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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?

Josh Brown Kramer
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    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 06 '15 at 04:03
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    By "converge faster to 0" do you mean "which has the smaller asymptotic variance" or something else? – Glen_b Feb 06 '15 at 04:37
  • @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 06 '15 at 05:00
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    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 Feb 06 '15 at 05:08

1 Answers1

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The mean and median are the same, in this particular case. It is known that the median is 64% efficient as the mean, so the mean is faster to converge. I can write more details but wikipedia deals with your question exactly.

Yair Daon
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    Do you have a citation? – Josh Brown Kramer Feb 06 '15 at 04:33
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    Laplace, P.S.de (1818) Deuxième supplément à la Théorie Analytique des Probabilités, Paris, Courcier -- Laplace gives the asymptotic distribution for both mean and median. See also the section on the variance of the median on Wikipedia – Glen_b Feb 06 '15 at 04:42
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    @Glen_b: (+1) the ultimate reference!!! – Xi'an Feb 06 '15 at 07:42
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    @Glen_b yeah that was an epic response, I laughed pretty hard. Thanks for that! – user541686 Feb 06 '15 at 09:58
  • @xi'an did you mean to write that the mean and median are the same quantity? – Yair Daon Feb 11 '15 at 04:38
  • My reference is probably not as impressive as the one @Glen_b provided. I remember that it is in Bulmer's Principles of Statistics. But I think for your case the refrence I added in my answer suffices. – Yair Daon Feb 11 '15 at 04:44
  • There is a confusion there: the theoretical mean and variances of the normal distribution are the same quantity. However, the empirical mean and median of a normal sample are not the same. – Xi'an Feb 12 '15 at 07:12