Integral from the Adversarial Spheres paper (maximum of the difference between a constant and a normal random variable)

Question

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:

1. Is this correct?

2. 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")
   

   

Answer 1

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.