There are $n$ ($n \ge 3$) iid random variables $\{ {c_i}\} _{i = 1}^n$ on the interval $[\underline c,\bar c]$ ($\underline c>0$). The cdf $F(\cdot)$ and pdf $f(\cdot)$ are unkown to us, but we konw that $\frac{{F(\cdot)}}{{f(\cdot)}}$ is increasing. With ${c_{(i)}}$ denoting the $i$-th order statistic (i.e., the $i$-th smallest among $\{ {c_i}\} _{i = 1}^n$), we have two complex random variables

${\kern 80pt}$
$X = {\left( {1 - \frac{{{c_{(3)}}}}{r}} \right)^2}$ and
$Y = \frac{{(1 + \theta)S_1^2}}{2} + \frac{{{{[{{({S_2} - \theta {S_1})}^ + }]}^2}}}{{2(1 - \theta )}}$,

where $\theta \in (0,1)$ and $r > \bar c + \frac{{F(\bar c)}}{{f(\bar c)}}$ are constants,
${S_j} = 1 - \frac{{{c_{(j)}}}}{r} - \frac{{F({c_{(j)}})}}{{r{\kern 1pt} f({c_{(j)}})}} (j=1,2)$, and
${(x)^+} = \left\{{\begin{array}{*{20}{c}} x&{x>0}\\0&{x\le0}\end{array}}\right.$.

${\kern 15pt}$ I have proved that $E[Y]>E[X]$. By numerical simulations with uniform distributions, exponential distributions and normal distributions, I observe that **as the sample size $n$ increases, $X$ and $Y$ will stochastically move closer to each other** (i.e., $E[Y-X]$ becomes closer to $0$ or $\frac{{E[Y]}}{{E[X]}}$ becomes closer to 1). **Now I want to prove this observation analytically**. The main difficulties are as follows.

- The random variable $Y$ is a piecewise function of two order statistics. The expectation value of $Y$ involves integrals with parts of the entire integral space. I didn't find good ways to deal with this problem.
- $X$ and $Y$ are non-iid random variables, because they are interdependent and with different intervals and different distributions.

${\kern 15pt}$ I am looking forward to your instructions. Thank you!