Expected maximum of Laplacian random variables Let $X={x_1, x_2,...,x_n}$ be Laplacian iid random variables with mean zero and scale $b$. Let $Y = \max(X)$.
Is there a closed-form expression for the expectation $E(Y)$? If not, is there any non-trivial lower bound that depends on $n$?
 A: The cumulative distribution $F(y)$ of $Y$ is the product of the cumulative distributions $f(x)$ of $X$, because
$$F(y)=P[(X_1<y)\cap(X_2<y)\cap\cdots\cap(X_n<y)]=\{f(y)\}^n.$$
Hence
$$F(y)=2^{-n}e^{(n/b)(y-\mu)}\;\;\text{if}\;\;y<\mu,$$
$$F(y)=\left[1-\tfrac{1}{2}e^{-(y-\mu)/b}\right]^n\;\;\text{if}\;\;y>\mu.$$
The expectation value follows upon integration,
$$\mathbb{E}[Y]=\mu+\int_{\mu}^\infty [1-F(y)]\,dy-\int_{-\infty}^\mu F(y)\,dy=\mu+bc_n.$$
(I allow for nonzero mean $\mu$.)
I don't have a closed form expression for the coefficients $c_n$, the values for $n=1,2,3,\ldots 10$ are

$$0,\frac{3}{4},\frac{9}{8},\frac{133}{96},\frac{305}{192},\frac{281}{160},\frac{3647}{1920},\frac{217687}{107520},\frac{153093}{71680},\frac{1442363}{645120}.$$

A: This is to complement Carlo Beenakker's answer by providing a lower bound on the integral in question:
$$\begin{aligned}J_n&:=\int_0^\infty[1-(1-e^{-y}/2)^n]\,dy \\
&>\int_0^\infty[1-\exp\{-ne^{-y}/2\}]\,dy \\
&=\gamma+\ln(n/2)+\int_{n/2}^\infty e^{-t}\frac{dt}t \\
&>\gamma+\ln(n/2)=:L_n,
\end{aligned}\tag{1}$$
where $\gamma=0.577\ldots$ is the Euler gamma; details on the evaluation of the second integral in the above display will be given at the end of this answer.

The lower bound $L_n$ on the integral $J_n$ is asymptotically exact, as we have the following upper bound on $J_n$:
$$\begin{aligned}J_n&=\int_0^1[1-(1-t/2)^n]\,\frac{dt}t \\
&<\int_0^1[1-\max(0,1-nt/2)]\,\frac{dt}t \\
&=\int_0^{2/n}[1-(1-nt/2)]\,\frac{dt}t+\int_{2/n}^1\frac{dt}t \\ 
&=1+\ln(n/2)\sim L_n.
\end{aligned}$$

Here is a graph of the ratio $J_n/L_n$:


Details on the evaluation of the second integral in display (1): With $m:=n/2$, the integral in question is
$$\begin{aligned}
K_m:=\int_0^\infty[1-\exp\{-me^{-y}\}]\,dy
=\int_0^1[1-\exp\{-m t\}]\,\frac{dt}t 
=\int_0^m[1-e^{-u}]\,\frac{du}u.  
\end{aligned}$$
By the fourth display on p. 31,
$$\begin{aligned}\gamma&=\int_0^\infty\Big(\frac1{1+u}-e^{-u}\Big)\frac{du}u \\
&=\int_0^m\Big(\frac1{1+u}-e^{-u}\Big)\frac{du}u
+\int_m^\infty\frac{du}{(1+u)u}
-\int_m^\infty\frac{e^{-u}du}u. 
\end{aligned}$$
So,
$$\begin{aligned}
&-K_m+\gamma+\ln m+\int_m^\infty e^{-t}\frac{dt}t \\
&=\int_0^m\Big(\frac1{1+u}-1\Big)\frac{du}u
+\int_m^\infty\frac{du}{(1+u)u}
+\ln m=0,
\end{aligned}$$
as claimed.
