# Uniform distribution on a circle

• Oct 23rd 2010, 03:59 PM
uberbandgeek6
Uniform distribution on a circle
Suppose that (X, Y ) is uniformly distributed on a circle with center at the origin and radius. What is P(3X + Y > 1)?
Hint: Sketch the region over which you must integrate. Also, an antiderivative of sqrt(1-y^2) is .5y*sqrt(1-y^2) +.5*arcsin(y)

First I found the pdf of (X,Y) by taking the double integral of a constant c from -sqrt(1-y^2) to sqrt(1-y^2) for dx and from -1 to 1 for dy, and then set it equal to 1 to find c = 1/pi. Then to find P(3X + Y > 1), I took the integral of 1/pi from (1/3)-(1/3)y to sqrt(1-y^2) for dx and from -4/5 to 1 for dy. This gave me an answer of 8.61, which can't possibly be right for a probability. Did I set this problem up incorrectly? Sorry if this explanation is hard to understand. I will clarify it if needed.
• Oct 23rd 2010, 09:21 PM
mr fantastic
Quote:

Originally Posted by uberbandgeek6
Suppose that (X, Y ) is uniformly distributed on a circle with center at the origin and radius. What is P(3X + Y > 1)?
Hint: Sketch the region over which you must integrate. Also, an antiderivative of sqrt(1-y^2) is .5y*sqrt(1-y^2) +.5*arcsin(y)

First I found the pdf of (X,Y) by taking the double integral of a constant c from -sqrt(1-y^2) to sqrt(1-y^2) for dx and from -1 to 1 for dy, and then set it equal to 1 to find c = 1/pi. Then to find P(3X + Y > 1), I took the integral of 1/pi from (1/3)-(1/3)y to sqrt(1-y^2) for dx and from -4/5 to 1 for dy. This gave me an answer of 8.61, which can't possibly be right for a probability. Did I set this problem up incorrectly? Sorry if this explanation is hard to understand. I will clarify it if needed.

I'd make the change of variable $X = R \cos \theta$ and $Y = R \sin \theta$ where R ~ U(0, 1) and $\theta$ ~ $U(0, 2\pi)$ and use the Change of Variable Theorem to get the joint pdf of X and Y.
• Oct 24th 2010, 10:04 AM
uberbandgeek6
That doesn't really seem necessary when I'm given the antiderivative of sqrt(1-y^2) in the problem. I think f(x,y) must be 1/pi because taking the integral of it from -sqrt(1-y^2) to sqrt(1-y^2) for dx and then from -1 to 1 for dy gives 1. Its just when I try to find P(3X + Y > 1) by taking the double integral of 1/pi from (1/3)-(1/3)y to sqrt(1-y^2) for dx and from -4/5 to 1 for dy that I get an answer > 1. Shouldn't this integral give me the answer I'm looking for?
• Oct 24th 2010, 10:56 PM
matheagle
IF this is a uniform over any region, I would NOT use calculus.
Just find the area of interest over the total area, i.e., use geometry.
IF I used calc I would only obtain the probability, not a new density.
• Oct 25th 2010, 04:46 AM
CaptainBlack
Quote:

Originally Posted by uberbandgeek6
Suppose that (X, Y ) is uniformly distributed on a circle with center at the origin and radius. What is P(3X + Y > 1)?
Hint: Sketch the region over which you must integrate. Also, an antiderivative of sqrt(1-y^2) is .5y*sqrt(1-y^2) +.5*arcsin(y)

Do what the hint is hinting at - draw a picture. Then as MathEagle says the required probability is the ratio of the area of the region of interest to that of the whole circle.

(also it is a disk not a circle, a circle is the line around the edge).

CB