<s>[INST] k-th smallest number of standard normal distribution There are $n$ random variables obey the standard normal distribution $N(0 ,1)$.
What is the expectation of the k-th smallest number?
My attempt
I tried to use conditional probability to solve the problem, but the situation of $n=4$ is too complex to handle.
for n = 2
$$p_1(x) = \frac{e^{-\frac{x^2}{2}} }{\sqrt{2 \pi }}\text{erfc}\left(\frac{x}{\sqrt{2}}\right)$$
$$p_2(x) = \frac{e^{-\frac{x^2}{2}} }{\sqrt{2 \pi }}\left(\text{erf}\left(\frac{x}{\sqrt{2}}\right)+1\right)$$
$E_1 = -\dfrac{1}{\sqrt{\pi }}, E_2 = \dfrac{1}{\sqrt{\pi }}$
for n = 3
$$p_1(x) = \frac{3 e^{-\frac{x^2}{2}} }{4 \sqrt{2 \pi }}\text{erfc}\left(\frac{x}{\sqrt{2}}\right)^2$$
$$p_2(x) = -\frac{3 e^{-\frac{x^2}{2}} }{2 \sqrt{2 \pi }}\left(\text{erf}\left(\frac{x}{\sqrt{2}}\right)^2-1\right)$$
$$p_3(x) = \frac{3 e^{-\frac{x^2}{2}} }{4 \sqrt{2 \pi }}\left(\text{erf}\left(\frac{x}{\sqrt{2}}\right)+1\right)^2$$
$E_1 = -\dfrac{3}{2 \sqrt{\pi }}, E_2 = 0, E_3 = \dfrac{3}{2 \sqrt{\pi }}$
Is there any good way to deal with this problem?
 [/INST] The density of the $k$-th smallest number among $n$ i.i.d standard normal random variables is given by
$$
f_{X_{(k)}}(x) = \frac{n!}{(r-1)!(n-r)!} \phi(x) \left[\Phi(x)\right]^{k-1}
\left[1-\Phi(x)\right]^{n-k},$$
where $\phi(x)$ and $\Phi(x)$ are PDF and CDF of standard normal distribution respectively. Then the task is to compute$\int x f_{X_{(k)}}(x) \,dx$. Thr approximation of this integral is done here.
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