# Expected (maximum minus minimum) of Laplacian random variables

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.

• A similar question was asked before (and answered by Iosif Pinelis then too): mathoverflow.net/questions/378647/…
– user44143
Commented Apr 16, 2023 at 18:40
• @MattF. : Interesting. :-) I certainly forgot about that. How did you find it? Commented Apr 16, 2023 at 18:45
• I figured this was written up somewhere so I googled it.
– user44143
Commented Apr 16, 2023 at 20:03
• Yeah, I have also noticed that website before, but they focus on the lower bound of our question. While we are searching for the upper bound! Anyway, thanks again Iosif Pinelis again, haha! Commented Apr 17, 2023 at 6:28

$$\newcommand\la\lambda$$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 (thanks to Matt F. for reminding us of it), formulas \eqref{10}, \eqref{20}, \eqref{30}, and the "rescaling" remark that $$\begin{equation*} l_n:=\gamma+\ln(n/2)-\frac1{n{2^n}}<\frac{EM_n}{2\la} where $$\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.

• Thanks for your answer! The closed form of $EM_n$ and the convergent upper bound $2\lambda\log n$ seem both correct. However, I am still confused how to derive the convergent upper bound $2\lambda\log n$? Could you please provide the proof of that? Commented Apr 17, 2023 at 3:06
• @white : At the end of the answer, I have now added a remark on asymptotically exact lower and upper bounds on $EM_n$. Commented Apr 17, 2023 at 14:17
• @white : Do you have a further response to the updated answer? Commented Apr 23, 2023 at 21:21
• Yeah, I think the response is quite clear! Thanks for your answer! Commented May 12, 2023 at 8:13