1 6 3 Tails Of A Standard Gaussian Distribution

1.

• Let $z>0$. With an integration by parts:

(1)

\begin{align} \mathbb{P}(|Z|\geq z) &= \sqrt{\frac{2}{\pi}}\int_z^{+\infty}\frac{x}{x}e^{-x^2/2}dx\\ &= \sqrt{\frac{2}{\pi}}[-x^{-1}e^{-x^2/2}]_0^\infty - \sqrt{\frac{2}{\pi}}\int_z^{+\infty}x^{-2}e^{-x^2/2}dx. \end{align}

So we have

(2)

\begin{align} \mathbb{P}(|Z|\geq z) = \sqrt{\frac{2}{\pi}}z^{-1}e^{-z^2/2} - \sqrt{\frac{2}{\pi}}\int_z^{+\infty}x^{-2}e^{-x^2/2}dx. \end{align}

• Let $f(z) = \int_z^{+\infty}e^{-x^2/2}dx$ and $g(z)=z^{-1}e^{-z^2/2}.$

We know that $f(z)>0$ and $g(z)>0$, and we just showed

(3)

\begin{align} g(z) - f(z) = \int_z^{+\infty}x^{-2}e^{-x^2/2}dx >0. \end{align}

We also have the following inequality:

(4)

\begin{align} \int_z^{+\infty}x^{-2}e^{-x^2/2}dx \leq z^{-2}f(z), \end{align}

so $0<g(z)-f(z)\leq z^{-2}f(z).$ By dividing by $f(z)$ and adding 1 we obtain $1<g(z)/f(z)\leq 1+z^{-2}$, then we get $(1+z^{-2})^{-1}g(z)\leq f(z) < g(z)$. Consequently we can see that for $z\rightarrow +\infty$, $f(z) = g(z)(1+O(z^{-2}))$, and finally

(5)

\begin{align} \boxed{\mathbb{P}(|Z|\geq z) = \sqrt{\frac{2}{\pi}}\frac{e^{-z^2/2}}{z}(1+O(z^{-2})).} \end{align}

2.

• Let $Z_1,\dots,Z_p \sim_\text{i.i.d.}\mathcal{N}(0, 1).$ and $\alpha>0$ For all $z>0$,

(6)

\begin{align} \mathbb{P}(\max_{j=1,\dots,p}|Z_j|\geq z) = 1-\mathbb{P}(\forall j\in\{1,\dots,p\},\ |Z_j|< z). \end{align}

Because of the independence of these random variables, we have:

(7)

\begin{align} \mathbb{P}(\forall j\in\{1,\dots,p\},\ |Z_i|< z) &= \Pi_{j=1}^p \mathbb{P}(|Z_j|< z),\\ &= \mathbb{P}(|Z_1|< z)^p,\\ &= (1-\mathbb{P}(|Z_1|\geq z))^p. \end{align}

So

(8)

\begin{align} \mathbb{P}(\max_{j=1,\dots,p}|Z_j|\geq z) = 1-(1-\mathbb{P}(|Z_1|\geq z))^p. \end{align}

• By replacing $z=\sqrt{\alpha\log p}$ and by (5), when $p\rightarrow\infty,$

(9)

\begin{align} \mathbb{P}(\max_{j=1,\dots,p}|Z_j|\geq \sqrt{\alpha\log p}) &= 1-\left(1-\sqrt{\frac{2}{\pi}}\frac{e^{-(\alpha\log p)/2}}{\sqrt{\alpha\log p}}\left(1+O((\log p)^{-1}\right)\right)^p,\\ &= 1-\left(1-\sqrt{\frac{2}{\alpha\pi}}\frac{p^{-\alpha/2}}{\sqrt{\log p}}\left(1+O((\log p)^{-1}\right)\right)^p. \end{align}

Let $\psi_\alpha(p) =\log(1-\mathbb{P}(\max_{j=1,\dots,p}|Z_j|\geq \sqrt{\alpha\log p}))$, we have the following development for $p\rightarrow\infty,$:

(10)

\begin{align} \psi_\alpha(p) &= -\sqrt{\frac{2}{\alpha\pi}}\frac{p^{-\alpha/2}}{\sqrt{\log p}}\left(1+O((\log p)^{-1})\right)+O\left(\frac{p^{-\alpha}}{\log p}\right),\\ &= -\sqrt{\frac{2}{\alpha\pi}}\frac{p^{-\alpha/2}}{\sqrt{\log p}}+ O\left(\frac{p^{-\alpha/2}}{(\log p )^{3/2}}\right), \end{align}

because $\frac{p^{-\alpha}}{\log p} = \omicron\left(\frac{p^{-\alpha/2}}{(\log p )^{3/2}}\right).$
As we know $\mathbb{P}(\max_{j=1,\dots,p}|Z_j|\geq \sqrt{\alpha\log p}) = 1-\exp(\psi_\alpha(p))$, we finally get:

(11)

\begin{align} \boxed{\mathbb{P}(\max_{j=1,\dots,p}|Z_j|\geq \sqrt{\alpha\log p}) = 1-\exp\left( -\sqrt{\frac{2}{\alpha\pi}}\frac{p^{1-\alpha/2}}{\sqrt{\log p}}+ O\left(\frac{p^{1-\alpha/2}}{(\log p )^{3/2}}\right) \right).} \end{align}

Solution proposed by Marin