For $i\in [n]$ , $g_i \sim N(1,1)$ is a Gaussian random variable. We assum that all the $g_i$ are independent.
Consider $u_i, x_i \in \mathbb{R}$
What is the best why to compute the expectation:
$$ E_g \big[ max\lbrace 0, \sum_{i=1}^n g_iu_ix_i\rbrace \big] = E_g \big[ max\lbrace 0, \langle g \odot u , x \rangle \rbrace \big]$$
Where $u\odot g$ is the element-wise product.
Can I look on $max\lbrace 0, \sum_{i=1}^n g_iu_ix_i\rbrace$ as a truncated random variable?
If the $g_i$ are independent, then $\sum \limits_{i = 1}^n g_i u_i x_i \sim \mathcal{N}(\mu, \sigma^2)$ with $\mu = \sum \limits_{i = 1}g_i u_i$ and $\sigma^2 = \sum \limits_{i = 1}^n(g_iu_i)^2$. So if $N$ is a standard normal variable, then $\sum \limits_{i = 1}^n g_i u_i x_i \stackrel{d}{=}\sigma N + \mu$. Now we can calculate the expectation:
$$\begin{align*} &\mathbb{E}[\max(0, \sigma N + \mu)] = \frac{1}{\sqrt{2\pi}}\int \max(0, \sigma x + \mu) e^{-x^2/2} \, dx = \frac{1}{\sqrt{2\pi}} \int_{-\mu/\sigma}^{\infty} (\sigma x + \mu)e^{-x^2/2} \, dx \\ ={}&\frac{1}{\sqrt{2\pi}} \left(\mu \Psi(\mu/\sigma) +\sigma e^{-\mu^2/(2\sigma^2)}\right) \end{align*}$$