Integral from the Adversarial Spheres paper (maximum of the difference between a constant and a normal random variable) I'm trying to follow a proof in the Adversarial Spheres preprint on arXiv. The proof requires the computation of the integral in Appendix F, page 14:
$$\mathbf{E}\left[\max\left(\sqrt{2}\left(\frac{\alpha}{\sqrt{n}}-Z\right),0\right)\right],\quad Z\sim\mathcal{N}\left(0,\frac{1}{n}\right)$$
where $\alpha>0$. According to the paper, this integral should be $O\left(\frac{\alpha}{\sqrt{n}}\right)$.Now, since 
$$\sqrt{n}Z=X\sim\mathcal{N}\left(0,1\right)$$
I can rewrite the expectation as 
$$\mathbf{E}\left[\max\left(\sqrt{2}\left(\frac{\alpha-X}{\sqrt{n}}\right),0\right)\right],\quad X\sim\mathcal{N}(0,1)$$
Using the properties of $\max(x,0)$ and expectation, I rewrite
$$\mathbf{E}_{X\sim\mathcal{N}(0,1)}\left[\max\left(\sqrt{2}\left(\frac{\alpha-X}{\sqrt{n}}\right),0\right)\right] = \sqrt{\frac{2}{n}} \mathbf{E}_{X\sim\mathcal{N}(0,1)}\left[\max\left(\alpha-X,0\right)\right]$$
I'm left with computing
$$\mathbf{E}_{X\sim\mathcal{N}(0,1)}\left[\max\left(\alpha-Y,0\right)\right]=\int_{-\infty}^{\infty}\max(\alpha-x,0)\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx=\int_{-\infty}^{\alpha}(\alpha-x)\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx$$
This integral can be split into
$$\alpha\int_{-\infty}^{\alpha}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx=\alpha\Phi(\alpha)$$
and
$$-\int_{-\infty}^{\alpha}\frac{x}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx=\int_{-\alpha}^{\infty}\frac{x}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx$$
Now, this last integral should be equal to the conditional expectation 
$$\mathbf{E}[X|-\alpha<X]$$
(right?). Thus I should have
$$\int_{-\alpha}^{\infty}\frac{x}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx=\frac{\phi(-\alpha)}{1-\Phi(-\alpha)}$$
Thus, finally I should have
$$\mathbf{E}_{Z\sim\mathcal{N}\left(0,\frac{1}{n}\right)}\left[\max\left(\sqrt{2}\left(\frac{\alpha}{\sqrt{n}}-Z\right),0\right)\right] = \sqrt{\frac{2}{n}}\left(\alpha\Phi(\alpha)+\frac{\phi(-\alpha)}{1-\Phi(-\alpha)}\right)$$
Questions:


*

*Is this correct?

*Can I conclude that, since $\Phi(\alpha)$ and $\frac{\phi(-\alpha)}{1-\Phi(-\alpha)}$ are bounded for $\alpha>0$, the integral is $O\left(\frac{\alpha}{\sqrt{n}}\right)$ as stated in the paper? Specifically, since $\Phi(\alpha) \to 1$ and $\frac{\phi(-\alpha)}{1-\Phi(-\alpha)} \to 0$ as $\alpha \to \infty$, I have that the integral is well-approximated by $\sqrt{\frac{2}{n}}\alpha$, as shown below:
a <- seq(0, 10, len = 1000)
y <- a*pnorm(a)+dnorm(-a)/(1-pnorm(-a))
plot(a, y, type = "l")
abline(a = 0, b = 1, col = "red")


 A: Adapting the answer by whuber to this question:
Expected value of x in a normal distribution, GIVEN that it is below a certain value
we have that for a standard normal $X$
$$\mathbf{E}[X|X \le \alpha] =  - \frac{\phi\left(\alpha\right)}{\Phi\left(\alpha\right)}$$
Now, setting $Y=-X$, and noting that $Y\sim\mathcal{N}(0,1) $, we have
$$\mathbf{E}[X|X \le \alpha]=\mathbf{E}[-Y|-Y \le \alpha] = - \mathbf{E}[Y|-\alpha \le Y]=-\frac{\phi\left(\alpha\right)}{\Phi\left(\alpha\right)}\implies\mathbf{E}[Y|-\alpha \le Y]=\frac{\phi\left(\alpha\right)}{\Phi\left(\alpha\right)}$$
Since
$$\frac{\phi\left(\alpha\right)}{\Phi\left(\alpha\right)}=\frac{\phi\left(-\alpha\right)}{1-\Phi\left(-\alpha\right)}$$
my formula
$$\mathbf{E}_{Z\sim\mathcal{N}\left(0,\frac{1}{n}\right)}\left[\max\left(\sqrt{2}\left(\frac{\alpha}{\sqrt{n}}-Z\right),0\right)\right] = \sqrt{\frac{2}{n}}\left(\alpha\Phi(\alpha)+\frac{\phi(-\alpha)}{1-\Phi(-\alpha)}\right)$$
is proved and it can be simplified to 
$$\mathbf{E}_{Z\sim\mathcal{N}\left(0,\frac{1}{n}\right)}\left[\max\left(\sqrt{2}\left(\frac{\alpha}{\sqrt{n}}-Z\right),0\right)\right] = \sqrt{\frac{2}{n}}\left(\alpha\Phi(\alpha)+\frac{\phi(\alpha)}{\Phi(\alpha)}\right)$$
thus the answer to my questions is 1) yes and 2) yes.
