Let $x > 0$ and let $Z \sim N(0, 1)$. How does $f(x) = \mathbb{E}[\max\{|Z| - x, 0\}^2]$ behave?

Question

Let $x > 0$ and let $Z \sim N(0, 1)$. Define $(\cdot)_+ = \max\{\cdot, 0\}$. Define $$ f(x) = \mathbb{E}[(|Z| - x)_+^2] \equiv \mathbb{E}[\max\{|Z| - x, 0\}^2]. $$ (We use the convention that the positive part binds before the square.) I am interested in getting sharp upper and lower bounds for this function.

Using the fact that $t \mapsto t_+^2$ is convex we have $$ f(x) \geq \Big(\sqrt{\frac{2}{\pi}} -x \Big)_+^2, $$ by Jensen's inequality.

Of course we also have $x_+ \leq |x|$, so that $$ f(x) \leq 1 + x^2 - 2 x \sqrt{\frac{2}{\pi}} = \Big(\sqrt\frac{2}{\pi} - x\Big)^2 + \frac{\pi - 2}{\pi}. $$ Hence, this verifies that $(\sqrt{2/\pi} - x)_+^2$ is of the correct order, up to multiplicative constants, provided that $x \in (0, x_0)$ where $x_0$ is sufficiently small.

However, these bounds will not be sharp when $x$ is growing. Any help?

Note that we have the identity $$ f(x) = (1+x^2) \mathrm{erfc}(x/\sqrt{2}) - 2 x \phi(x) $$ where $\phi(x)$ is the standard Normal density and $\mathrm{erfc}$ is the complementary error function.

Answer 1

If I did not commit calculation mistakes, we can express $f(x)$ as a function of the error function: \begin{align*} f(x) &= 2\int_x^\infty (t-x)^2\frac{e^{-t^2/2}}{\sqrt{2\pi}} dt \\ &= (1+x^2) \mathrm{erfc}(x/\sqrt{2}) - \sqrt{\frac{2}{\pi}}x e^{-x^2/2}. \end{align*} Now, you can refer to the literature (for example, here) to choose suitable bounds for the error function.

For example, using 7.8.2, we get that \begin{align*} f(x) \ge \left(\sqrt{x^2+4}-x \left(x^2-\sqrt{x^2+4} x+3\right)\right) \frac{e^{-\frac{x^2}{2}} }{\sqrt{2 \pi}}. \end{align*} Using 7.8.4, we end up with \begin{align*} f(x) \le \frac{x^2-x\sqrt{x^2+8}+4}{\sqrt{x^2+8}+3 x} \sqrt{\frac{2}{\pi }} e^{-\frac{x^2}{2}} . \end{align*}

The upper bound is sharp as $x\to\infty$ in the sense that the ratio between $f$ and the upper bound converges to $1$.

Using the series expansion of the error function, we can also conclude that \begin{align*} f(x) = \left(\frac{2}{x^3} - \frac{12}{x^5} + \mathcal{O}(x^{-7})\right) \sqrt{\frac{2}{\pi }} e^{-\frac{x^2}{2}} \end{align*} as $x\to\infty$.