Distribution of value closest to 0 Consider $K$ independent Laplace variables $X_i$ ($1 \leq i \leq K$) with mean 0 and scale $\lambda$. Let $X′$ be the variable taking the value of the Laplace variable whose absolute value is the minimum among all $X_i$'s. Due to the randomness of $X_i$'s, $X′$ may not always equal a fixed $X_i$. I would like to know what the CDF of $X′$ is. Does it also follow Laplace distribution? How to prove or disprove that? Many thanks!
 A: Consider a set of random variables $Y_i\stackrel{\text{iid}}{\sim}\text{Exp}(\lambda)$ (this is the scale parameterization not the rate parameterization, so it has scale parameter $\lambda$). Let $B_i\stackrel{\text{iid}}{\sim}\text{Bernoulli}(\frac12)$, where the $Y$'s and $B$'s are all mutually independent. 
Let $Z_i=2B_i-1$, so the $Z$'s are random $+1$'s and $-1$'s ("random signs").
Let $X_i=Z_iY_i$. Clearly $X_i\stackrel{\text{iid}}{\sim}\text{Laplace}(0,\lambda)$.
Let $Y_{(1)}$ be the first order statistic of $Y_1,Y_2,...,Y_n$.


*

*It's a simple matter to show that $Y_{(1)}\sim\text{Exp}(\lambda/n)$.
$P(Y_{(1)}>y)=P(Y_{1}>y,Y_{2}>y,...,Y_{n}>y)$
$\qquad=\exp(-y/\lambda)\cdot \exp(-y/\lambda)\cdot....\cdot\exp(-y/\lambda)$
$\qquad=\exp(-ny/\lambda)$
Hence $Y_{(1)}\sim\text{Exp}(\lambda/n)$.

*Let $W_1=X_k$ be the particular Laplace observation corresponding to $Y_{(1)}$; that is $Y_{(1)}=Y_k$.
Then $W_1=Z_k\cdot Y_{(1)}$. Since the $Z$'s are independent of the $Y$'s, this attaches a random sign to $Y_{(1)}$, making $W_1\sim\text{Laplace}(0,\lambda/n)$.
