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.
For simplicity $\lambda=1.$ Let $N=\mathrm{card}\{k; X_k>0\}\sim B(n,1/2).$ If $x<0$ then $$\Pr(\min X_k>x)=\sum_{j=0}^n\Pr(\min X_k>x; N=n-j)=\frac{1}{2^n}\sum_{j=0}^nC^j_n(1-e^x)^j=(1-\frac{e^x}{2})^n.$$ If $x>0$ then $\Pr(\min X_k>x)=\frac{1}{2^n}e^{-nx}.$ The difficulty is now to compute by parts the integral $$I_n=\int_{-\infty}^0x\frac{d}{dx}(1-\frac{e^x}{2})^n \, dx$$ since the trick is to choose a suitable primitive. Here we go: $$I_n=\left[x(\left(1-\frac{e^x}{2})^n-1\right)\right]_{-\infty}^0-\int_{-\infty}^0\left(1-\frac{e^x}{2})^n-1\right)dx=0+\int_0^{1/2}\frac{1-y^n}{1-y}dy=\int_0^{1/2}(1+y+\cdots+y^{n-1})dy=\sum_{j=1}^n\frac{1}{2^jj}.$$ We are ready for computing $$E(\min X_k)=-I_n-\int_0^{\infty}x\frac{d}{dx}\frac{1}{2^n}e^{-nx}dx=-I_n+\frac{1}{2^nn}=-\sum_{j=1}^{n-1}\frac{1}{2^jj}.$$ (it seems correct for $n=1$!). Finally, by symmetry $\max X_k\sim- \min X_k$ and therefore
$$E(\max X_k-\min X_k)=2\sum_{j=1}^{n-1}\frac{1}{2^jj}.$$Well, I would be interested in a more clever solution. It could use the fact that a Laplace distribution can be seen as the mixing of a standard exponential distribution and its opposite, something like $X=X_+-X_-.$ Does no seem shorter.