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$X_{n}$ has binomial distribution with $p=q=\frac{1}{2}$ and $n$ trials. How does one show $\lim_{n \to \infty} E(|2X_{n}-n|) = \sqrt{\frac{2n}{\pi}}$? (This is the expected value of the absolute displacement of a symmetric random walk). I have tried using the de Moivre-Laplace theorem but can't figure out how exactly to make the argument.

Edit: trying to do this only using deMoivre-Laplace or the Stirling approximation.

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1 Answer 1

up vote 1 down vote accepted

$X_n$ is the sum of $n$ iid Bernoulli random variables each with mean $1/2$ and standard deviation $1/2$ so by the central limit theorem the distribution of $\frac{X_n - n/2}{\sqrt{n}/2} = \frac{2X_n - n}{\sqrt{n}}$ converges in distribution to a standard normal distribution.

The half-normal distribution made by folding (or taking the absolute value of) a standard normal distribution has a mean of $\sqrt{2/\pi}$, leading to your requested result.

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can you give an argument using only deMoivre-Laplace? –  Red Rover Mar 7 '12 at 8:21
@Red: You can use the de Moivre–Laplace theorem to replace the words "central limit theorem" in my first paragraph. You still need the second. –  Henry Mar 7 '12 at 9:02
Could you elaborate on your argument? Since the absolute value function isn't bounded, I don't see how convergence in distribution immediately gives the desired result. –  Jeff Hussmann Sep 25 '12 at 21:40

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