# Which converges faster, mean or median?

If I draw i.i.d. variables from N(0,1), will the mean or the median converge faster? How much faster?

To be more specific, let $x_1, x_2, \ldots$ be a sequence of i.i.d. variables drawn from N(0,1). Define $\bar{x}_n = \frac{1}{n}\sum_{i=1}^n x_i$, and $\tilde{x}_n$ to be the median of $\{x_1, x_2, \ldots x_n\}$. Which converges to 0 faster, $\{\bar{x}_n\}$ or $\{\tilde{x}_n\}$?

For concreteness on what it means to converge faster: does $\lim_{n \to \infty} Var(\bar{X}_n)/Var(\tilde{X}_n)$ exist? If so, what is it?

• Are you asking about the convergence in probability of a point estimate with respect to the population parameter? Or are you asking about the convergence in distribution of a random variable? – Ryan Simmons Feb 6 '15 at 4:03
• By "converge faster to 0" do you mean "which has the smaller asymptotic variance" or something else? – Glen_b -Reinstate Monica Feb 6 '15 at 4:37
• @Glen_b To some extent this is motivated by a real problem : the median is more robust against outliers, so it seems like the sample median should converge more rapidly than the mean as the sample size grows. I don't really know what the best way of expressing the rate of convergence is in this situation. For concreteness, I could ask whether $\lim_{n \to \infty} Var(\bar{X}_n)/Var(\tilde{X}_n)$ exists, and if so, what it is. – Josh Brown Kramer Feb 6 '15 at 5:00
• If the data are truly sampled from a normal distribution, outliers are extremely rare - so rare that the impact on the mean leaves the sample mean as the most efficient estimate of the population mean. But you don't need a vary heavy tail to make the median competitive. That ratio you mention will indeed be about 0.63 – Glen_b -Reinstate Monica Feb 6 '15 at 5:08