How to find the minimum-maximum of $ N( 0, 1) $ Given that two random variables $J,P∼N(0,1)$ are independent. Show how to compute the following and provide the answer:
(i) $E[\min(J,P)]$ and $E[\min(J^2,P^2)]$ and
(ii) $E[\max(J,P)]$ and $[\max(P^2,J^2)].$
A similar question was posted about 6 years ago, click here to see it. I followed the steps over there to try to solve this question. I understand there may be several ways of solving this question and I welcome any alternative methods in solving the question. However, here’s my understanding so far which may not be entirely correct:
I will let $D = \min(J,P)$ and $N=\max(J,P).$
It is a fact that $D+N = J+P.$ Then we can able to present the following:
$$ E(D)+E(N) = E(J) + E(P) = 0; \quad E(D)= -E(N)$$
Note that
$$ 2E(D) = E(D)-E(N) = E(D-N)=E(|J - P|), \ \text{(expectation of half-normal random variable)}$$
$$2E(D) = \frac{2}{\sqrt{\pi}}; \quad E(D) = \frac{1}{\sqrt{\pi}}, \quad E(N) = -\frac{1}{\sqrt{\pi}}.$$
$W = \sum_{i=1}^k Z_i^2$ is a $\chi^2$ distribution, where $Z_i \sim N(0,1)$.  Let $A = \min(J^2,P^2)$ and $B= \max(J^2,P^2) .$
$$ E(A) + E(B) =E(J^2) + E(P^2) =2$$
$$E(A) = 2-E(B)$$
$$E(A)=E(B)=1.$$

For the last part, you already know that $\mathbb{E}(A+B)=2$, and all you have to do is find the value of $\mathbb{E}(B-A)$, which is not zero this time.
Indeed, $\mathbb{E}(B-A) = \mathbb{E}(|J^2 - P^2|)$. Given $\theta$ uniformly distributed on $[0, 2\pi[$ independent from $R$ with density $f(r) = r e^{-\frac{1}{2}r^2}$, $r \ge 0$, use polar coordinates to write $(J,P) \sim (R\cos\theta, R\sin\theta)$, and then: $$\mathbb{E}(B-A) = \mathbb{E}(R^2)\mathbb{E}\big(|\cos(\theta)^2-\sin(\theta)^2|\big) = 2 \cdot \mathbb{E}(|\cos(2\theta)| = \frac{4}{\pi}$$
Hence $\mathbb{E}(A) = 1 - \frac{2}{\pi}$ and $\mathbb{E}(B) = 1 + \frac{2}{\pi}$.
