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Given that $X,Y \sim N(0, 1)$ and IID, what is $P(X > 3Y \mid Y > 0)$? I think the answer is $1/12$.

If we look at the $X-Y$ plane of the joint distribution of $X,Y$, we see that conditioning on $Y > 0$, gets rid of the bottom half. The area under the lie $Y = X/3$ is $1/12$ of the area of the top half. However, when I try to use monte carlo, I am seeing about $0.102$ as I increase the number of trials.

Is my solution wrong?

My other question regarding this is, what if $X,Y$ are not independent? How would you compute the conditional probability in this case?

RobPratt
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student010101
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    Where are you getting 1/12? If you draw the line $Y=3X$, it marks out an angle $\theta = \arctan(1/3)$ with the $X$ axis. Since your conditional distribution is radially symmetric on the upper half plane, that makes your desired region a proportion $\theta/\pi \simeq 0.102$. – preferred_anon Oct 06 '20 at 14:27
  • @preferred_anon I got 1/12 because, let's consider the line $Y = X$. This marks out a 45 degree angle, which would give us $1/4$ of the total area of the top half. Then $Y = X/3$ is just a third of that. Is this wrong? – student010101 Oct 06 '20 at 14:28
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    Yes, it is. Imagine it was $Y=2X$. Does that mark out half of the top area? What about $Y=5X$? etc. – preferred_anon Oct 06 '20 at 14:29
  • @preferred_anon ahh you're right the relationship isn't linear between the area and the line. – student010101 Oct 06 '20 at 14:30
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    Indeed - to find out the true relationship, draw a semicircle and mark your line $Y=3X$ as a radius. Then the area of the smaller sector divided by the total area of the semicircle is the probability you want. – preferred_anon Oct 06 '20 at 14:33
  • As for independence, just knowing that $X$ and $Y$ are dependent and have marginal normal distributions is not enough to know how they are distributed. In one extreme you could have $X=4Y$ and then the probability is $1$. If you had $X=2Y$ then the probability is $0$. I'm pretty sure you can get any number in between by picking a suitable ratio. – preferred_anon Oct 06 '20 at 14:34
  • @preferred_anon Do you mean we wouldn't know if they are bivariate normal? We would only know if the joint distribution is bivariate normal if any linear combination of $X,Y$ is normal right, which we wouldn't know if they are not independent. – student010101 Oct 06 '20 at 14:37
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    Similar: https://math.stackexchange.com/q/3671399/321264. – StubbornAtom Oct 06 '20 at 15:15

1 Answers1

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The conditional probability can be rewritten as: $$P(X > 3Y | Y > 0) = \frac{P(X > 3Y \wedge Y > 0)}{P(Y > 0)} = 2P(X > 3Y \wedge Y > 0),$$

since obviously $P(Y>0)= \frac{1}{2}.$

Now, we need to integrate:

$$P(X > 3Y \wedge Y > 0) = \int_{0}^{+\infty}\int_{3y}^{+\infty} \frac{1}{2\pi}e^{-\frac{1}{2}(x^2 + y^2)}dx dy$$

We can use polar coordinates, where:

$$\begin{cases} x = r \cos\theta\\ y = r \sin \theta \end{cases},$$

and

$$dxdy = r d\theta dr.$$ $$P(X > 3Y \wedge Y > 0) = \int_{0}^{+\infty} \int_{0}^{\arctan\frac{1}{3}} \frac{1}{2\pi}e^{-\frac{1}{2}r^2}rd\theta dr = \\ = \frac{1}{2\pi}\left(\int_{0}^{+\infty}e^{-\frac{1}{2}r^2}rdr \right)\left(\int_{0}^{\arctan\frac{1}{3}} d\theta\right) =\\= \frac{1}{2\pi}\cdot 1\cdot \arctan\frac{1}{3} = \frac{1}{2\pi}\arctan\frac{1}{3} \simeq 0.0512$$

Finally, the probability you are looking for is:

$$P(X > 3Y | Y > 0) = \frac{1}{\pi}\arctan\frac{1}{3} \simeq 0.1024.$$

the_candyman
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