Expectation of minimum of normally distributed random variables Let $(X,Y)$ be normally distributed and such that
$\;\;\;\;\mathrm{Cov}(X,Y)=\varrho$, and $\mathrm{Var}(X)=\mathrm{Var}(Y)=1$.
For which $\varrho$ does the following equality hold?
$\;\;\;\;\min (E(X),E(Y)) = E(\min(X,Y))$
For $\varrho=1$? For $\varrho=0$? For $\varrho=-1$? For any other $\varrho$?
 A: You can calculate $E(X)$ and $E(Y)$ by fixing the mean vector of $(X,Y)$, $\mu$. 
Note that if $Z=\min(X,Y)$, then $P(Z>z) = P(\min(X,Y)>z) = P(X >z\text{ and }Y>z) = \int_{x,y > z} \frac{1}{2 \pi \sqrt{|K|}} e^{- ( (x;y) - \mu)^T K^{-1} ( (x;y) - \mu)}$ where $K=[var(X),cov(X,Y);cov(X,Y), var(Y)]$. You can do a Cholesky decomposition on $K$ to get a nice change of coordinates which decorrelates $X$ and $Y$ to evaluate the integral getting the distribution of the minimum. Then, find its mean, and compare. 
A: Assume without loss of generality that $E(X)\leqslant E(Y)$. Then, $X\geqslant\min(X,Y)$ almost surely hence the condition $E(X)=E(\min(X,Y))$ implies that $X=\min(X,Y)$ almost surely, that is, that $X\leqslant Y$ almost surely. 
Every two-dimensional normal distribution is symmetric with respect to its mean $(\mu_X,\mu_Y)$, whether the covariance is invertible or not, hence $P(X\gt Y)=0$ implies $P(X-2\mu_X\lt Y-2\mu_Y)=0$. This shows that $X\leqslant Y\leqslant X+c$ almost surely, for some finite $c$. This can only happen if the support of the distribution of $(X,Y)$ is either a point or a line $\{(x,y)\mid y=x+c'\}$ for some $c'$ in $[0,c]$, which means that $\varrho=1$. 
Conversely, when $\varrho=1$, $Y=X+\mu_Y-\mu_X$ hence the result holds.
A: Let's enter random variable $T=X-Y$,for any 2-D distribution occurs: $$V(X-Y)=V(X)+V(Y)-2\cdot COV(X,Y)$$
Remembering that $V(X)=V(Y)=1$ we obtain: $$V(T)=V(X-Y)=2\cdot(1-\rho)$$
It's obvious that if $\rho=1$ then $V(T)=0$ , which means that $T$ is no more random variable, but deterministic value equal to $E(T)$. It results even if Bienaymé-Chebyshev inequality ,described here and here and is truth for any distribution with real valued random variable (there is no sense defining function $\min$ or $\max$ for numbers which are not real). Let $E(T)=C$, in such a case, we have to find such a $C$ that:$$\min(E(X),E(X-C))=E(\min(X,X-C))$$
$E(X-C)=E(X)-C$ for any distribution. If we want to find what is greater $X$ , or $X-C$ (we know the value of $C$ , it's our parameter),then there is no need to know value of $X$ . So, after short consideration there is easy to check that: $\min(E(X),E(X-C))$ is equal to $E(\min(X,X-C))$ for any distribution with real valued random variable $X$ and any real parameter $C$. Accepting $\rho=1$, we find solution.
Let's consider other values of $\rho$, $-1\leq\rho<1$ (*). Let $f(x,y)$ be probability density function of random variables $X$,$Y$ and $g(t,x)$ probability density function for variables $T$,$X$.
$$E(\min(X,Y))=\int _{-\infty }^{+\infty }\int _{y\leq x}y f(x,y)dydx+\int _{-\infty }^{+\infty }\int _{y>x}x f(x,y)dydx$$
$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int _0^{+\infty }\int _{-\infty }^{+\infty }(x-t) g(t,x)dxdt+\int _{-\infty }^0\int _{-\infty }^{+\infty }x g(t,x)dxdt$$
$$=E(X)-\int_0^{+\infty } t f_T(t) \, dt\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$
According to definition $f_T(t)=\int_{-\infty}^{+\infty}g(t,x)dx$ is probability density function of normal distribution ($T=X+(-1)\cdot Y$) . Notice that $\int_0^{+\infty } t f_T(t) \, dt\neq 0$, ($\rho\neq 1\ \rightarrow V(T)\neq 0$), so it must be: $\min(E(X),E(Y))=E(Y)$,which implies
$$E(T)=\int_0^{+\infty } t f_T(t) \, dt$$
and is impossible when $V(T)\neq 0$.
