Can someone explain me how one can show $\Bbb{P}(\sqrt{s}|N|>\sqrt{1-s}|N'|)=2\arcsin(\sqrt{s})/\pi$?

Question

Let $s\in [0,1]$ and let $N, N'$ be two independent random variable distributed as $\mathcal{N}(0,1)$. I have given the following computation which I don't get:

$$\begin{align}\Bbb{P}(\sqrt{s}|N|>\sqrt{1-s}|N'|)&=\Bbb{P}\left[ (N,N')\in \{(\pm r\cos(\theta),r\sin(\theta)): |\theta|<\arcsin(\sqrt{s}), r>0\}\right]\\&=4\arcsin(\sqrt{s})/(2\pi)\\&=2\arcsin(\sqrt{s})/\pi\end{align}$$

Nw my problem is that I first of all don't get how to get the first equality, it would be nice if someone could explain me what happens there and how to get there, then also the second one is not really clear.

Answer 1

Easy peasy. Let's rewrite the question so we can see it a little bit more clear.

First of all, observe that if $s\in (0,1)$, then $\sqrt s \in (0,1)$. The cases $s=0,s=1$ are trivial and follow from definition. Therefore $$\mathbb P\left(\sqrt s |N| > \sqrt{1-s}|N'|\right) = \mathbb P \left(\sqrt s \frac{|N|}{|N'|} > \sqrt{1-s}\right) = \mathbb P \left( \left|\frac{N}{N'}\right| < \frac{\sqrt{1-s}}{\sqrt s}\right)$$ The quotient of two independent random variables distributed as $\mathcal N (0,1)$ is known to be distributed as a Cauchy distribution, I'll denote it by $\mathcal C$,as well as I will denote $\alpha = \frac{\sqrt{1-s}}{\sqrt s}$. So we have $$Y = \frac{N}{N'} \sim \mathcal C (0,1)$$ So $$\mathbb P (|Y| < \alpha) = \mathbb P (-\alpha < Y < \alpha) = \mathbb P(Y < \alpha) - \mathbb P(Y < -\alpha)$$ Note: The absolute value of a Cauchy distribution is known as a Folded Cauchy distribution.

It is known that, for a Cauchy distribution $Y \sim \mathcal C(0,1)$ we have that the CDF is $$\mathbb P(Y < x) = F(x;(0,1)) = \frac{1}{\pi} \arctan(x) + \frac{1}{2}$$ In our case

• $\mathbb P(Y < \alpha) = \frac{1}{\pi} \arctan(\alpha) + \frac{1}{2}$
• $\mathbb P(Y < -\alpha) = \frac{1}{\pi} \arctan(-\alpha) + \frac{1}{2}$ Therefore $$\mathbb P(|Y| < \alpha) = \frac{1}{\pi} \arctan(\alpha) + \frac{1}{2} - \frac{1}{\pi} \arctan(-\alpha) - \frac{1}{2} = \frac{\arctan(\alpha)-\arctan(-\alpha)}{\pi} $$ Almost there. Now use the trig identities
• $\arctan(x)-\arctan(-x) = 2 \arctan(x)$
• $\arcsin(x) = \arctan \left( \frac{x}{\sqrt{1-x^2}}\right) $ if $x^2 < 1$ [in our case, $x=\sqrt{ 1- s}$]

Therefore $$\mathbb P(|Y| < \alpha) = \frac{\arctan(\alpha)-\arctan(-\alpha)}{\pi} = \frac{2\arctan(\alpha)}{\pi} = \frac{2\arctan\left(\frac{\sqrt{1-s}}{\sqrt s}\right)}{\pi}$$ $$=\frac{2\arcsin(\sqrt s)}{\pi} $$

Answer 2

We have that $(N,N')$ can be uniquely written as $(\pm r\cos\theta,r\sin\theta)$ where $r\geqslant 0$ and $\theta\in]-\pi/2,\pi/2]$ (drawing a circle may help you to vizualise why). The condition $\sqrt{s}|N|>\sqrt{1-s}|N'|$ translates to $\sqrt{s}|\cos\theta|>\sqrt{1-s}|\sin\theta|$ i.e $|\theta|<\arctan\sqrt{\frac{s}{1-s}}$. Using the identity $\sin(\arctan(x))=\frac{|x|}{\sqrt{x^2+1}}$, we get the condition $|\theta|<\arcsin(\sqrt{s})$, which gives the first equality. As for the second equality, since $N$ and $N'$ are distributed as $\mathcal{N}(0,1)$ then $\theta$ is distributed as $\mathcal{U}(]-\pi/2,\pi/2[)$, since there is no condition on $r$ and there are two possibilities for the sign in front of $r\cos\theta$, this gives the second equality.