The efficiency cost of averaging only half your data

Question

To save computation, an analyst estimates a Normal mean $\mu$ by averaging only every other observation, using $n/2$ of the $n$ available data points:

$\hat\mu_{\text{half}} = \frac{2}{n}\sum_{i \text{ odd}} X_i.$

The full-data competitor is the ordinary sample mean $\bar X$ over all $n$ points.

What is the asymptotic relative efficiency of the half-sample estimator versus the full sample mean?

Accepted Answer

Both are unbiased. Averaging any subset of the i.i.d. observations gives \mathbb{E}[\hat\mu{	ext{half}}] = \mu = \mathbb{E}[\bar X], so this is purely a variance (efficiency) question. Compare variances. The full mean averages n points: \operatorname{Var}(\bar X) = \frac{\sigma^2}{n}. The half-sample estimator averages n/2 points: \operatorname{Var}(\hat\mu{	ext{half}}) = \frac{\sigma^2}{n/2} = \frac{2\sigma^2}{n}. The efficiency. 	ext{ARE} = \frac{\operatorname{Var}(\bar X)}{\operatorname{Var}(\hat\mu{	ext{half}})} = \frac{\sigma^2/n}{2\sigma^2/n} = \frac{1}{2}. The shortcut is 50\% effi…

The efficiency cost of averaging only half your data

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