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.