# $X，Y$ i.i.d. $N(0,\sigma^2)$. What's the probability of $P(X>|2Y|)$?

We know that $X-2Y \sim N(0,5\sigma^2)$, but how to compute $P(X>|2Y|)$?

• The distribution of X-2Y is not useful for answering the question. Look at the joint density of X and integrate over the region where X>2Y for Y positive and X>-2Y when Y is negative, Also add the self-study tag to your list of tags. – Michael R. Chernick Feb 26 '18 at 5:47
• @MichaelChernick: At first I got stuck at the integral part, and now I figured it out. Btw, I came across this question on the Internet and found it interesting. Does it belong to self-study question? – Needle Feb 27 '18 at 20:19
• It still should be self study as it is a type of class exercise. – Michael R. Chernick Feb 27 '18 at 20:27

Here is a hint. Since $X$ and $Y$ are independent, their joint density is $f_{XY}(x,y)=f_X(x)f_Y(y)=\frac{1}{2\pi\sigma^2}\exp(-\frac{x^2+y^2}{2\sigma^2})$. Further, because of symmetry, $$\text{Prob}[X>|2Y|] = 2\text{Prob}[X>2Y,Y>0] = 2 \int_0^{+\infty} \text{d}y \int_{2y}^{+\infty} \text{d}x f_{XY}(x,y)\,,$$ which you can easily solve if you transform to polar coordinates.

Another way of looking at it is to think of $f_{XY}(x,y)$ as a perfectly round cake out of which you take a wedge with angle $2\text{arctan}(\frac{1}{2})$. The probability mass of that wedge should therefore be $\text{arctan}(\frac{1}{2})/\pi \approx 0.4636/3.1415=0.1476$.

• I can't understand the second way. Could you give more explanation? – Needle Feb 27 '18 at 4:56
• @Needle: Try to imagine what the area is in the $(x,y)$ plane where the event $X>|2Y|$ is true. It is a wedge with its tip at $(0,0)$ and an outward angle of $\alpha=2\text{arctan}(\frac{1}{2})$. The density $f_{XY}$ is rotation symmetric around $(0,0)$. Integrating $f_{XY}$ over the whole plane gives probability 1. Integrating over the wedge alone gives $\text{Prob}[X>|2Y|]$. – StijnDeVuyst Feb 27 '18 at 8:53

$P[X>|2Y|] = P[X>0]*P[X^2 > 4Y^2 | X > 0] = 0.5*P[\frac{Y^2}{X^2}<0.25]$

Now, note that $X^2$ and $Y^2$ are both squares of standard normal variates which means they follow $\chi^2_{(1)}$ distribution. Now you can use the result that if U and V are independent and follow chi squared distribution with degrees of freedom $a$ and $b$ respectively then the ratio $\frac{U}{V}$ follows F-Distribution with parameters $a$ and $b$ respectively.

Now to find $P[\frac{Y^2}{X^2}<0.25]$ use the tables or if you're not allowed to do that just integrate the pdf of F-Distribution over the specified range. The integration should not be much time consuming since the degrees of freedom are (1,1) hence the pdf will reduce to a relatively simple function.

• @Glen_b Yes I missed it, Can you give it a check now? – Vishaal Sudarsan Feb 26 '18 at 9:14
• I believe it's correct now. – Glen_b Feb 26 '18 at 9:24