Expected difference for two largest samples I'm trying to get an understanding as to how much difference between the two largest samples drawn from some distribution $f(x)$ scales with $n$. That is, as we increase the size of our sample, how big can we expect the difference between the two largest elements we draw? Supposing it scales as $O(n^{-\alpha})$, is $\alpha$ more or less than $1$?
Suppose that $x \in [0,1]$ for simplicity (the important fact is that it is bounded).
So far I think I have (making use of basic results here):
$$
\mathbb{E}[X_{(n)}-X_{(n-1)}] = \mathbb{E}[X_{(n)}]-\mathbb{E}[X_{(n-1)}] = \int_0^1(1-F_{X_{(n)}}(x))dx- \int_0^1(1-F_{X_{(n-1)}}(x))dx
$$
$$
\Rightarrow \int_0^1(F_{X_{(n-1)}}(x)-F_{X_{(n)}}(x))dx = \int_0^1 nF_X(x)^{n-1}(1-F_X(x))+F_X(x)^n-F_X(x)^n dx
$$
$$
\Rightarrow n\int_0^1 F_X(x)^{n-1}(1-F_X(x))dx
$$
Assuming that is right, I'm not sure where I can go from here. From what I can see, the distribution I'm looking at is roughly Gaussian bounded in a small subset of $[0,1]$, so I could in principle invoke some more assumptions on $F_X(x)$, but I wonder if there's a general result that is known/can be derived.
 A: This is a partial answer that shows an upper bound of $O(\log(n)/n)$ under smoothness assumptions on $F$ near $F^{-1}(1)$. I'd be interested in arguments without such an assumption. My gut says that the uniform density is probably the worst case, and a generic $O(1/n)$ upper bound should be within reach. Perhaps a calculus of variations approach?

Notice that for $u \in [0,1]$, it holds that $ ku^{k-1}(1-u) < 1/k^2$ if $u > 1-1/k^2$ or if $u < 1/k^{3/k-1} = \exp\left(-3 \log(k)/(k-1) \right).$ Since $1-x < e^{-x}$, the same also holds if $u < 1- 3\log(k)/(k-1)$.
Now let $u_n := F^{-1}(1-1/n)$ and let $v_n := F^{-1}(1 - 3\log(n)/n)$. Let $$I_n(F) := \int_0^1 nF^{n-1}(1-F) \mathrm{d}x.$$
By the above reasoning, 
$$ I_n \le \frac{1}{n^2} + \int_{v_n}^{u_n} nF^{n-1}(1-F).$$ The 1 in $1/n^2$ can of course be replaced by the length of the support when it is not $[0,1]$.
Note that the integrand is $O(1)$, so we don't have to worry about that too much, as long as it holds that $u_n - v_n$ is not too big. Since both $u_n$ and $v_n$ are very close to $1$, it stands to reason that we need to worry about $F^{-1}$ near $1$. 
It is natural to try to check the validity of the (one-sided) Taylor expansion below: $$ F^{-1}(1-\varepsilon) = F^{-1}(1) - {(F^{-1})'(1)}{\varepsilon} + O(\varepsilon^2).$$
If the above holds, then we have $I_n \le 1/n^2 + O(u_n - v_n) = O(\log(n)/n)$ where the $O$ hides $(F^{-1})'(1)$ and other constant factors.
Since $(F^{-1})'(u) = 1/F'(F^{-1}(u)) = 1/f(F^{-1}(u)),$ where $f$ is the density, one can put forth conditions which enable something like the above - for instance, if $f$ is non-zero at $F^{-1}(1)$ (the maximum point of the support) and smooth in some right neighbourhood, then we're in business. Alternately, it need not be smooth, but must be bounded away from $0$ in some right neighbourhood of $F^{-1}(1)$ (using the Lagrange form remainder).
Under conditions such as the above, we get $I_n(F) = O(\log(n)/n).$ This is nearly tight since the uniform density induces such a 'nice' $F$ and has $I_n(\mathrm{Unif}) = 1/(n+1)$. 
