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Let $X_1,\dots,X_n$ are i.i.d. samples from uniform distribution on $(0,1)$. Let $\hat F_n$ be their modified empirical distribution function defined by $$ \hat F_n(x)=\frac1{n+2}\left(1+\sum_{i=1}^n1_{\{X_i\le x\}}\right) $$ for every $x$, where $1_A$ is the characteristic function of the event $A$.

Define $T_n$ by $$ T_n=\int_0^1\left\{\Phi^{-1}(\hat F_n(u))-u\right\}du $$ where $\Phi$ is the cummulative standard normal distribution.

I wish to compute $E(T_n^k)$ for $k\ge 1$. For $k=1$, $E(T_n)$ can be computed as follows:

Let $X_{(1)}\le \dots\le X_{(n)}\le X_{(n+1)}=1$ be the ordered statistics of our observation. Then \begin{align} E(T_n)&=\sum_{i=0}^n\int_0^1\left\{\Phi^{-1}\left(\frac{i+1}{n+2}\right)-u\right\}P\left(X_{(i)}\le u\le X_{(i+1)}\right)du\\ &=\sum_{i=0}^n\int_0^1\left\{\Phi^{-1}\left(\frac{i+1}{n+2}\right)-u\right\}{n\choose i}u^i(1-u)^{n-i}du. \end{align}

But then, how do I compute the second moment of $T_n$, i.e, $E(T_n^2)$? Could anyone help me? Any help will be vastly apprecaiated.

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  • $\begingroup$ You could try to copy faithfully the method exposed there: math.stackexchange.com/q/787596 $\endgroup$
    – Did
    May 21, 2014 at 17:45
  • $\begingroup$ Not ye @Did. I've just realized that CLT cannot be applied for order statistics, can it? So, I wonder your method posted there. $\endgroup$
    – Jlamprong
    May 21, 2014 at 17:55
  • $\begingroup$ It can very much--please read the post (or continue to wonder). $\endgroup$
    – Did
    May 21, 2014 at 18:11
  • $\begingroup$ I have commented on the post. $\endgroup$
    – Jlamprong
    May 21, 2014 at 18:16

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