If $r \cdot v X \sim N(0, 1)$ then $E\left(\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{X} e^{-z^2/2} dz\right)$ equals to
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For any continuous random variable, the expected value of its own CDF is always $1/2$. This is because the CDF transforms the variable into a $U(0,1)$ distribution, which has a mean of $0.5$.