# 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 $$n$$?

• Do you have a response to the answers below? Commented Apr 17, 2023 at 13:51

## 2 Answers

The cumulative distribution $$F(y)$$ of $$Y$$ is the product of the cumulative distributions $$f(x)$$ of $$X$$, because $$F(y)=P[(X_1 Hence $$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 $$0,\frac{3}{4},\frac{9}{8},\frac{133}{96},\frac{305}{192},\frac{281}{160},\frac{3647}{1920},\frac{217687}{107520},\frac{153093}{71680},\frac{1442363}{645120}.$$

• Thanks, but I wonder if a close form solution exist. Commented Dec 11, 2020 at 11:28
• you need the integral $\int_0^\infty [1-(1-\tfrac{1}{2}e^{-y})^n]\,dy$, which I have not been able to evaluate in closed form for arbitrary $n$. But for specific integer $n$ it evaluates readily to a rational. Commented Dec 11, 2020 at 11:40
• Thanks, is there any explicit lower bound that can be specified? Commented Dec 11, 2020 at 11:57

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.