# Probability that π>3π where π,π are π(0,1) random variables

• member 428835
In summary, there is a problem involving a joint probability distribution that can be solved using an integral. However, there is also a geometric approach that is much easier. The angles involved in quadrant II and I can be used to find the solution, but it is unclear how they relate to the normal distribution. One possible approach is to convert the integrals to polar coordinates, which reveals the rotational symmetry of the distribution. Another approach is to use the fact that the ratio of normal random variables follows a Cauchy distribution. Both methods can be used to solve the problem.
member 428835
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.

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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.

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joshmccraney said:
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.

Mark44 said:
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.

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

vela
joshmccraney said:
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.

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

Makes tons of sense, thanks all!

## 1. What is the meaning of "π(0,1) random variables" in this context?

In this context, π(0,1) refers to a normal distribution with a mean of 0 and a standard deviation of 1. This means that the values of π and π are likely to be closer to 0 and less likely to be further away, following a bell-shaped curve.

## 2. How is the probability calculated for π>3π?

The probability of π>3π can be calculated by finding the area under the normal distribution curve between the values of 3π and infinity. This can be done using a mathematical formula or by using a statistical software.

## 3. What does it mean if the probability of π>3π is high?

If the probability of π>3π is high, it means that there is a high likelihood of π being greater than 3 times π. This could indicate that π and π are positively correlated, meaning that as π increases, π also tends to increase.

## 4. Is there a way to increase the probability of π>3π?

No, the probability of π>3π is determined by the values of π and π and their relationship. It cannot be increased or decreased unless the values of π and π are changed.

## 5. How is the probability of π>3π useful in a scientific context?

The probability of π>3π can be useful in predicting the likelihood of certain events or outcomes. It can also be used in statistical analysis to understand the relationship between two variables and make informed decisions based on the probability of certain outcomes.

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