Distribution of minimum absolute value

Question

Consider $K$ independent Laplace variables $X_k, k=1,\ldots,K$, with mean 0 and scale $\lambda$ (so that their PDF is $f(x)=\frac{1}{2\lambda}e^{-\frac{|x|}{\lambda}}$. Let $Y$ be the variable taking the value of the Laplace variable whose absolute value is the minimum among all $X_k$. That is,

$$Y=X_{k^*}$$ $$k^*=arg\min_k{|X_k|}$$

I would like to know what the CDF of $Y$ is. Does it also follow a Laplace distribution? How might I prove or disprove that? Many thanks!

Answer 1

Consider $n$ independent, identically distributed random variables $X_i$ drawn from the CDF $\Phi(x)$. I.e., $\mathbb{P}(X_i \leq x) = \Phi(x)$. By independence, we compute that the probability that all $n$ of these iid random variables are $\leq x$ is the product that each of them individually is $\leq x$, which equals $\Phi(x)^n$. But this means that the maximum of the set $\{X_i\}$ is $\leq x$ too. $$ \mathbb{P}(\max\{X_i : i = 1,\ldots,n\} \leq x) = \prod_{i=1}^n \mathbb{P}(X_i \leq x) = \Phi(x)^n\,. $$ This is a standard result. Similarly, one can show $$ \mathbb{P}(\min\{X_i : i = 1,\ldots,n\} \leq x) = 1 - \mathbb{P}(\min\{X_i : i = 1,\ldots,n\} > x) = 1 - (1-\Phi(x))^n\,. $$

From here, we can apply the above results to your problem. The density for the absolute value of the Laplace random variable $X_i$ is just the exponential density on the half-line (see further commentary in the edit below): $$ \phi_X(x) = \frac{1}{\lambda}\,e^{-x/\lambda},\quad x \in \mathbb{R}^+\,. $$ It follows that your CDF is $\Phi_X(x) = 1 - \exp(-x/\lambda)$. Consequently, the density for the maximum of $n$ such random variables is $$ \phi_{\max\{X\}}(x) = \frac{d}{dx}\,\Phi_X(x)^n = \frac{d}{dx}\,\big(1 - e^{-x/\lambda})^n = \frac{n}{\lambda}\,e^{-x/\lambda}\,(1 - e^{-x/\lambda})^{n-1}\,, $$ and the density for the minimum is $$ \phi_{\min\{X\}}(x) = \frac{d}{dx}\,\Big(1 - \big(1-\Phi_X(x)\big)^n\Big) = \frac{d}{dx}\,(1 - e^{-nx/\lambda}) = \frac{n}{\lambda}\,e^{-nx/\lambda}\,. $$

This seems to be the result you're expecting.

Edit: Used above is the fact that if $X$ is distributed according to the density $\phi_X(x) = \frac{1}{2\lambda}\,\exp(-|x|/\lambda)$ for $x \in \mathbb{R}$ then the density for $|X|$ is $\phi_{|X|}(x) = \phi_X(x) + \phi_X(-x) = \frac{1}{\lambda}\exp(-x/\lambda)$ for $x \in \mathbb{R}^+$.

Answer 2

Here's some R code to experiment with; it seems reasonable to think the answer is in fact Laplace-distributed:

    k <- 5
    n <- 10000
    mm = 0
    ll = 5
        X <- matrix(rdoublex(k*n, mu=mm, 
                    lambda=ll), ncol=n)
    
    Y <- sapply(1:n, function(y)X[, 
           y][which.min(abs(X[, y]))])
    
    h <- hist(Y, probability=TRUE)
    lines(h$breaks, ddoublex(h$breaks, mu=mm, 
           lambda=ll), lwd=3, col="red")

However, playing around a bit suggests the answer may be $Y \sim L(0,\frac{\lambda}{K})$. Definitely not a proof though.

For a little more insight into how things are working, I suggest you work out exactly what's going on for $K=2$. To that end, here's a plot of the value $Y$ takes for every $(X_1,X_2)$ pair, and whether the implied $Y$ is positive (light green) or negative (orange).