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When we draw $n$ samples of Laplace-distributed random variable such that $n=2k+1$ and the location parameter is zero, the median $x$ (or the $k$-th order statistic) has the following p.d.f.:


where the p.d.f. of the underlying Laplace distribution is given as $f(y)=\frac{1}{2b}e^{-|y|/b}$.

The formula for p.d.f. of the median stems from the usual method of characterizing the distributions of order statistics and is found as equation (2.5.10) in Kotz's volume on Laplace distribution. There is another formula for the case when $n$ is even, but we shall not be concerned with it for now.

I am interested in the variance of the sample median. Since $f_m(x)$ is symmetric about $x=0$, I can get rid of the absolute value and write it as follows:


What I need is to reduce the following definite integral to a more manageable form:


For fixed $k$ this is fairly easy integral to solve. However, I am interested in asymptotics of variance $\sigma_m^2$ as $n\rightarrow\infty$ and $b$ is a linear function of $\sqrt{n}$, and thus need a solution for an arbitrary $k$.

Any ideas?

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You've tried binomial expansion already? – J. M. Aug 18 '12 at 3:30
Just to understand the question correctly: $b= \alpha \sqrt{n}$? – Fabian Aug 18 '12 at 3:36
@J.M. Binomial expansion didn't come to mind for some reason. Great idea, will try it in the morning. – M.B.M. Aug 18 '12 at 4:13
@Fabian Yes, that's correct (it may be more complicated in the problem I am actually trying to solve, but for now I can just assume $b=\alpha\sqrt{n}$) – M.B.M. Aug 18 '12 at 4:15
up vote 2 down vote accepted

Ok, I think I understand now. You want to find $$\sigma^2=\frac{n!}{(k!)^2}2^{-n}\frac{1}{b}\int_{0}^{\infty}\!dx\,x^2e^{-(k+1)x/b}(2-e^{-x/b})^k$$ where $n=2k+1$ and $b \sim \sqrt{n}$ for $n,k \to \infty$. Let us first use the change of variable $y=x/b$. We then find $$\sigma^2= b^2\frac{n!}{(k!)^2}2^{-n}\int_{0}^{\infty}\!dy\,y^2e^{-(k+1)y}(2-e^{-y})^k.$$

We will use the method of steepest decent to figure out the asymtptotic behavior of $\sigma^2$ for $k,n \to \infty.$ Let us first rewrite the integral as $$ \int_0^\infty\!dy\,y^2e^{-(k+1)y}(2-e^{-y})^k = \int_0^\infty\!dy\, y^2 e^{-y} e^{- k f(y)}, \qquad f(y) =y-\log(2-e^{-y}). $$ The function $f(y)$ assumes its maximum at $y=0$. Thus, the integral is asymptotically dominated for values of $y$ close to $0$. To this end, we expand $f(y) = y^2 +O(y^3)$ and we have $$\int_0^\infty\!dy\, y^2 e^{-y} e^{- k f(y)} \sim \int_0^\infty\!dy\, (y^2 -y^3 +O(y^4)) e^{-k y^2} = \frac{\sqrt\pi}{4 k^{3/2}} - \frac{1}{2k^2} +O(k^{-5/2}). $$

Additionally, we use that (see central binomial coefficient) $$\frac{n!}{(k!)^2} \sim2 \binom{2k}{k} \sim 2 \frac{4^k}{\sqrt{\pi k}} = \frac{2^n}{\sqrt{\pi k}}.$$

Together, we have (with $b \sim \alpha \sqrt{n} \sim 2 \alpha \sqrt{k} $) $$\sigma^2 \sim \underbrace{4 \alpha^2 k}_{b^2} \frac{2^n}{\sqrt{\pi k}} 2^{-n} \frac{\sqrt\pi}{4 k^{3/2}} = \frac{\alpha^2}{k}. $$

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By the scaling arguments, it should be proportional to $b^2$. – Sasha Aug 18 '12 at 3:49
@Sasha: you are right (for the variance), but there is a $1/b$ which the OP pulled out of his integral -> see his post. I added the post to make it clear that I refer to the last integral in his post. – Fabian Aug 18 '12 at 3:50
@Fabian Right, unfortunately,that dependence on $k$ is pretty important. Maybe J.M.'s suggestion of using binomial expansion will work to solve that. – M.B.M. Aug 18 '12 at 4:17
@M.B.M.: I didn't understand in the beginning that $k$ depends on $n$ as well. I changed my answer. I hope it answers your question. – Fabian Aug 18 '12 at 8:19
Looking now a bit more careful at where the integral came from, I believe I just made a pretty long calculation to confirm the central limit theorem in the end... – Fabian Aug 18 '12 at 8:22

Let's first set $b=1$, and recover it later using scaling. For $X \sim \mathrm{Lap}(0,1)$ $$ f_{X_{k:2k+1}}(x) = \frac{k+1}{2^{2k+1}} \binom{2k+1}{k+1} \mathrm{e}^{-(k+1)|x|} \left(2-\mathrm{e}^{-|x|}\right)^k = \frac{k+1}{2^{2k+1}} \binom{2k+1}{k+1} \sum_{m=0}^k \binom{k}{m} 2^{k-m} (-1)^m \exp\left(-(m+k+1)|x|\right) $$ Thus $$ \mathbb{Var}\left(X_{k:2k+1}\right) = \frac{k+1}{2^{2k+1}} \binom{2k+1}{k+1} \sum_{m=0}^k \binom{k}{m} 2^{k-m} (-1)^m \frac{4}{(k+1+m)^3} = \\ \binom{2k+1}{k+1} \frac{2^{1-k}}{(k+1)^2} \cdot {}_4F_3\left( \left.\begin{array}{cccc} -k & k+1 & k+1 & k+1 \\ & k+2 & k+2 & k+2 \end{array} \right| \frac{1}{2} \right) $$ enter image description here

The sequence above satisfies an inhomogeneous recurrence equation of rank $3$: enter image description here

This recurrence equation allows to determine the large $k$ asymptotic behavior. There are fewer undetermined coefficients, because $X_{k:2k+1}$ converges to a degenerate random variable for large $k$, meaning that the variance must vanish in this limit: $$ \mathbb{Var}\left(X_{k:2k+1}\right) = \frac{c_1}{k}\left(1+ \mathcal{o}\left(k^{-1}\right) \right) + \frac{c_2}{k^{3/2}}\left(1+ \mathcal{o}\left(k^{-1}\right) \right) $$ The following plot indicates that $c_1>0$: enter image description here

Sequence acceleration suggests that value of $c_1$ is close to $\frac{1}{2}$: enter image description here

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