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I have two formulas about variance estimation of linear combination of order statistics given order statistics $X_{1}\leqslant X_{2} \leqslant ...\leqslant X_{n} $ from a random sample of arbitrary distribution. My simulation in R already shows that the two are asymptotically equivalent for lots of cases.But I still need to prove them formally. The two formulas are

$F_{1}(X)$=$\frac{1}{n}\sum_{i=2}^{n}\sum_{j=2}^{n}\left( a_{0}+a_{1}\left ( \frac{i-1}{n} \right )\right)\left ( a_{0}+a_{1}\left ( \frac{j-1}{n} \right )\right)\left ( \frac{min(i-1,j-1)}{n}-\frac{(i-1)(j-1)}{n^2} \right )\left ( X_{i}-X_{i-1} \right )\left(X_{j}-X_{j-1} \right ),$

$F_{2}(X)$=$\frac{1}{n}\sum_{k=1}^{n}\left ( a_{0}X_{k}+a_{1}X_{k}\frac{k}{n}+a_{1}\sum_{i=k}{n}X_{i}\frac{1}{n}-a_{0}\frac{1}{n}\sum_{i=1}^{n}X_{i} -2a_{1}\frac{1}{n}\sum_{i=1}^{n}X_{i}\frac{i}{n}\right )^2.$

where $a_{0}, a_{1}$ are just arbitrary fixed coeficients.

It is very time consuming to compare them term by term but I already did a large part of it. Anyone else may have some other ideas about how to prove the two formulas are asymptotically the same?

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  • $\begingroup$ Did you try to show that their ratio converges to a constant; that is to show $\lim_{n \rightarrow \infty} \frac{F_1(n)}{F_2(n)} = K$ where $K \neq 0$ is a constant? $\endgroup$ – Sobi Dec 11 '15 at 17:31
  • $\begingroup$ @Sobi, hi Sobi, my simulations in R shows that they converge to 1. But can numerical study account for formal argument of proof ? I find it not so easy to compare term by term. I have tried to compare the coefficients of them term by term , for which I mean coefficients of $ X_{i}^2 $ ....etc. I am just not sure what accounts for a formal proof. thx $\endgroup$ – lzstat Dec 11 '15 at 20:35
  • $\begingroup$ Correction to my first comment: you need to show that $\lim_{n \rightarrow \infty} \frac{F1(n)}{F2(n)}=1$. There might be different ways to do this, but, any formal proof should imply that the ratio goes to 1. I also can't see an easy way to prove the problem, but, showing experimentally that the two sequences are equivalent does not count as a formal proof. Good luck! $\endgroup$ – Sobi Dec 11 '15 at 22:13
  • $\begingroup$ A minor point but should the summations start at 2 rather than 1 in $F_{1}(X)$ because of the use of $X_{i-1}$ and $X_{j-1}$ ? $\endgroup$ – JimB Dec 20 '15 at 18:20
  • $\begingroup$ @JimBaldwin, yes , you are correct because of the approximation. $\endgroup$ – lzstat Dec 21 '15 at 1:58
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The formulas for the coefficients $d_{ki}$ for $F_2(X)$ were presented at https://mathematica.stackexchange.com/questions/102454/how-could-i-expand-the-formula-in-terms-of-x-i2-and-x-ix-j-using-ma/102478#102478. Those are as follows:

$d_{ki}={{-2}\over{n^2}} a_0^2+({{2}\over{n^2}}-{{4(k+i)}\over{n^3}})a_0 a_1 + ({{4k}\over{n^3}}-{{8ki}\over{n^4}})a_1^2$

The corresponding coefficients $c_{ki}$ for $F_1(X)$ (using a bit of summation algebra) are as follows:

$c_{ki}={{-2}\over{n^2}} a_0^2+({{2}\over{n^2}}-{{4(k+i-1)}\over{n^3}})a_0 a_1 + ({{2(2k-1)}\over{n^3}}-{{2(1-2k-2i+4k i)}\over{n^4}})a_1^2$

And both sets of coefficients need to be divided by $n$.

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  • $\begingroup$ could you leave me with your email address? thx $\endgroup$ – lzstat Dec 23 '15 at 2:38
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@JimBaldwin, For the problem of index i, you are correct. Today I compared $ c_{k} $ with $d_{k} $ , the coefficients of squared terms $X_{k}^{2}$. With the help of Jim and careful rearrangement, we have

$c_{k}$=$(\frac{1}{n}-\frac{1}{n^2})a_{0}^{2}+(\frac{2k}{n^2}+\frac{2-4k}{n^3})a_{0}a_{1}+(\frac{k^2+k-1}{n^3}+\frac{-4k^2+4k-1}{n^4})a_{1}^{2}$;

$d_{k}$=$(\frac{1}{n}-\frac{1}{n^2})a_{0}^{2}+(\frac{2k+2}{n^2}+\frac{-4k}{n^3})a_{0}a_{1}+(\frac{k^2+3k}{n^3}+\frac{4k^2}{n^4})a_{1}^{2}$;

It can be shown that that $\sum_{k=1}^{n}c_{k}X_{k}^{2}-\sum_{k=1}^{n}d_{k}X_{k}^{2}=o_{p}(1)$, under one of my assumption that $X=O_{p}(n^{\frac{1}{3}})$

Next step is left to show the same relationship holds for the cross terms. I am still trying to combine the cross term in $F_{1}(X)$ in terms of $\sum_{k=1}^{n-1}\sum_{i=k+1}^{n}c_{ki}X_{k}X_{i}$ in this elegant manner.

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