+1-1/n\). Now set \(T_n=\theta+Y_n/\sqrt n\). Then \(\sqrt n(T_n-\theta)=Y_n\xrightarrow{d}N(0,1)\), but \(\operatorname{Var}(T_n)=(n+1-1/n)/n\to 1\), not \(1/n\). The rare values disappear in the limiting distribution but still dominate the second moment.
To prove this, convergence in distribution gives convergence of expectations for bounded continuous functions. For fixed \(K>0\), the truncated function \(x\mapsto \min(x^2,K)\) is bounded and continuous, so \(E[\min(Z_n^2,K)]\to E[\min(Z^2,K)]\). Uniform integrability makes the difference between \(E[Z_n^2]\) and \(E[\min(Z_n^2,K)]\) uniformly small for large \(K\). Since \(E[Z^2]=V<\infty\), the corresponding truncated expectation for \(Z\) tends to \(V\) as \(K\to\infty\). Combining these facts yields \(E[Z_n^2]\to V\).
Also, \(E[Z_n]\to E[Z]=0\). Indeed, the same uniform-integrability condition implies uniform integrability of \(|Z_n|\): for \(M>0\), \(E[|Z_n|\mathbf{1}\{|Z_n|>M\}]\le E[Z_n^2\mathbf{1}\{Z_n^2>M^2\}]/M\). Truncate \(x\) to a bounded continuous function, use convergence in distribution for the truncated expectations, and then let the truncation level increase. Therefore \(\operatorname{Var}(Z_n)=E[Z_n^2]-(E[Z_n])^2\to V\). Finally, \(Z_n=\sqrt n(T_n-\theta)\) implies \(\operatorname{Var}(Z_n)=n\operatorname{Var}(T_n)\). \(\square\)
A Location-Model Comparison
A location model has observations whose distribution is shifted by the unknown parameter: \(X_i=\theta+\varepsilon_i\), where the errors have a fixed distribution not depending on \(\theta\). Here is a comparison that uses a specific location distribution and makes the normalization explicit.
Worked Example: Using Half the Observations in a Laplace Location Model
Suppose \(X_i=\theta+\varepsilon_i\), where the independent errors have density \(f(e)=\tfrac12 e^{-|e|}\). This centered Laplace distribution has mean \(0\) and variance \(2\). Compare the full-sample mean \(T_{A,n}=\bar X_n\) with the mean \(T_{B,n}\) of the first \(m=\lfloor n/2\rfloor\) observations.
The full-sample mean is unbiased for \(\theta\), with variance \(2/n\). The central limit theorem gives \(\sqrt n(T_{A,n}-\theta)\xrightarrow{d}N(0,2)\), so \(V_A=2\). The first-\(m\) mean is also unbiased, with variance \(2/m\). Since \(m/n\to 1/2\), its scaled limiting variance is \(n(2/m)=2n/m\to4\), and the central limit theorem gives \(\sqrt n(T_{B,n}-\theta)\xrightarrow{d}N(0,4)\). Thus \(V_B=4\).
The efficiency ratio is \(V_B/V_A=4/2=2\); the asymptotic-variance ratio is \(V_A/V_B=1/2\). In this example the exact variance ratio is also \((2/m)/(2/n)=n/m\to2\). Both estimators estimate the same location parameter, but one discards about half the observations, so its limiting variance constant is twice as large.
Common Pitfalls and Practical Use
An efficiency comparison is a compact summary of first-order precision, but it has a defined scope. It concerns particular estimators, a specified target, a common normalization, and the assumed sampling model. It does not establish that one estimator is preferable for every purpose: finite-sample bias, robustness, computational cost, and model sensitivity may also matter.
- Leaving the ratio direction implicit. State whether the ratio is \(V_B/V_A\) or \(V_A/V_B\), and say which estimator a value above one favors.
- Calling \(V/n\) the actual variance without conditions. It is the variance of the asymptotic normal approximation. To claim \(n\operatorname{Var}(T_n)\to V\), establish suitable moment control, such as uniform integrability of the squared normalized errors.
- Comparing different targets or normalizations. A ratio of limiting variances is meaningful only when the estimators target the same parameter and their limits use the same scaling.
- Ignoring the sample-size interpretation’s assumptions. The equal-precision calculation uses first-order normal approximations; it is not a guarantee of equal finite-sample performance.
When the assumptions are appropriate, asymptotic relative efficiency turns a pair of limiting variance constants into an interpretable comparison. The essential discipline is to label the convention and distinguish asymptotic approximation from actual moment behavior.
Check Your Understanding
Use the efficiency convention \(\operatorname{ARE}(A:B)=V_B/V_A\) unless a question specifies otherwise.
- If \(V_A=3\) and \(V_B=6\), calculate both the efficiency ratio of \(A\) relative to \(B\) and the asymptotic-variance ratio.
- Why does convergence of \(\sqrt n(T_n-\theta)\) to \(N(0,V)\) not by itself establish that \(\operatorname{Var}(T_n)\) is approximately \(V/n\)?
- Suppose two estimators have \(V_A=5\) and \(V_B=2\). Which has the smaller asymptotic variance, and what is \(\operatorname{ARE}(A:B)\)?
- In a location model with error variance \(\tau^2\), an estimator averages the first \(m_n\) observations, where \(m_n/n\to\rho\in(0,1)\). What is its limiting variance under the \(\sqrt n\) normalization?
- What additional type of condition can justify convergence of \(n\operatorname{Var}(T_n)\) to the limiting normal variance?