Let $m(z)= \max \lbrace 0, z \rbrace$ and $x = (s,t)$ be a random vector having the bivariate standard normal distribution. Let also $u=(u_1,u_2),v=(v_1,v_2)$ be constant vectors. Compute the following expectation w.r.t to $u,v$:
$$ \mathbb{E}_x [m(u^Tx)m(v^Tx)] $$
My thoughts so far:
Let $f(y)$ be the (single variable) standard normal distribution, and let $D$ be the "conic" region where both inputs $u^Tx$ and $v^Tx$ are positive, namely:
$$ D = \lbrace (s,t) \vert u_1s + u_2t \geq 0, v_1s + v_2t \geq 0\rbrace$$
It seems to me that the expected value will simply then be the (double) integral:
$$\mathbb{E}_x = \int_D (u_1s + u_2t)(v_1s + v_2t)f(s)f(t)dsdt $$
which after a few calculations should give us something like the following:
$$\mathbb{E}_x = u_1v_1 \mathbb{E}_x(s^2 \vert D) + (u_1v_2 + u_2v_1) \mathbb{E}_x (s \vert D) \mathbb{E}_x(t \vert D) + u_2v_2 \mathbb{E}_x (t^2 \vert D)$$
which is in fact an expression in terms of $u,v$, is this correct?
If so, then other than basically "translating" the original statement, I don't feel that I have done much in terms of calculations. Can this be simplified further, without knowing anything else about $u,v$. And if so how?
Edit: There is a great suggestion in the comments by kimchi lover (to use polar coordinates and then leverage the symmetry) which works great.
Bonus question: What happens in the general case (n-dimensional vectors instead of 2-dimensional ones)?