$\mathbb{E}[\operatorname{sign}\langle v,z\rangle]$ for $v$ fixed, $z_i\sim N(0,1)$ I am trying to evaluate $\mathbb{E}[\operatorname{sign}\langle v,z\rangle]$ for $v\in\mathbb{R}^n$ fixed and $z_i\sim N(0,1)\ \forall\ i\in[n]$. 
The $\operatorname{sign}$ part is what is confusing me. Clearly,
$$
\mathbb{E}[\operatorname{sign}\langle v,z\rangle]
=
\mathbb{E}\Big[\operatorname{sign}\sum_{i=1}^nv_iz_i\Big]
=
\mathbb{P}\Big[\sum_{i=1}^nv_iz_i > 0\Big]
- \mathbb{P}\Big[\sum_{i=1}^nv_iz_i < 0\Big].
$$
But I don't know how to simplify further (i.e. to get the expectation just on the $z_i$). I am wondering how to pass the $\operatorname{sign}$ operator through the expectation.

Observe that $z$ and $-z$ have the same distribution.
Consequently $\operatorname{sign}\langle v,z\rangle$ has the same distribution as $\operatorname{sign}\langle v,-z\rangle=-\operatorname{sign}\langle v,z\rangle$, so that:
$$\mathbb{E}[\operatorname{sign}\langle v,z\rangle]=\mathbb{E}[\operatorname{sign}\langle v,-z\rangle]=\mathbb{E}[-\operatorname{sign}\langle v,z\rangle]=-\mathbb{E}[\operatorname{sign}\langle v,z\rangle]$$
This implies that: $$\mathbb{E}[\operatorname{sign}\langle v,z\rangle]=0$$
