Given an iid sample $X_1, \cdots, X_n$ drawn from a sufficiently nice (finite expectation, maybe L2 integrable, etc.) distribution, I want I want to know the population CDF of the mid-range or center point

$$ C(x) = \mathbb P \left[\frac {\max X_i+\min X_i}2 \leq 2x\right] = \mathbb P(\max X_i+\min X_i \leq 2x)$$

or at least some interesting facts and inequalities about it. This isn't a homework problem so nailing an answer for an exact question isn't important. Maybe this is easy with extra assumptions? I will obviously report the results of my continued research here for everyone to share.

These are my goals. Let me share my best work so far.

The density $C'(x)$ has to be the convolution of the densities for $\min X_i, \max X_i$ whose CDF is respectively $$ m(x) = [\mathbb P(X<x)]^n $$

and for symmetric reasons

$$ M(x) = [1-\mathbb P(X<x)]^n $$

What I have in my Batman utility belt is the Laplace transform. (Edit: as per comments below, I passed over the fact that $\min X_i, \max X_i$ are not independent r.v.s so the convolution method doesn't apply. Please refer to the nice accepted answer) It's nice because $\mathcal L(M'(x)) = s\mathcal L(M)(s)$ so

$$ m'*M' = \mathcal L^{-1} \left\{[s\mathcal L(m)(s)] [s\mathcal L(M)(s)]\right\} = \mathcal L^{-1} \{ s^2 \left([\mathcal Lm](s) \cdot [\mathcal LM](s)\right) $$

and um, frantically going over the list of common facts about the Laplace transform, we see that

$$ \mathcal L^{-1} \{s^2 [\mathcal Lf(s)] - f'(0) \} = f''(x)$$

which would be awesome if I was trying to inverse Laplace $s^2(\mathcal Lm + \mathcal LM) = s^2 \mathcal (L+M)$ rather than $s^2[\mathcal L m]\cdot [\mathcal LM]$. Which is close enough to be frustrating.

I'm at a loss by now. Am I even going in the right direction?

  • 1
    $\begingroup$ You might want to look at the Wikipedia page on order statistics, and search questions here that contain the term $\endgroup$
    – Glen_b
    Mar 29, 2019 at 16:21
  • 3
    $\begingroup$ $X_{(1)}$ and $X_{(n)}$ are not independent random variables and so the density of their sum is not the convolution of their marginal densities. $\endgroup$ Mar 29, 2019 at 16:22
  • 3
    $\begingroup$ In the absolutely continuous case, there is a formula for the joint density of any pair of order statistics on Wikipedia $\endgroup$ Mar 29, 2019 at 16:27
  • $\begingroup$ I jumped too fast to the convolution approach. I have to edit the post later to see what's salvageable. $\endgroup$
    – user8948
    Mar 29, 2019 at 17:03
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    $\begingroup$ There seem to be too many "2"s in your displayed equation. // Also, several math stat books on my shelf show how to find the distribution of the range (max - min), and getting the dist'n of the midrange ought to be similar. // I know of no way to invert a Laplace transform (moment-generating function), but there is an inversion formula for characteristic functions. $\endgroup$
    – BruceET
    Mar 29, 2019 at 18:40

1 Answer 1


As currently stated, the equation at the top of the question appears to have an extra (or missing) $2$ somewhere, so I will tackle the probability $$ P\left(\min_i X_i + \max_i X_i \leq a\right) $$ for an arbitrary $a \in \mathbb{R}$ that you can scale as needed for your context.

Proposition 1. Let $X_1, \ldots, X_n$ be an i.i.d. sample from an absolutely continuous distribution with probability density $f$ and cumulative distribution function $F$. Let $X_{(1)} = \min_i X_i$ and $X_{(n)} = \max_i X_i$. Then the joint density of $(X_{(1)}, X_{(n)})$, call it $g$, is given by $$ g(x, y) = n (n - 1) \left(F(y) - F(x)\right)^{n - 2} f(x) f(y) $$ for all $(x, y) \in \mathbb{R}^2$ with $x \leq y$ ($g(x, y) = 0$ otherwise).

Proof. We first compute $$ P(X_{(1)} \geq x, X_{(n)} \leq y) $$ for all $x, y \in \mathbb{R}$. This probability is clearly zero if $x > y$, so assume $x \leq y$. Notice that $$ \{X_{(1)} \geq x\} = \{X_1 \geq x, \ldots, X_n \geq x\} $$ and $$ \{X_{(n)} \leq y\} = \{X_1 \leq y, \ldots, X_n \leq y\}, $$ whence $$ \begin{aligned} P(X_{(1)} \geq x, X_{(n)} \leq y) &= P(x \leq X_1 \leq y, \ldots, x \leq X_n \leq y) \\ &= P(x \leq X_1 \leq y) \cdots P(x \leq X_n \leq y) \\ &= \left(F(y) - F(x)\right)^n. \end{aligned} $$ Here we used the independence and identically-distributed-ness of $X_1, \ldots, X_n$ and the fact that their distributions are absolutely continuous. To summarize, $$ P(X_{(1)} \geq x, X_{(n)} \leq y) = \begin{cases} \left(F(y) - F(x)\right)^n & \text{if $x \leq y$} \\ 0 & \text{otherwise.} \end{cases} $$

The joint density of $(X_{(1)}, X_{(n)})$ is now obtained by differentiating: $$ g(x, y) = -\frac{\partial^2}{\partial x \partial y} P(X_{(1)} \geq x, X_{(n)} \leq y) $$ for a.e. $(x, y) \in \mathbb{R}^2$. When $x > y$, both sides are zero, and when $x \leq y$, then a.e. we have $$ \begin{aligned} g(x, y) &= -\frac{\partial^2}{\partial x \partial y} P(X_{(1)} \geq x, X_{(n)} \leq y) \\ &= -\frac{\partial^2}{\partial x \partial y} \left(F(y) - F(x)\right)^n \\ &= n (n - 1) \left(F(y) - F(x)\right)^{n - 2} f(x) f(y), \end{aligned} $$ as claimed.

Proposition 2. With the same assumptions and notation as in Proposition 1, we have $$ P(X_{(1)} + X_{(n)} \leq a) = n (n - 1) \int_{-\infty}^{a / 2} f(x) \left(\int_x^{a - x} \left(F(y) - F(x)\right)^{n - 2} f(y) \, dy\right) \, dx. $$

Proof. We have $$ \begin{aligned} P(X_{(1)} + X_{(n)} \leq a) &= \int_{-\infty}^\infty \int_{-\infty}^{a - x} g(x, y) \, dy \, dx \end{aligned} $$ Since $g(x, y) = 0$ if $x > y$, this integral reduces to $$ \int_{-\infty}^{a/2} \int_x^{a - x} g(x, y) \, dy \, dx $$ (draw the picture of the intersection of the regions $x \leq y$ and $x + y \leq a$ to see why). Now just plug in the density formula from Proposition 1.


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