Finding $P(|Y|\ge \frac{1}{\sqrt{3}}|X|)$ when $X,Y$ are i.i.d $N(0,1)$

Question

Let $X,Y\sim N(0,1)$ be two independent random variables.

Find: $P(|Y|\ge \frac{1}{\sqrt{3}}|X|)$

My way so far:

$P(|Y|\ge \frac{1}{\sqrt{3}}|X|)=P(\sqrt{3}\cdot |Y|\ge |X|)=P(3Y^2\ge X^2)=P(\frac{3Y^2}{X^2}\ge 1)$

And since $X^2,Y^2\sim Gamma(\frac{1}{2},\frac{1}{2}), \frac{3Y^2}{X^2}\sim Beta(1.5,-1)$ but it's not possible because $-1<0$. Where is my mistake?

Answer 1

I am not sure about the ratio distribution of two gamma-square distributions, so let me go back to the basic. Since we know the joint distribution, as the form of p.d.f., we can simply write down the formula computing the probability in question as

\begin{align*} \mathbf{P}\left(|Y| \geq \tfrac{1}{\sqrt{3}}|X|\right) = \iint\limits_{|y|\geq \frac{1}{\sqrt{3}}|x|} f_{X,Y}(x,y)\, \mathrm{d}x\mathrm{d}y = \iint\limits_{|y|\geq \frac{1}{\sqrt{3}}|x|} \frac{1}{2\pi}e^{-(x^2+y^2)/2}\, \mathrm{d}x\mathrm{d}y. \end{align*}

Now we see that both the domain of integration and the integrand have nice symmetry which can be easily described in terms of polar coordinates. So, applying the polar-coordinate $(x,y)=r(\cos\theta,\sin\theta)$,

\begin{align*} \mathbf{P}\left(|Y| \geq \tfrac{1}{\sqrt{3}}|X|\right) &= \left( \int_{0}^{\infty} re^{-r^2/2}\, \mathrm{d}r \right) \left( \int_{|\sin\theta|\geq \frac{1}{\sqrt{3}}|\cos\theta|} \frac{1}{2\pi} \, \mathrm{d}\theta\right) \\ &= \left[ -e^{-r^2/2} \right]_{r=0}^{r=\infty} \cdot \frac{\text{[length of the set $\{\theta \in [0, 2\pi] : |\tan\theta| \geq 1/\sqrt{3}\}$]}}{2\pi} \\ &= \frac{2}{3}. \end{align*}

Answer 2

I'm not sure how do you came up with that equality between distributions.. I can offer a simpler answer just by decomposing the modules.

$|Y| > \frac{|X|}{\sqrt{3}}$ is

• $Y < \frac{X}{\sqrt{3}} $ if y < 0 and x < 0
• $Y > \frac{-X}{\sqrt{3}} $ if y > 0 and x < 0
• $-Y > \frac{X}{\sqrt{3}} $ if y < 0 and x > 0
• $Y > \frac{X}{\sqrt{3}} $ if y > 0 and x > 0

So you divided your problem in 4 simple probabilities

And for the law of total probablity:

$ P( |Y| > \frac{|X|}{\sqrt{3}} ) =$

$ P (Y < \frac{X}{\sqrt{3}} | ~ y < 0, x < 0) . P( y < 0, x < 0 ) + $

$P (Y > \frac{-X}{\sqrt{3}}| ~ y > 0 , x < 0) . P(y > 0 , x < 0) + $

$P (-Y > \frac{X}{\sqrt{3}} | ~ y < 0 , x > 0) . P(y < 0 , x > 0) + $

$P (Y > \frac{X}{\sqrt{3}} | ~ y > 0 , x > 0) . P (y > 0 , x > 0)) $

as each cuadrant has the same probability = 0.25

$ P( |Y| > \frac{|X|}{\sqrt{3}} ) = $

$P (Y < \frac{X}{\sqrt{3}} | ~ y < 0, x < 0) . 0.25 +$

$P (Y > \frac{-X}{\sqrt{3}}| ~ y > 0 , x < 0) . 0.25 +$

$P (-Y > \frac{X}{\sqrt{3}} | ~ y < 0 , x > 0) . 0.25 +$

$P (Y > \frac{X}{\sqrt{3}} | ~ y > 0 , x > 0) .0.25$

Its look messy, but it has a realy simple solution does not requires any transformation.