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Given $n$ independent normally distributed random variables $X_1,X_2,...,X_n \sim N(\mu,\sigma)$. For any $k\leq n$, let $X_{(k)}$ be the k-th order statistics (i.e., the k-th smallest value). What is the expected sum $\mathbb{E}[\sum_{r=1}^k X_{(r)}] $?

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Closed-form expressions for arbitrary $n,k$ are not available. For small $k$ you can use an integral expression to evaluate $\mathbb{E}[\sum_{r=1}^k X_{(r)}]=\sum_{r=1}^k\mathbb{E}[X_{(r)}]$, with $$\mathbb{E}[X_{(r)}]=\frac{n!}{(r-1)!(n-r)!}\int_{-\infty}^\infty[1-\Phi(x)]^{r-1}[\Phi(x)]^{n-r}\frac{x}{\sqrt{2\pi}}e^{-x^2/2}\,dx.$$ (Here $\Phi(x)$ is the CDF of the normal distribution.)

For example, for $n=5$ and $k=1,2,\ldots n$, you would find $$\mathbb{E}\left[\sum_{r=1}^k X_{(r)}\right]=\frac{5}{2\pi^3}\times\left(\pi -6 \arctan\sqrt{2},- 3\pi +6 \arctan\sqrt{2},- 3\pi +6 \arctan\sqrt{2},\pi -6 \arctan\sqrt{2},0\right).$$

The large-$n$ asymptotics is given in terms of the functional inverse ${\rm erf}^{-1}$ of the error function, $$\mathbb{E}\left[\sum_{r=1}^k X_{(r)}\right]\rightarrow\sqrt{2}\sum_{r=1}^k\text{erf}^{-1}\left(\frac{2r}{n+1}-1\right),\;\;\text{for}\;\;n\rightarrow\infty,$$ which is cumbersome to evaluate. As an alternative simple formula for large $n$ I can offer the parabolic approximation $$\mathbb{E}\left[\sum_{r=1}^k X_{(r)}\right]\approx -\frac{\pi k}{2 n} (n-k).$$

The plots compare the exact result (gold) with the parabolic approximation (blue) and the $\text{erf}^{-1}$ asymptote (green) for $n=10$, $50$, $100$, and $500$. As you can see, the parabolic approximation is accurate within 10% even for relatively small $n$.

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