# Expected gap between two consecutive order statistics

Consider a random variable $X\sim_X(x)$ with $\operatorname{supp}{p_X}=[a,b]\subset{\mathbb R}$, and let $X_1, X_2, \ldots, X_n$ be $n$ $i.i.d.$ samples from $p_X(x)$, then we have their associated order statistics $X_{(1:n)}, X_{(2:n)}, \ldots, X_{(n:n)}$ such that $X_{(1:n)}\leqslant X_{(2:n)}\leqslant \cdots\leqslant X_{(n:n)}$. Define the spacing between any two consecutive order statistics as $W_{(r:n)}=X_{(r+1:n)}-X_{(r:n)} (1\leqslant r\leqslant n-1)$, then in "Note on Francis Galton's problem" by K. Pearson (1902), it has been shown that $${\mathbb E}\left[W_{(r:n)}\right]=\frac{n!}{(n-r)!r!}\int_{-\infty}^{\infty}F(x)^{n-r}[1-F(x)]^r \, dx, \hspace{2mm} 1\leqslant r\leqslant n-1,$$ where $F(x)=\int_{-\infty}^xp_X(y) \, dy$ is the distribution function of $X$.

So, $\forall r, n$ such that $1\leqslant r\leqslant n-1$, $2\leqslant n$, do we have $${\mathbb E}\left[W_{(r:n)}\right] \geqslant {\mathbb E}\left[W_{(r:n+1)}\right],$$ i.e., the expected gap between any two consecutive order statistics will decrease or remain the same as $n$ increases?

Intuitively, this is correct since the more samples one draws, the denser the order statistics will be filled in the support. I also checked several specific distributions (e.g., uniform, Gaussian, etc), and the statement holds, but I have not figured out a good way to prove this for the general case. Hope someone can provide some inspiration.

• Nice question. For counterexamples, I would look at the asymptotics when $n\to\infty$ through a Taylor expansion of $F$ around the point $x^*$ such that $F(x^*)=\frac12$. Differentiable densities $f$ such that $f(x^*)^2f'(x^*)$ is large might contradict the claim. – Did May 29 '13 at 7:09
• Instead of saying "Let $X_1,\ldots,X_n$ be i.i.d. samples", you should say "Let $X_1,\ldots,X_n$ be an i.i.d. sample". The sample is the whole thing. $X_i$ is an observation, not a sample. – Michael Hardy May 29 '13 at 14:40
• If the cumulative distribution function $F$ of the distribution of each observation $X_i$ is continuous, then $F(X_{(1:n)}), \ldots, F(X_{(n:n)})$ are the order statistics from the uniform distribution on the interval $(0,1)$. I'm not sure whether that can be readily exploited here. – Michael Hardy May 29 '13 at 14:48
• – BruceET Feb 5 '18 at 18:58

This is not true. Consider a Bernoulli distribution with $p=\frac12$. (Technically, this doesn't have support on an interval, but we can approximate it arbitrarily closely with support on an interval.)

Then with $S_n=\sum_{k=1}^nX_n$ we have

$$\mathsf E\left[W_{(r:n)}\right]=\mathsf P\left(X_{(r:n)}=0\land X_{(r+1:n)}=1\right)=\mathsf P(S_n=n-r)=2^{-n}\binom nr\;.$$

But

$$2^{-(n+1)}\binom{n+1}r\gt2^{-n}\binom nr$$

whenever

$$\frac{n+1}{n-r+1}\gt2\;,$$

that is, for all $r\gt\frac{n+1}2$.

In this case, the support cannot be filled in; increasing $n$ merely changes the probability that these particular order statistics are separated by the constant gap. It could be that this effect can't occur if the distribution is unimodal.