Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The question is as follows: If $X_i,\ldots,X_n$ are iid $N(\theta, \sigma^2)$, compare the asymptotic variance of the sample mean and sample median.


Would the asymptotic variance for the mean be the limit as $n$ approaches infinity of the quantity $\sigma^2/n$? Could we use the Cramer Rao Information bound as an approximation to the asymptotic variance? Also, does it matter what the parent distribution is? Wouldn't the asymptotic variance of the mean always be $\sigma^2/n$ due to the central limit theorem?

share|cite|improve this question
Ordering? Then not answering. // You might realize that the limit as n approach(es) infinity of the quantity sigma^2/n has another, simpler, description, which makes your question rather moot. – Did Nov 29 '11 at 8:03
up vote 2 down vote accepted

When we have an estimator $T$ of some parameter $\theta$, when one defines the so-called asymptotic variance $V$ what is actually meant is the number such that $\sqrt n (T - \theta) \stackrel{d}{\longrightarrow} N(0, V)$ (more generally one can replace $\sqrt n$ with a general $k_n$ such that you are converging in distribution to a normal; see e.g. Casella and Berger). Note that this lets us talk about the "asymptotic variance" of estimators that don't even necessarily have an have a finite first or second moment.

For the normal distribution, the asymptotic variance of $\sqrt n (\bar X - \mu)$ under the above definition is $\sigma^2$, so a large sample approximation of the variance of $\bar X$ is $\sigma^2 / n$ - this happens to also the exact variance of $\bar X$. If $M_n$ is the sample median, we can show that $$\sqrt n (M_n - \mu) \stackrel{d}{\longrightarrow} N\left(0,\frac{1}{4f(\mu)^2}\right) \qquad (\dagger)$$

and so the asymptotic variance of $\sqrt n (M_n - \mu)$ is $\frac{1}{4 f(\mu)^2}$ - $f$ here is the density of the underlying distribution. This result holds more generally than just for normal iid random variables, so I'm writing it in the more general form with the density $f$ unspecified.

The usual comparison between two asymptotic variances is the asymptotic relative efficiency (ARE), which is the ratio of the asymptotic variance of $\bar X$ to the asymptotic variance of $M_n$, which is $$4 \sigma^2 f(\mu)^2 = \frac {4 \sigma ^ 2}{2\pi \sigma^2} = \frac 2 \pi \approx .64$$ so for the normal distribution $\bar X$ is doing better in terms of asymptotic variance than $M_n$. This doesn't hold for all distributions.

I'm feeling a bit lazy to write out how one arrives at $(\dagger)$, so I'll leave that to someone else if they want to slog through that.

As far as linking this to the Cramer-Rao lower bound: estimators that achieve the Cramer-Rao lower bound asymptotically, which $\bar X$ does in this case, are called efficient. Estimators that acheive the Cramer-Rao lower bound are optimal in some sense - in terms of asymptotic variance you can actually do better than the Cramer-Rao lower bound, but the estimators that beat the Cramer-Rao lower bound have some subtle problems associated with them, so typically if we have an estimator that achieves the CR lower bound we are satisfied with it so far as the large sample properties of the estimator are concerned.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.