Suppose that I have two independent random variables $g_1,g_2\sim N(0,1)$.
From these variables I construct a vector: $g=(g_1,g_2)$.
Also, I have a unit vector $u=(u_1,u_2)$.
It follows that the inner product $\langle u,g\rangle=u_1g_1+u_2g_2$ follows a normal distribution with the parameters $0$ and $u_1^2+u_2^2=1$, i.e., $\langle u,g\rangle\sim N(0,1)$.
How do I find the expected value $E(g_1 \operatorname{sign}{\langle u,g\rangle})$?
One approach is to claim that gaussian variable is rotation invariant, so we can take $u=(1,0)$ and in this case $E(g_1 \operatorname{sign}{\langle u,g\rangle})=E(g_1 \operatorname{sign}(g_1)$ which can be easily calculated.
But what is the justification of such approach? Is there some "direct" way?
Thank you.
Suppose $u_1 \neq 0$. Then $h\equiv u_1^{-1}g_1- \langle u, g \rangle$ is independent of $\langle u, g \rangle$ because these variables are jointly normal and their covariance is, by a routine calcualtion, $0$. Note that $g_1=u_1h+u_1\langle u, g \rangle$.
Hence, $E(g_1 sign \langle u, g \rangle)=E(u_1h sign \langle u, g \rangle)+E(u_1\langle u, g \rangle sign \langle u, g \rangle)$. By independence, $E(u_1h sign \langle u, g \rangle)=u_1 Eh E( sign \langle u, g \rangle)=0$. Also, $E(u_1\langle u, g \rangle sign \langle u, g \rangle)=u_1E|\langle u, g \rangle)|$ and this quantity is just $E|X|$ where $X \sim N(0,1)$. SO the answer is $u_1E|X|=u_1\sqrt {\frac 2 {\pi}}$.
The case $u_1=0$ is trivial.