# Distribution of minimum absolute value

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!

• You should really write all the things you've done into your question so that people know you've put some effort in and can get a better idea of where you're stuck. Jul 7, 2015 at 14:33

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 $x \in \mathbb{R}^+$.

• you seem to have sidestepped the absolute value issue, could you elaborate on that Jul 9, 2015 at 1:37
• This is exactly what I am looking for. It perfectly satisfies my curiosity. Thanks! Jul 9, 2015 at 21:39
• Glad to hear it! You could always upvote the answer and/or select it, too...
– josh
Jul 10, 2015 at 0:49

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).

• Thanks a lot, Michael. $L(0, \frac{\lambda}{K})$ is also our speculation, and is the conclusion we are expecting. We will try the case of $K=2$. Jul 7, 2015 at 3:24
• I worked out the K=2 case finally, and your (our) intuition was right. It was messy; if you're stuck on I could offer some pointers. Jul 7, 2015 at 14:41