Probability that 𝑌>3𝑋 where 𝑋,𝑌 are 𝑁(0,1) random variables

[joshmccraney]

Homework Statement:

Probability that $Y>3X$ given $Y>0$ where $X,Y$ are $N(0,1)$ random variables

Relevant Equations:

Nothing comes to mind

After plotting the above (not shown) I believe one way (the hard way) to solve this problem is to compute the following integral where $f(x) = e^{-x^2/2}/\sqrt{2\pi}$: $$\frac{\int_0^\infty \int_{3X}^\infty f(X)f(Y)\, dydx + \int_{-\infty}^0 \int_0^\infty f(X)f(Y)\, dydx}{\int_{-\infty}^\infty \int_0^\infty f(X)f(Y)\, dydx}$$
But evidently there's a geometric approach that's much easier. Any help here?

To elaborate, I can see how quadrant II has angle $\pi$ and quadrant I has angle $\arctan (1/3)$ and the total region we consider is $\pi$ angle, so that the total angles considered divided by the angle allowed is the solution $(\pi/2+\arctan(1/3))/\pi$ but I don't understand how these angle additions relate to the normal distribution above.

 

[joshmccraney]

But I don't understand the reasoning here. I can picture the angles (as explained above) but fail to see the relation.

Edit by Mark44 -- I've deleted the image as it didn't match the problem here.

 

[Mark44]

But I don't understand the reasoning here. I can picture the angles (as explained above) but fail to see the relation.

I've deleted my original drawing, as I mistakenly read N(0, 1) as if X and Y were constrained to the interval [0, 1]. I was also treating the distribution as being uniform, which is not the case. I'll have to think about this some more.

 

[joshmccraney]

I've deleted my original drawing, as I mistakenly read N(0, 1) as if X and Y were constrained to the interval [0, 1]. I was also treating the distribution as being uniform, which is not the case. I'll have to think about this some more.

Copy all. It is interesting the integral above agrees with the angle addition I've outlined, but for the life of me I can't see why. For sure the joint distribution is cylindrically symmetric about (0,0), but beyond this I don't see how angle addition is related.

 

[Orodruin]

Consider that the fact that the joint probability distribution in the xy-plane is rotationally symmetric.

 

[vela]

After plotting the above (not shown) I believe one way (the hard way) to solve this problem is to compute the following integral where $f(x) = e^{-x^2/2}/\sqrt{2\pi}$: $$\frac{\int_0^\infty \int_{3X}^\infty f(X)f(Y)\, dydx + \int_{-\infty}^0 \int_0^\infty f(X)f(Y)\, dydx}{\int_{-\infty}^\infty \int_0^\infty f(X)f(Y)\, dydx}$$
But evidently there's a geometric approach that's much easier. Any help here?

To elaborate, I can see how quadrant II has angle $\pi$ and quadrant I has angle $\arctan (1/3)$ and the total region we consider is $\pi$ angle, so that the total angles considered divided by the angle allowed is the solution $(\pi/2+\arctan(1/3))/\pi$ but I don't understand how these angle additions relate to the normal distribution above.

If you convert the integrals to polar coordinates, they're not that bad, and you'll see the rotational symmetry @Orodruin mentioned explicitly.

 

[Orodruin]

Particularly note the integration boundaries in polar coordinates and note what happens to the radial integrals in both denominator and numerator.

 

[BWV]

The ratio of normal random variables follows a Cauchy distribution - that would be one way to approach it

https://mathworld.wolfram.com/NormalRatioDistribution.html

 

[Orodruin]

The ratio of normal random variables follows a Cauchy distribution - that would be one way to approach it

https://mathworld.wolfram.com/NormalRatioDistribution.html

Sure, you can use that and it will work. The polar coordinates solution is significantly simpler though.

 

[joshmccraney]

Makes tons of sense, thanks all!