# $\mathbb{E}[\operatorname{sign}\langle v,z\rangle]$ 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: $$\mathbb{E}[\operatorname{sign}\langle v,z\rangle]=0$$