Suppose there are $n$ IID random variables denoted as $X=(X\_1,\dots, X\_n)$, they follow Laplace distribution with parameter $\lambda$, denoted as $Lap(\lambda)$. That is,
$$f(x)=\frac{1}{2\lambda}\exp (-\frac{|x|}{\lambda})$$
$$
F(x)= \begin{cases}\frac{1}{2} \exp \left(\frac{x}{\lambda}\right) & \text { if } x<0 \\ 1-\frac{1}{2} \exp \left(-\frac{x}{\lambda}\right) & \text { if } x \geq 0\end{cases}
$$


Let $M=\max(X)-\min(X)$, how can we compute the expectation of $M$, that is $E(M)$, or could we provide an upper bound of $E(M)$?


I have tried to compute the cdf and pdf of $M\_1=\max(X), M\_2=\min(X)$. Then we have
$$F\_{M\_1}(x)=(F(x))^n, F\_{M\_2}=1-(1-F(x))^n$$


However, the expectation computation seems difficult, and I did not figure it out.


Besides, I found this problem is related to *Extreme Value Theory*, but I am not a statistic student and I don't know much about that.



Expected (maximum minus minimum) of Laplacian random variables
$\newcommand\la\lambda$By rescaling, it is enough to consider the case $\la=1$ (multiplying the resulting expectation by $\la$ at the end). By symmetry, $Y:=\max X$ and $-\min X$ are identically distributed. So, for $M\_n:=M$ we have
\begin{equation\*}
 EM\_n=2EY. \tag{10}\label{10}
\end{equation\*}
Next, $Y=\max(0,Y)-\max(0,-Y)$ and hence
\begin{equation\*}
 EY=I\_1-I\_2, \tag{20}\label{20}
\end{equation\*}
where
\begin{equation\*}
\begin{aligned}
 I\_1&:=E\max(0,Y)=\int\_0^\infty dx\,P(Y>x)=\int\_0^\infty dx\,(1-(1-F(x)^n),\\ 
 I\_2&:=E\max(0,-Y)=\int\_0^\infty dx\,P(-Y>x)=\int\_0^\infty dx\,F(-x)^n. 
\end{aligned}
\end{equation\*}
Here I use a standard trick:
\begin{equation\*}
 \int\_0^\infty dx\,P(Y>x)=\int\_0^\infty dx\,E\,1(Y>x) \\ 
 =E\int\_0^\infty dx\,1(Y>x)
 =E\int\_0^{\max(0,Y)} dx=E\max(0,Y). 
\end{equation\*}


Further,
\begin{equation\*}
 I\_2=\int\_0^\infty dx\,(e^{-x}/2)^n=\frac{2^{-n}}n \tag{30}\label{30}
\end{equation\*}
and
\begin{equation\*}
\begin{aligned}
 I\_1&=\int\_0^\infty dx\,(1-(1-e^{-x}/2)^n) \\ 
 &=-\int\_0^\infty dx\,\sum\_{j=1}^n\binom nj(-e^{-x}/2)^j \\ 
 &=-\sum\_{j=1}^n\binom nj\frac{(-1/2)^j}j 
 =\frac{n}{2} \, \_3F\_2\left(1,1,1-n;2,2;\frac{1}{2}\right), 
\end{aligned}
\end{equation\*}
where $\_3F\_2$ is the hypergeometric function.


Thus, for $\la=1$,
\begin{equation\*}
 EM\_n=n \, \_3F\_2\Big(1,1,1-n;2,2;\frac{1}{2}\Big)-\frac1{n2^{n-1}}
\end{equation\*}
and
\begin{equation\*}
 EM\_n=\la\Big(n \, \_3F\_2\Big(1,1,1-n;2,2;\frac{1}{2}\Big)-\frac1{n2^{n-1}}\Big)
\end{equation\*}
for any real $\la>0$.




---


It follows from the [previous answer](https://mathoverflow.net/a/378667/36721) (thanks to Matt F. for reminding us of it), formulas \eqref{10}, \eqref{20}, \eqref{30}, and the "rescaling" remark that
\begin{equation\*}
 l\_n:=\gamma+\ln(n/2)-\frac1{n{2^n}}<\frac{EM\_n}{2\la}<u\_n:=1+\ln(n/2)-\frac1{n{2^n}}, 
\end{equation\*}
where $\gamma=0.577\ldots$ is the Euler gamma.


It is clear now that  

\begin{equation\*}
 \frac{EM\_n}{2\la}\sim l\_n\sim u\_n\sim\ln(n/2)\sim\ln n,
\end{equation\*}
so that the upper bound $u\_n$ and the lower bound $l\_n$ on $\frac{EM\_n}{2\la}$ are each asymptotically exact.


